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Magnetic Field of Current — JEE Previous Year Questions

Every Magnetic Field of Current question asked in JEE Main and JEE Advanced across the last 186 papers — 211 questions, each with its correct answer. Free to read, no account needed.

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211

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147/186

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79%

All 211 Magnetic Field of Current questions

Most recent papers first.

Q1·PhysicsNumericalJEE Advanced 2026
A hollow, right circular cone of base radius RRR and height hhh, with its tip at the origin is rotating about the ZZZ-axis with an angular velocity ω\omegaω, as shown in the figure. The cone carries a total charge QQQ uniformly distributed on its curved surface. The magnitude of magnetic field at a point (0,0,z)(0, 0, z)(0,0,z), where z≫Rz \gg Rz≫R and z≫hz \gg hz≫h, is nμ04πQR2ωz3\frac{n\mu_0}{4\pi}\frac{QR^2\omega}{z^3}4πnμ0​​z3QR2ω​. The value of nnn is:

Correct answer: 0.5

Step-by-step solution →
Q2·PhysicsMultiple correctJEE Advanced 2026
In a vacuum chamber, a particle of charge 1 μ\muμC and mass 1 mg is projected with a velocity (i^+2j^)\left(\hat{i} + 2\hat{j}\right)(i^+2j^​) ms−1^{-1}−1 from the XZXZXZ plane at time t=0t = 0t=0 in an electric field of 1i^1\hat{i}1i^ Vm−1^{-1}−1. At t=0.2t = 0.2t=0.2 s, the electric field is switched off and a magnetic field of 6j^6\hat{j}6j^​ T is switched on. The acceleration due to gravity is −10j^-10\hat{j}−10j^​ ms−2^{-2}−2. Correct option(s) is/are:
  1. (A)The vertical distance of the particle from the XZXZXZ plane at t=0.3t = 0.3t=0.3 s is 15 cm.
  2. (B)The vertical distance of the particle from the XZXZXZ plane at t=0.4t = 0.4t=0.4 s is 10 cm.
  3. (C)The radius of the trajectory of the particle for t>0.2t > 0.2t>0.2 s is 20 cm.
  4. (D)The particle will be in the XZXZXZ plane at t=0.35t = 0.35t=0.35 s.

Correct answer: (A), (C)

Step-by-step solution →
Q3·PhysicsNumericalJEE Main 2026
A 5 mg particle carrying a charge of 5π×10−65\pi \times 10^{-6}5π×10−6 C is moving with velocity of (3i^+2k^)×10−2(3\hat{i}+2\hat{k})\times 10^{-2}(3i^+2k^)×10−2 m/s in a region having magnetic field B⃗=0.1k^\vec{B} = 0.1\hat{k}B=0.1k^ Wb/m2^22. It moves a distance of α\alphaα meter along k^\hat{k}k^ when it completes 5 revolutions. The value of α\alphaα is ________.

Correct answer: 2

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
A current carrying circular loop of radius 2 cm with unit normal n^=k^+i^2\hat{n} = \dfrac{\hat{k}+\hat{i}}{\sqrt{2}}n^=2​k^+i^​ is placed in a magnetic field, B⃗=B0(3i^+2k^)\vec{B} = B_0(3\hat{i}+2\hat{k})B=B0​(3i^+2k^). If B0=4×10−3B_0 = 4\times 10^{-3}B0​=4×10−3 T and current I=1002I = 100\sqrt{2}I=1002​ A, the torque experienced by the loop is ________ Wb·A. (π=3.14\pi = 3.14π=3.14)
  1. (A)16×10−5 k^16\times 10^{-5}\,\hat{k}16×10−5k^
  2. (B)5024×10−7 k^5024\times 10^{-7}\,\hat{k}5024×10−7k^
  3. (C)5024×10−7 i^5024\times 10^{-7}\,\hat{i}5024×10−7i^
  4. (D)5024×10−7 j^5024\times 10^{-7}\,\hat{j}5024×10−7j^​

Correct answer: (D)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm. The magnetic field at the mid point between the two wires is __________ μT. (μ04π=10−7 N/A2)\left(\frac{\mu_{0}}{4\pi} = 10^{-7} \text{ N/A}^{2}\right)(4πμ0​​=10−7 N/A2)
  1. (A)30
  2. (B)300
  3. (C)150
  4. (D)0.0

Correct answer: (B)

Step-by-step solution →
Q6·PhysicsSingle correctJEE Main 2026
A small cube of side 1 mm is placed at the centre of a circular loop of radius 10 cm carrying a current of 2 A. The magnetic energy stored inside the cube is α×10−14\alpha \times 10^{-14}α×10−14 J. The value of α\alphaα is _______. (μo=4π×10−7\mu_o = 4\pi \times 10^{-7}μo​=4π×10−7 Tm/A, π=3.14\pi = 3.14π=3.14)
  1. (A)6.28
  2. (B)6.28×10−66.28 \times 10^{-6}6.28×10−6
  3. (C)628
  4. (D)6.28×10−46.28 \times 10^{-4}6.28×10−4

Correct answer: (A)

Step-by-step solution →
Q7·PhysicsNumericalJEE Main 2026
The charged particle moving in a uniform magnetic field of (3i^+2j^)(3\hat{i} + 2\hat{j})(3i^+2j^​) T has an acceleration (4i^−x2j^)\left(4\hat{i} - \frac{x}{2}\hat{j}\right)(4i^−2x​j^​) m/s2^{2}2. The value of xxx is _____.

Correct answer: 12

Step-by-step solution →
Q8·PhysicsSingle correctJEE Main 2026
A particle of charge q and mass m is projected from origin with an initial velocity v⃗=(v02x^+v02y^)\vec{v} = \left(\dfrac{v_{0}}{\sqrt{2}}\hat{x} + \dfrac{v_{0}}{\sqrt{2}}\hat{y}\right)v=(2​v0​​x^+2​v0​​y^​). There exists a uniform magnetic field B⃗=B0z^\vec{B} = B_{0}\hat{z}B=B0​z^ and a space varying electric field E⃗=E0e−λxx^\vec{E} = E_{0}e^{-\lambda x}\hat{x}E=E0​e−λxx^ within the region 0 ≤ x ≤ L. After travelling a distance such that x-coordinate has changed from x = 0 to x = L, the change in the kinetic energy is ________.
  1. (A)qE0λ[1−e−λL]\dfrac{qE_{0}}{\lambda}[1 - e^{-\lambda L}]λqE0​​[1−e−λL]
  2. (B)(v0qB02λ)[2−e−2λL]\left(\dfrac{v_{0}qB_{0}}{2\lambda}\right)[2 - e^{-2\lambda L}](2λv0​qB0​​)[2−e−2λL]
  3. (C)qE0λ[1+e−λL]\dfrac{qE_{0}}{\lambda}[1 + e^{-\lambda L}]λqE0​​[1+e−λL]
  4. (D)q(E0+v0B0λ)[1−e−λL/2]q\left(\dfrac{E_{0} + v_{0}B_{0}}{\lambda}\right)[1 - e^{-\lambda L/2}]q(λE0​+v0​B0​​)[1−e−λL/2]

Correct answer: (A)

Step-by-step solution →
Q9·PhysicsNumericalJEE Main 2026
A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the direction of the magnetic field. The torque acting on the coil is α × 10−410^{-4}10−4 N.m. The value of α is ______. (π=3.14\pi = 3.14π=3.14)

Correct answer: 314

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
An insulated wire is wound so that it forms a flat coil with N=200N = 200N=200 turns. The radius of the innermost turn is r1=3r_1 = 3r1​=3 cm, and of the outermost turn r2=6r_2 = 6r2​=6 cm. If 20 mA current flows in it then the magnetic moment will be α×10−2\alpha \times 10^{-2}α×10−2 A.m2^22. The value of α\alphaα is ______.
  1. (A)4.4
  2. (B)2.64
  3. (C)3.25
  4. (D)1.2

Correct answer: (B)

Step-by-step solution →
Q11·PhysicsNumericalJEE Main 2026
1 μC charge moving with velocity v⃗=(i^−2j^+3k^)\vec{v} = \left( \hat{i} - 2\hat{j} + 3\hat{k} \right)v=(i^−2j^​+3k^) m/s in the region of magnetic field B⃗=(2i^+3j^−5k^)\vec{B} = \left( 2\hat{i} + 3\hat{j} - 5\hat{k} \right)B=(2i^+3j^​−5k^) T. The magnitude of force acting on it is α×10−6\sqrt{\alpha} \times 10^{-6}α​×10−6 N. The value of α is ________.

Correct answer: 171

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
A particle having charge 10−910^{-9}10−9 C moving in xxx-yyy plane in fields of 0.4j^0.4\hat{j}0.4j^​ N/C and 4×10−3k^4 \times 10^{-3}\hat{k}4×10−3k^ T experiences a force of (4i^+2j^)×10−10(4\hat{i} + 2\hat{j}) \times 10^{-10}(4i^+2j^​)×10−10 N. The velocity of the particle at that instant is ______ m/s.
  1. (A)50i^+100j^50\hat{i} + 100\hat{j}50i^+100j^​
  2. (B)100i^+50j^100\hat{i} + 50\hat{j}100i^+50j^​
  3. (C)−50i^+100j^-50\hat{i} + 100\hat{j}−50i^+100j^​
  4. (D)50i^−100j^50\hat{i} - 100\hat{j}50i^−100j^​

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsSingle correctJEE Main 2026
Two identical long current carrying wires are bent into the shapes shown in the following figures. If the magnitude of magnetic fields at the centres P and Q of a semicircular arc are B1B_{1}B1​ and B2B_{2}B2​ respectively, then the ratio B1B2\frac{B_{1}}{B_{2}}B2​B1​​ is ________.
  1. (A)2+π1+π\frac{2+\pi}{1+\pi}1+π2+π​
  2. (B)1+π1−π\frac{1+\pi}{1-\pi}1−π1+π​
  3. (C)2+π1−π\frac{2+\pi}{1-\pi}1−π2+π​
  4. (D)1+π2−π\frac{1+\pi}{2-\pi}2−π1+π​

Correct answer: (A)

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
A long cylindrical conductor with large cross section carries an electric current distributed uniformly over its cross-section. Magnetic field due to this current is : A. maximum at either ends of the conductor and minimum at the midpoint B. maximum at the axis of the conductor C. minimum at the surface of the conductor D. minimum at the axis of the conductor E. same at all points in the cross-section of the conductor Choose the correct answer from the options given below :
  1. (A)D Only
  2. (B)A, D Only
  3. (C)B, C Only
  4. (D)E Only

Correct answer: (A)

Step-by-step solution →
Q15·PhysicsSingle correctJEE Main 2026
The magnetic field at the centre of a current carrying circular loop of radius RRR is 16 μT. The magnetic field at a distance x=3Rx = \sqrt{3}Rx=3​R on its axis from the centre is ________ μT.
  1. (A)222\sqrt{2}22​
  2. (B)4
  3. (C)2
  4. (D)8

Correct answer: (C)

Step-by-step solution →
Q16·PhysicsSingle correctJEE Main 2026
Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire QQQ is ______ . (μ0=4π×10−7\mu_{0} = 4\pi \times 10^{-7}μ0​=4π×10−7 T.m/A)
  1. (A)6×10−76 \times 10^{-7}6×10−7 N towards PPP
  2. (B)6×10−66 \times 10^{-6}6×10−6 N towards RRR
  3. (C)6×10−76 \times 10^{-7}6×10−7 N towards RRR
  4. (D)6×10−66 \times 10^{-6}6×10−6 N towards PPP

Correct answer: (B)

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2026
Two identical circular loops P and Q each of radius r are lying in parallel planes such that they have common axis. The current through P and Q are I and 4I respectively in clockwise direction as seen from O. The net magnetic field at O is:
  1. (A)3μoI42r\frac{3\mu_o I}{4\sqrt{2}r}42​r3μo​I​ toward P
  2. (B)μoI42r\frac{\mu_o I}{4\sqrt{2}r}42​rμo​I​ toward P
  3. (C)μoI42r\frac{\mu_o I}{4\sqrt{2}r}42​rμo​I​ towards Q
  4. (D)3μoI42r\frac{3\mu_o I}{4\sqrt{2}r}42​r3μo​I​ towards Q

Correct answer: (D)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2026
The current passing through a conducting loop in the form of equilateral triangle of side 434\sqrt{3}43​ cm is 2A. The magnetic field at its centroid is α×10−5\alpha \times 10^{-5}α×10−5 T. The value of α\alphaα is ______. (Given : μo=4π×10−7\mu_{o} = 4\pi \times 10^{-7}μo​=4π×10−7 SI units)
  1. (A)232\sqrt{3}23​
  2. (B)3\sqrt{3}3​
  3. (C)333\sqrt{3}33​
  4. (D)32\frac{\sqrt{3}}{2}23​​

Correct answer: (C)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2026
A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :
  1. (A)a = g
  2. (B)a > g
  3. (C)a = 0
  4. (D)0 < a < g

Correct answer: (A)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2026
An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of the circular part is r. The magnetic field at the centre O of the circular loop is :
  1. (A)μ02πIr(π+1)i^\dfrac{\mu_0}{2\pi}\dfrac{\text{I}}{\text{r}}\left(\pi+1\right)\hat{\text{i}}2πμ0​​rI​(π+1)i^
  2. (B)−μ02πIr(π−1)i^-\dfrac{\mu_0}{2\pi}\dfrac{\text{I}}{\text{r}}\left(\pi-1\right)\hat{\text{i}}−2πμ0​​rI​(π−1)i^
  3. (C)μ02πIr(π−1)i^\dfrac{\mu_0}{2\pi}\dfrac{\text{I}}{\text{r}}\left(\pi-1\right)\hat{\text{i}}2πμ0​​rI​(π−1)i^
  4. (D)−μ02πIr(π+1)i^-\dfrac{\mu_0}{2\pi}\dfrac{\text{I}}{\text{r}}\left(\pi+1\right)\hat{\text{i}}−2πμ0​​rI​(π+1)i^

Correct answer: (B)

Step-by-step solution →
Q21·PhysicsNumericalJEE Advanced 2025
A conducting solid sphere of radius R and mass M carries a charge Q. The sphere is rotating about an axis passing through its center with a uniform angular speed ω. The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as αQ2M\alpha\frac{Q}{2M}α2MQ​ . The value of α\alphaα is ______

Correct answer: 1.66 or 1.67

Step-by-step solution →
Q22·PhysicsSingle correctJEE Main 2025
Figure shows a current carrying square loop ABCD of edge length is 'aaa' lying in a plane. If the resistance of the ABC part is rrr and that of ADC part is 2r2r2r, then the magnitude of the resultant magnetic field at centre of the square loop is:
  1. (A)3πμ0I2 a\dfrac{3\pi\mu_0 I}{\sqrt2\,a}2​a3πμ0​I​
  2. (B)μ0I2πa\dfrac{\mu_0 I}{2\pi a}2πaμ0​I​
  3. (C)2 μ0I3πa\dfrac{\sqrt2\,\mu_0 I}{3\pi a}3πa2​μ0​I​
  4. (D)2μ0I3πa\dfrac{2\mu_0 I}{3\pi a}3πa2μ0​I​

Correct answer: (C)

Step-by-step solution →
Q23·PhysicsSingle correctJEE Main 2025
A particle of charge qqq, mass mmm and kinetic energy EEE enters in magnetic field perpendicular to its velocity and undergoes a circular arc of radius rrr. Which of the following curves (shown in the figure) represents the variation of rrr with EEE?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q24·PhysicsSingle correctJEE Main 2025
The percentage increase in magnetic field (B) when space within a current carrying solenoid is filled with magnesium (magnetic susceptibility χmg=1.2×10−5\chi_{mg}=1.2\times10^{-5}χmg​=1.2×10−5) is:
  1. (A)65×10−3\dfrac{6}{5}\times10^{-3}56​×10−3%
  2. (B)56×10−5\dfrac{5}{6}\times10^{-5}65​×10−5%
  3. (C)56×10−4\dfrac{5}{6}\times10^{-4}65​×10−4%
  4. (D)53×10−5\dfrac{5}{3}\times10^{-5}35​×10−5%

Correct answer: (A)

Step-by-step solution →
Q25·PhysicsSingle correctJEE Main 2025
Uniform magnetic fields of different strengths (B1B_1B1​ and B2B_2B2​), both normal to the plane of the paper exist as shown in the figure. A charged particle of mass mmm and charge qqq, at the interface at an instant, moves into the region 2 with velocity vvv and returns to the interface. It continues to move into region 1 and finally reaches the interface. What is the displacement of the particle during this movement along the interface? (Consider the velocity of the particle to be normal to the magnetic field and B2>B1B_2>B_1B2​>B1​)
  1. (A)mvqB1(1−B2B1)×2\dfrac{mv}{qB_1}\left(1-\dfrac{B_2}{B_1}\right)\times2qB1​mv​(1−B1​B2​​)×2
  2. (B)mvqB1(1−B1B2)\dfrac{mv}{qB_1}\left(1-\dfrac{B_1}{B_2}\right)qB1​mv​(1−B2​B1​​)
  3. (C)mvqB1(1−B2B1)\dfrac{mv}{qB_1}\left(1-\dfrac{B_2}{B_1}\right)qB1​mv​(1−B1​B2​​)
  4. (D)mvqB1(1−B1B2)×2\dfrac{mv}{qB_1}\left(1-\dfrac{B_1}{B_2}\right)\times2qB1​mv​(1−B2​B1​​)×2

Correct answer: (D)

Step-by-step solution →
Q26·PhysicsIntegerJEE Main 2025
A particle of charge 1.6 μ\muμC and mass 16 μ\muμg is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to the magnetic field. The time required for the particle to return to original location for the first time is ______ s. (π=3.14\pi=3.14π=3.14)

Correct answer: 10

Step-by-step solution →
Q27·PhysicsIntegerJEE Main 2025
A loop ABCDA, carrying current I=12I=12I=12 A, is placed in a plane, consists of two semi-circular segments of radius R1=6πR_1=6\piR1​=6π m and R2=4πR_2=4\piR2​=4π m (as shown in the figure). The magnitude of the resultant magnetic field at the centre O is k×10−7k\times10^{-7}k×10−7 T. The value of kkk is __________. (Given μ0=4π×10−7 Tm A−1\mu_0=4\pi\times10^{-7}\,\text{Tm A}^{-1}μ0​=4π×10−7Tm A−1)

Correct answer: 1

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2025
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: If oxygen ion (O−)(O^-)(O−) and Hydrogen ion (H+)(H^+)(H+) enter normal to the magnetic field with equal momentum, then the path of O−O^-O− ion has a smaller curvature than that of H+H^+H+. Reason R: A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)A is true but R is false.
  2. (B)Both A and R are true but R is NOT the correct explanation of A.
  3. (C)A is false but R is true.
  4. (D)Both A and R are true and R is the correct explanation of A.

Correct answer: (A)

Step-by-step solution →
Q29·PhysicsIntegerJEE Main 2025
A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to a uniform magnetic field of strength 0.15 T. The magnetic force on the wire is __________ mN.

Correct answer: 48

Step-by-step solution →
Q30·PhysicsSingle correctJEE Main 2025
Let B1B_1B1​ be the magnitude of magnetic field at the centre of a circular coil of radius RRR carrying current III. Let B2B_2B2​ be the magnitude of magnetic field at an axial distance xxx from the centre. For x:R=3:4x:R=3:4x:R=3:4, B2B1\dfrac{B_2}{B_1}B1​B2​​ is:
  1. (A)4 : 5
  2. (B)16 : 25
  3. (C)64 : 125
  4. (D)25 : 16

Correct answer: (C)

Step-by-step solution →
Q31·PhysicsSingle correctJEE Main 2025
In a moving coil galvanometer, two moving coils M1M_1M1​ and M2M_2M2​ have the following particulars: R1=5 ΩR_1=5\,\OmegaR1​=5Ω, N1=15N_1=15N1​=15, A1=3.6×10−3 m2A_1=3.6\times10^{-3}\,m^2A1​=3.6×10−3m2, B1=0.25 TB_1=0.25\,TB1​=0.25T; R2=7 ΩR_2=7\,\OmegaR2​=7Ω, N2=21N_2=21N2​=21, A2=1.8×10−3 m2A_2=1.8\times10^{-3}\,m^2A2​=1.8×10−3m2, B2=0.5 TB_2=0.5\,TB2​=0.5T. Assuming the other constants are the same, the ratio of voltage sensitivity of M1M_1M1​ and M2M_2M2​ is:
  1. (A)1:11:11:1
  2. (B)1:41:41:4
  3. (C)1:31:31:3
  4. (D)1:21:21:2

Correct answer: (A)

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2025
Consider a long straight wire of a circular cross-section (radius a) carrying a steady current I. The current is uniformly distributed across this cross-section. The distances from the centre of the wire’s cross-section at which the magnetic field [inside the wire, outside the wire] is half of the maximum possible magnetic field, any where due to the wire, will be
  1. (A)[a4,3a2]\left[\dfrac{a}{4},\dfrac{3a}{2}\right][4a​,23a​]
  2. (B)[a2,2a]\left[\dfrac{a}{2},2a\right][2a​,2a]
  3. (C)[a2,3a]\left[\dfrac{a}{2},3a\right][2a​,3a]
  4. (D)[a4,2a]\left[\dfrac{a}{4},2a\right][4a​,2a]

Correct answer: (B)

Step-by-step solution →
Q33·PhysicsIntegerJEE Main 2025
The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9×10−42.9\times 10^{-4}2.9×10−4 Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is ______ π\piπ cm.

Correct answer: 8

Step-by-step solution →
Q34·PhysicsSingle correctJEE Main 2025
An infinite wire has a circular bend of radius a, and carrying a current I as shown in figure. The magnitude of magnetic field at the origin O of the arc is given by:
  1. (A)μ0I4πa[π2+1]\frac{\mu_0 I}{4\pi a}\left[\frac{\pi}{2}+1\right]4πaμ0​I​[2π​+1]
  2. (B)μ0I4πa[3π2+1]\frac{\mu_0 I}{4\pi a}\left[\frac{3\pi}{2}+1\right]4πaμ0​I​[23π​+1]
  3. (C)μ0I2πa[π2+2]\frac{\mu_0 I}{2\pi a}\left[\frac{\pi}{2}+2\right]2πaμ0​I​[2π​+2]
  4. (D)μ0I4πa[3π2+2]\frac{\mu_0 I}{4\pi a}\left[\frac{3\pi}{2}+2\right]4πaμ0​I​[23π​+2]

Correct answer: (B)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2025
Consider a long thin conducting wire carrying a uniform current I. A particle having mass "M" and charge "q" is released at a distance "a" from the wire with a speed v0v_0v0​ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [μ0\mu_0μ0​ is vacuum permeability]
  1. (A)a[1−mv02qμ0I]a\left[1-\frac{mv_0}{2q\mu_0 I}\right]a[1−2qμ0​Imv0​​]
  2. (B)a2\frac{a}{2}2a​
  3. (C)a[1−mv0qμ0I]a\left[1-\frac{mv_0}{q\mu_0 I}\right]a[1−qμ0​Imv0​​]
  4. (D)ae−4πmv0/qμ0Iae^{-4\pi mv_0/q\mu_0 I}ae−4πmv0​/qμ0​I

Correct answer: (D)

Step-by-step solution →
Q36·PhysicsIntegerJEE Main 2025
A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75 ns. The number of turns per metre in the solenoid is __________. [Take mass of electron =9×10−31=9\times10^{-31}=9×10−31 kg, charge of electron ∣qe∣=1.6×10−19|q_e|=1.6\times10^{-19}∣qe​∣=1.6×10−19 C, μ0=4π×10−7 N/A2\mu_0=4\pi\times10^{-7}\ \mathrm{N/A^2}μ0​=4π×10−7 N/A2, 111 ns =10−9=10^{-9}=10−9 s]

Correct answer: 250

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2025
Given below are two statements. Assertion (A) : An electron in a certain region of uniform magnetic field is moving with constant velocity in a straight line path. Reason (R) : The magnetic field in that region is along the direction of velocity of the electron. In the light of the above statements, choose the correct answer from the options given below :
  1. (A)(A) is false but (R) is true
  2. (B)Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. (C)Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  4. (D)(A) is true but (R) is false

Correct answer: (B)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2025
N equally spaced charges each of value q, are placed on a circle of radius R. The circle rotates about its axis with an angular velocity ω\omegaω as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, IA−IBI_A-I_BIA​−IB​, for the given Amperian loops is :
  1. (A)N22πqω\dfrac{N^2}{2\pi}q\omega2πN2​qω
  2. (B)2πNqω\dfrac{2\pi}{N}q\omegaN2π​qω
  3. (C)N2πqω\dfrac{N}{2\pi}q\omega2πN​qω
  4. (D)Nπqω\dfrac{N}{\pi}q\omegaπN​qω

Correct answer: (C)

Step-by-step solution →
Q39·PhysicsIntegerJEE Main 2025
A current of 5A exists in a square loop of side 12\dfrac{1}{\sqrt2}2​1​ m. Then the magnitude of the magnetic field B at the centre of the square loop will be p×10−6p\times10^{-6}p×10−6 T, where value of p is _______. [Take μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7 T m A−1^{-1}−1].

Correct answer: 8

Step-by-step solution →
Q40·PhysicsSingle correctJEE Main 2025
A long straight wire of a circular cross-section with radius "a" carries a steady current I. The current I is uniformly distributed across this cross-section. The plot of magnitude of magnetic field B with distance r from the centre of the wire is given by (choose the correct graph):
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q41·PhysicsSingle correctJEE Main 2025
Consider a moving coil galvanometer (MCG) : A : The torsional constant in moving coil galvanometer has dimensions [ML2T−2][ML^2T^{-2}][ML2T−2]. B : Increasing the current sensitivity may not necessarily increase the voltage sensitivity. C : If we increase number of turns (N) to its double (2N), then the voltage sensitivity doubles. D : MCG can be converted into an ammeter by introducing a shunt resistance of large value in parallel with galvanometer. E : Current sensitivity of MCG depends inversely on number of turns of coil. Choose the correct answer from the options given below :
  1. (A)A, B only
  2. (B)A, D only
  3. (C)B, D, E only
  4. (D)A, B, E only

Correct answer: (A)

Step-by-step solution →
Q42·PhysicsIntegerJEE Main 2025
Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5A and 4A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is x×10−5x\times10^{-5}x×10−5 T. The value of x is ______. Take permeability of free space as μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7 SI units.

Correct answer: 1

Step-by-step solution →
Q43·PhysicsIntegerJEE Main 2025
A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of 2×1052\times10^52×105 ms−1^{-1}−1. When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is x×104x\times10^4x×104 N/C, the value of x is ______. Take the mass of the proton =1.6×10−27=1.6\times10^{-27}=1.6×10−27 kg.

Correct answer: 2

Step-by-step solution →
Q44·PhysicsMultiple correctJEE Advanced 2024
A positive, singly ionized atom of mass number AMA_MAM​ is accelerated from rest by the voltage 192V . Thereafter, it enters a rectangular region of width w with magnetic field B⃗0=0.1k^\vec{B}_0 = 0.1\hat{k}B0​=0.1k^ Tesla, as shown in the figure. The ion finally hits a detector at the distance x below its starting trajectory. [Given: Mass of neutron/proton = (5/3)×10−27kg(5/3) \times 10^{-27}kg(5/3)×10−27kg , charge of the electron = 1.6×10−19C1.6 \times 10^{-19}C1.6×10−19C .] Which of the following option(s) is(are) correct?
  1. (A)The value of x for H+H^+H+ ion is 4 cm .
  2. (B)The value of x for an ion with AM=144A_M = 144AM​=144 is 48cm .
  3. (C)For detecting ions with 1≤AM≤1961 \leq A_M \leq 1961≤AM​≤196 , the minimum height (x1−x0)(x_1 - x_0)(x1​−x0​) of the detector is 55cm .
  4. (D)The minimum width w of the region of the magnetic field for detecting ions with AM=196A_M = 196AM​=196 is 56cm .

Correct answer: (A), (B)

Step-by-step solution →
Q45·PhysicsSingle correctJEE Advanced 2024
A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass m and radius r and it is in a uniform vertical magnetic field B0B_0B0​ , as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity g , on two conducting supports at P and Q . When a current I is passed through the loop, the loop turns about the line PQ by an angle θ given by
  1. (A)tan θ =πrlB0/(mg)= \pi r l B_0/(mg)=πrlB0​/(mg)
  2. (B)tan θ =2πrlB0/(mg)= 2\pi r l B_0/(mg)=2πrlB0​/(mg)
  3. (C)tan θ =πrlB0/(2mg)= \pi r l B_0/(2mg)=πrlB0​/(2mg)
  4. (D)tan θ =mg/(πrlB0)= mg/(\pi r l B_0)=mg/(πrlB0​)

Correct answer: (A)

Step-by-step solution →
Q46·PhysicsSingle correctJEE Advanced 2024
An infinitely long wire, located on the z-axis, carries a current III along the +z-direction and produces the magnetic field B⃗\vec{B}B. The magnitude of the line integral ∫B⃗⋅dl→\int \vec{B} \cdot \overrightarrow{dl}∫B⋅dl along a straight line from the point (−3a,a,0)\left(-\sqrt{3}a, a, 0\right)(−3​a,a,0) to (a,a,0)(a, a, 0)(a,a,0) is given by [μ0\mu_{0}μ0​ is the magnetic permeability of free space.]
  1. (A)7μ0I/247\mu_{0}I / 247μ0​I/24
  2. (B)7μ0I/127\mu_{0}I / 127μ0​I/12
  3. (C)μ0I/8\mu_{0}I / 8μ0​I/8
  4. (D)μ0I/6\mu_{0}I / 6μ0​I/6

Correct answer: (A)

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2024
Given below are two statements: Statement (I): When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account. Statement (II): Ampere's circuital law does not depend on Biot-Savart's law. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Both Statement I and Statement II are false.
  2. (B)Statement I is true but Statement II is false.
  3. (C)Statement I is false but Statement II is true.
  4. (D)Both Statement I and Statement II are true.

Correct answer: (B)

Step-by-step solution →
Q48·PhysicsNumericalJEE Main 2024
A square loop of edge length 2 m carrying current of 2 A is placed with its edges parallel to the x-y axis. A magnetic field is passing through the x-y plane and expressed as B⃗=B0(1+4x)k^\vec{B}=B_{0}(1+4x)\hat{k}B=B0​(1+4x)k^, where B0=5B_{0}=5B0​=5 T. The net magnetic force experienced by the loop is ______ N.

Correct answer: 160

Step-by-step solution →
Q49·PhysicsSingle correctJEE Main 2024
A proton and a deuteron (q=+eq=+eq=+e, m=2.0m=2.0m=2.0 u) having same kinetic energies enter a region of uniform magnetic field B⃗\vec{B}B, moving perpendicular to B⃗\vec{B}B. The ratio of the radius rdr_{d}rd​ of the deuteron path to the radius rpr_{p}rp​ of the proton path is:
  1. (A)1:1
  2. (B)1:21:\sqrt{2}1:2​
  3. (C)2:1\sqrt{2}:12​:1
  4. (D)1:2

Correct answer: (C)

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2024
A long straight wire of radius aaa carries a steady current I. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a2\dfrac{a}{2}2a​ and 2a2a2a from axis of the wire is:
  1. (A)1 : 4
  2. (B)4 : 1
  3. (C)1 : 1
  4. (D)3 : 4

Correct answer: (C)

Step-by-step solution →
Q51·PhysicsNumericalJEE Main 2024
An electron with kinetic energy 5 eV5\,eV5eV enters a region of uniform magnetic field of 3 μT3\,\mu T3μT perpendicular to its direction. An electric field is applied perpendicular to the direction of velocity and magnetic field. The value of EEE, so that electron moves along the same path, is ___ NC−1NC^{-1}NC−1. (Given, mass of electron =9×10−31=9\times10^{-31}=9×10−31 kg, electric charge =1.6×10−19=1.6\times10^{-19}=1.6×10−19 C)

Correct answer: 4

Step-by-step solution →
Q52·PhysicsNumericalJEE Main 2024
A coil having 100 turns, area of 5×10−35 \times 10^{-3}5×10−3 m2^{2}2, carrying current of 1 mA is placed in uniform magnetic field of 0.200.200.20 T such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through 90∘90^{\circ}90∘ is ______ μ\muμJ.

Correct answer: 100

Step-by-step solution →
Q53·PhysicsSingle correctJEE Main 2024
An element Δl=Δx i^\Delta l=\Delta x\,\hat{i}Δl=Δxi^ is placed at the origin and carries a large current I=10I=10I=10 A. The magnetic field on the yyy-axis at a distance of 0.5 m from the element Δx\Delta xΔx of 1 cm length is:
  1. (A)4×10−84\times10^{-8}4×10−8 T
  2. (B)8×10−88\times10^{-8}8×10−8 T
  3. (C)12×10−812\times10^{-8}12×10−8 T
  4. (D)10×10−810\times10^{-8}10×10−8 T

Correct answer: (A)

Step-by-step solution →
Q54·PhysicsNumericalJEE Main 2024
A circular coil having 200 turns, 2.5×10−42.5\times10^{-4}2.5×10−4 m2^{2}2 area and carrying 100 μ100\,\mu100μA current is placed in a uniform magnetic field of 1 T. Initially the magnetic dipole moment (M⃗)(\vec{M})(M) was directed along B⃗\vec{B}B. The amount of work required to rotate the coil through 90∘90^{\circ}90∘ from its initial orientation such that M⃗\vec{M}M becomes perpendicular to B⃗\vec{B}B is _______ μ\muμJ.

Correct answer: 5

Step-by-step solution →
Q55·PhysicsSingle correctJEE Main 2024
The electrostatic force (F1⃗)(\vec{F_1})(F1​​) and magnetic force (F2⃗)(\vec{F_2})(F2​​) acting on a charge qqq moving with velocity v⃗\vec{v}v can be written as:
  1. (A)F1⃗=q(V⃗⋅E⃗)\vec{F_1}=q(\vec{V}\cdot\vec{E})F1​​=q(V⋅E), F2⃗=q(B⃗⋅V⃗)\vec{F_2}=q(\vec{B}\cdot\vec{V})F2​​=q(B⋅V)
  2. (B)F1⃗=qB⃗\vec{F_1}=q\vec{B}F1​​=qB, F2⃗=q(B⃗×V⃗)\vec{F_2}=q(\vec{B}\times\vec{V})F2​​=q(B×V)
  3. (C)F1⃗=qE⃗\vec{F_1}=q\vec{E}F1​​=qE, F2⃗=q(V⃗×B⃗)\vec{F_2}=q(\vec{V}\times\vec{B})F2​​=q(V×B)
  4. (D)F1⃗=qE⃗\vec{F_1}=q\vec{E}F1​​=qE, F2⃗=q(B⃗×V⃗)\vec{F_2}=q(\vec{B}\times\vec{V})F2​​=q(B×V)

Correct answer: (C)

Step-by-step solution →
Q56·PhysicsNumericalJEE Main 2024
A 2 A current-carrying straight metal wire of resistance 1 Ω1\ \Omega1 Ω, resistivity 2×10−6 Ω m2\times10^{-6}\ \Omega\,\text{m}2×10−6 Ωm, area of cross-section 10 mm210\ \text{mm}^210 mm2 and mass 500 g is suspended horizontally in mid air by applying a uniform magnetic field B⃗\vec{B}B. The magnitude of B is ___ ×10−1\times10^{-1}×10−1 T (given g=10 m/s2g = 10\ \text{m/s}^2g=10 m/s2).

Correct answer: 5

Step-by-step solution →
Q57·PhysicsNumericalJEE Main 2024
A solenoid of length 0.5 m has a radius of 1 cm and is made up of mmm number of turns. It carries a current of 5 A. If the magnitude of the magnetic field inside the solenoid is 6.28×10−36.28\times10^{-3}6.28×10−3 T, then the value of mmm is __________.

Correct answer: 500

Step-by-step solution →
Q58·PhysicsSingle correctJEE Main 2024
In a coaxial straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:
  1. (A)inside the outer conductor
  2. (B)in between the two conductors
  3. (C)outside the cable
  4. (D)inside the inner conductor

Correct answer: (C)

Step-by-step solution →
Q59·PhysicsNumericalJEE Main 2024
Two parallel long current carrying wire separated by a distance 2r2r2r are shown in the figure. The ratio of magnetic field at A to the magnetic field produced at C is x7\tfrac{x}{7}7x​. The value of xxx is ______.

Correct answer: 5

Step-by-step solution →
Q60·PhysicsSingle correctJEE Main 2024
An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then:
  1. (A)the electron will be accelerated along the axis.
  2. (B)the electron will continue to move with uniform velocity along the axis of the solenoid.
  3. (C)the electron will be deflected to move along the circular path about axis.
  4. (D)the electron will experience a force at 45∘45^\circ45∘ to the axis and execute a helical path.

Correct answer: (B)

Step-by-step solution →
Q61·PhysicsNumericalJEE Main 2024
The magnetic field existing in a region is given by B⃗=0.2(1+2x)k^\vec B=0.2(1+2x)\hat kB=0.2(1+2x)k^ T. A square loop of edge 50 cm50\,cm50cm carrying 0.5 A0.5\,A0.5A current is placed in x-y plane with its edges parallel to the x-y axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is ___ mN.

Correct answer: 50

Step-by-step solution →
Q62·PhysicsNumericalJEE Main 2024
A regular polygon of 666 sides is formed by bending a wire of length 4π4\pi4π metre. If electric current of 4π3 A4\pi\sqrt3\,A4π3​A is flowing through the sides of the polygon, the magnetic field at the centre of the polygon would be x×10−7 Tx\times10^{-7}\,Tx×10−7T. The value of xxx is ___

Correct answer: 72

Step-by-step solution →
Q63·PhysicsNumericalJEE Main 2024
A moving coil galvanometer has 100 turns and each turn has an area 2.0 cm2^22. The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x×10−5x\times 10^{-5}x×10−5 N-m/rad. The value of x is __________.

Correct answer: 4

Step-by-step solution →
Q64·PhysicsSingle correctJEE Main 2024
A galvanometer has a resistance of 50 Ω50\,\Omega50Ω and it allows maximum current of 5 mA5\,mA5mA. It can be converted into voltmeter to measure upto 100 V100\,V100V by connecting in series a resistor of resistance:
  1. (A)5975 Ω5975\,\Omega5975Ω
  2. (B)20050 Ω20050\,\Omega20050Ω
  3. (C)19950 Ω19950\,\Omega19950Ω
  4. (D)19500 Ω19500\,\Omega19500Ω

Correct answer: (C)

Step-by-step solution →
Q65·PhysicsNumericalJEE Main 2024
Two circular coils P and Q of 100 turns each have same radius of π\piπ cm. The currents in P and Q are 1 A and 2 A respectively. P and Q are placed with their planes mutually perpendicular with their centers coincide. The resultant magnetic field induction at the center of the coils is x\sqrt{x}x​ mT, where x=x=x= ______. [Use μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7 T A m−1^{-1}−1]

Correct answer: 20

Step-by-step solution →
Q66·PhysicsSingle correctJEE Main 2024
A uniform magnetic field of 2×10−32\times10^{-3}2×10−3 T acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is in Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is
  1. (A)2×10−42\times10^{-4}2×10−4 N–m along positive Z-direction
  2. (B)2×10−42\times10^{-4}2×10−4 N–m along negative Z-direction
  3. (C)2×10−42\times10^{-4}2×10−4 N–m along positive X-direction
  4. (D)2×10−42\times10^{-4}2×10−4 N–m along positive Y-direction

Correct answer: (B)

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2024
A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B=B0j^B=B_0\hat{j}B=B0​j^​ as shown in figure. The magnetic force on the wire if it has a current i is :
  1. (A)−iBRj^-iBR\hat{j}−iBRj^​
  2. (B)2iBRj^2iBR\hat{j}2iBRj^​
  3. (C)iBRj^iBR\hat{j}iBRj^​
  4. (D)−2iBRj^-2iBR\hat{j}−2iBRj^​

Correct answer: (D)

Step-by-step solution →
Q68·PhysicsNumericalJEE Main 2024
An electron moves through a uniform magnetic field B⃗=B0i^+2B0j^\vec{B}=B_0\hat{i}+2B_0\hat{j}B=B0​i^+2B0​j^​ T. At a particular instant of time, the velocity of electron is u⃗=3i^+5j^\vec{u}=3\hat{i}+5\hat{j}u=3i^+5j^​ m/s. If the magnetic force acting on electron is F⃗=5ek^\vec{F}=5e\hat{k}F=5ek^ N, where e is the charge of electron, then the value of B0B_0B0​ is ______ T.

Correct answer: 5

Step-by-step solution →
Q69·PhysicsNumericalJEE Main 2024
The current of 5A flows in a square loop of sides 1 m is placed in air. The magnetic field at centre of the loop is X2×10−7X\sqrt{2}\times10^{-7}X2​×10−7 T. The value of XXX is ______.

Correct answer: 40

Step-by-step solution →
Q70·PhysicsSingle correctJEE Main 2024
Two insulated circular loop AAA and BBB of radius aaa carrying a current of III in the anti clockwise direction as shown in figure. The magnitude of the magnetic induction at the centre will be:
  1. (A)2 μ0Ia\dfrac{\sqrt2\,\mu_0 I}{a}a2​μ0​I​
  2. (B)μ0I2a\dfrac{\mu_0 I}{2a}2aμ0​I​
  3. (C)μ0I2 a\dfrac{\mu_0 I}{\sqrt2\,a}2​aμ0​I​
  4. (D)2μ0Ia\dfrac{2\mu_0 I}{a}a2μ0​I​

Correct answer: (C)

Step-by-step solution →
Q71·PhysicsNumericalJEE Main 2024
A charge of 4.0 μ\muμC is moving with a velocity of 4.0×1064.0\times10^{6}4.0×106 ms−1^{-1}−1 along the positive y-axis under a magnetic field B⃗\vec{B}B of strength (2k^)(2\hat{k})(2k^) T. The force acting on the charge is xi^x\hat{i}xi^ N. The value of xxx is ___.

Correct answer: 32

Step-by-step solution →
Q72·PhysicsSingle correctJEE Main 2024
The deflection in moving coil galvanometer falls from 25 divisions to 5 division when a shunt of 24 Ω24\,\Omega24Ω is applied. The resistance of galvanometer coil will be:
  1. (A)12 Ω12\,\Omega12Ω
  2. (B)96 Ω96\,\Omega96Ω
  3. (C)48 Ω48\,\Omega48Ω
  4. (D)100 Ω100\,\Omega100Ω

Correct answer: (B)

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2024
Two particles X and Y having equal charges are being accelerated through the same potential difference. Thereafter they enter normally in a region of uniform magnetic field and describe circular paths of radii R1R_{1}R1​ and R2R_{2}R2​ respectively. The mass ratio of X and Y is:
  1. (A)(R2R1)2\left(\dfrac{R_{2}}{R_{1}}\right)^{2}(R1​R2​​)2
  2. (B)(R1R2)2\left(\dfrac{R_{1}}{R_{2}}\right)^{2}(R2​R1​​)2
  3. (C)(R1R2)\left(\dfrac{R_{1}}{R_{2}}\right)(R2​R1​​)
  4. (D)(R2R1)\left(\dfrac{R_{2}}{R_{1}}\right)(R1​R2​​)

Correct answer: (B)

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2024
A galvanometer having coil resistance 10 Ω10\,\Omega10Ω shows a full scale deflection for a current of 3 mA. If it to measure a current of 8 A, the value of the shunt should be:
  1. (A)3×10−3 Ω3\times10^{-3}\,\Omega3×10−3Ω
  2. (B)4.85×10−3 Ω4.85\times10^{-3}\,\Omega4.85×10−3Ω
  3. (C)3.75×10−3 Ω3.75\times10^{-3}\,\Omega3.75×10−3Ω
  4. (D)2.75×10−3 Ω2.75\times10^{-3}\,\Omega2.75×10−3Ω

Correct answer: (C)

Step-by-step solution →
Q75·PhysicsNumericalJEE Main 2024
The magnetic field at the centre O of a wire loop formed by two semicircular wires of radii R1=2πR_1=2\piR1​=2π m and R2=4πR_2=4\piR2​=4π m carrying current I=4I=4I=4 A as that figure given below is α×10−7\alpha\times10^{-7}α×10−7 T. The value of α\alphaα is ________. (Centre O is common for all segments)

Correct answer: 3.00

Step-by-step solution →
Q76·PhysicsNumericalJEE Main 2024
Two long, straight wires carry equal currents in opposite directions as shown in figure. The separation between the wires is 5.0 cm. The magnitude of the magnetic field at a point P midway between the wires is __________ μ\muμT. (Given: μ0=4π×10−7\mu_0=4\pi\times 10^{-7}μ0​=4π×10−7 TmA−1^{-1}−1)

Correct answer: 160

Step-by-step solution →
Q77·PhysicsSingle correctJEE Main 2024
A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E⃗\vec EE and B⃗\vec BB represent the electric and magnetic fields respectively, then the region of space may have: (A) E=0E=0E=0, B=0B=0B=0 (B) E≠0E\neq 0E=0, B≠0B\neq 0B=0 (C) E≠0E\neq 0E=0, B=0B=0B=0 (D) E=0E=0E=0, B≠0B\neq 0B=0. Choose the most appropriate answer from the options given below:
  1. (A)(A), (B) and (C) only
  2. (B)(A), (C) and (D) only
  3. (C)(A), (B) and (D) only
  4. (D)(B), (C) and (D) only

Correct answer: (C)

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2024
A current of 200 μ\muμA deflects the coil of a moving coil galvanometer through 60∘60^\circ60∘. The current to cause deflection through π10\dfrac{\pi}{10}10π​ radian is :
  1. (A)60 μ60\,\mu60μA
  2. (B)120 μ120\,\mu120μA
  3. (C)30 μ30\,\mu30μA
  4. (D)180 μ180\,\mu180μA

Correct answer: (A)

Step-by-step solution →
Q79·PhysicsSingle correctJEE Main 2023
For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54 Ω\OmegaΩ showing a full scale deflection for 1 mA as in figure. (A) for voltmeter R≈50R \approx 50R≈50 kΩ\OmegaΩ (B) for ammeter r≈0.2r \approx 0.2r≈0.2 Ω\OmegaΩ (C) for ammeter r≈6r \approx 6r≈6 Ω\OmegaΩ (D) for voltmeter R≈5R \approx 5R≈5 kΩ\OmegaΩ (E) for voltmeter R≈500R \approx 500R≈500 Ω\OmegaΩ Choose the correct answer from the options given below :
  1. (A)(C) and (E)
  2. (B)(C) and (D)
  3. (C)(A) and (C)
  4. (D)(A) and (B)

Correct answer: (C)

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2023
An electron is moving along the positive x-axis. If the uniform magnetic field is applied parallel to the negative z-axis, then A. The electron will experience magnetic force along positive y-axis B. The electron will experience magnetic force along negative y-axis C. The electron will not experience any force in magnetic field D. The electron will continue to move along the positive x-axis E. The electron will move along circular path in magnetic field. Choose the correct answer from the options given below:
  1. (A)B and E only
  2. (B)A and E only
  3. (C)C and D only
  4. (D)B and D only

Correct answer: (A)

Step-by-step solution →
Q81·PhysicsNumericalJEE Main 2023
A straight wire AB of mass 40 g and length 50 cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ____ A. (Take g=10g = 10g=10 ms−2^{-2}−2)

Correct answer: 2

Step-by-step solution →
Q82·PhysicsSingle correctJEE Main 2023
The current sensitivity of moving coil galvanometer is increased by 25%. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:
  1. (A)+25%
  2. (B)−50%- 50\%−50%
  3. (C)Zero
  4. (D)−25%- 25\%−25%

Correct answer: (A)

Step-by-step solution →
Q83·PhysicsSingle correctJEE Main 2023
An electron is allowed to move with constant velocity along the axis of a current carrying straight solenoid. A. The electron will experience magnetic force along the axis of the solenoid. B. The electron will not experience magnetic force. C. The electron will continue to move along the axis of the solenoid. D. The electron will be accelerated along the axis of the solenoid. E. The electron will follow a parabolic path inside the solenoid. Choose the correct answer from the options given below:
  1. (A)B, C and D only
  2. (B)B and C only
  3. (C)A and D only
  4. (D)B and E only

Correct answer: (B)

Step-by-step solution →
Q84·PhysicsNumericalJEE Main 2023
A straight wire carrying a current of 141414 A is bent into a semicircular arc of radius 2.22.22.2 cm as shown in the figure. The magnetic field produced by the current at the centre (O)(O)(O) of the arc is _______ ×10−4\times10^{-4}×10−4 T.

Correct answer: 2

Step-by-step solution →
Q85·PhysicsSingle correctJEE Main 2023
Given below are two statements: Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double. Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio. In the light of the above statement, choose the correct answer from the options given below:
  1. (A)Statement I is true but Statement II is true
  2. (B)Both Statement I and Statement II are true
  3. (C)Both Statement I and Statement II are false
  4. (D)Statement I is true but Statement II is false

Correct answer: (D)

Step-by-step solution →
Q86·PhysicsNumericalJEE Main 2023
The ratio of the magnetic field at the centre of a current carrying coil of radius rrr to the magnetic field at a distance rrr from the centre of the coil on its axis is x:1\sqrt{x}:1x​:1. The value of x is ____.

Correct answer: 8

Step-by-step solution →
Q87·PhysicsSingle correctJEE Main 2023
A charge particle moving in magnetic field BBB, has the components of velocity along BBB as well as perpendicular to BBB. The path of the charge particle will be
  1. (A)helical path with the axis perpendicular to the direction of magnetic field BBB
  2. (B)straight along the direction of magnetic field BBB
  3. (C)helical path with the axis along magnetic field BBB
  4. (D)circular path

Correct answer: (C)

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2023
Certain galvanometers have a fixed core made of non magnetic metallic material. The function of this metallic material is
  1. (A)to oscillate the coil in magnetic field for longer period of time
  2. (B)to bring the coil to rest quickly
  3. (C)to produce large deflecting torque on the coil
  4. (D)to make the magnetic field radial

Correct answer: (B)

Step-by-step solution →
Q89·PhysicsNumericalJEE Main 2023
A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude π2×10−3\dfrac{\pi}{2}\times10^{-3}2π​×10−3 T. The angle between the direction of magnetic field and velocity of proton is 60°. The pitch of the helical path taken by the proton is ____ cm. (Take mass of proton =1.6×10−27=1.6\times10^{-27}=1.6×10−27 kg and charge on proton =1.6×10−19=1.6\times10^{-19}=1.6×10−19 C)

Correct answer: 40

Step-by-step solution →
Q90·PhysicsNumericalJEE Main 2023
Two identical circular wires of radius 20 cm20\,cm20cm and carrying current 2 A\sqrt2\,A2​A are placed in perpendicular planes as shown in figure. The net magnetic field at the centre of the circular wire is _____ ×10−8 T\times10^{-8}\,T×10−8T. (Take π=3.14\pi=3.14π=3.14)

Correct answer: 628

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2023
A long straight wire of circular cross-section (radius aaa) is carrying steady current III. The current III is uniformly distributed across this cross-section. The magnetic field is:
  1. (A)Zero in the region r<ar<ar<a and inversely proportional to rrr in the region r>ar>ar>a
  2. (B)Inversely proportional to rrr in the region r<ar<ar<a and uniform throughout the region r>ar>ar>a
  3. (C)Directly proportional to rrr in the region r<ar<ar<a and inversely proportional to rrr in the region r>ar>ar>a
  4. (D)Uniform in the region r<ar<ar<a and inversely proportional to distance rrr from the axis, in the region r>ar>ar>a

Correct answer: (C)

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2023
A charge particle of 2 μC2\,\mu C2μC accelerated by a potential difference of 100100100 V enters a region of uniform magnetic field of magnitude 444 mT at right angle to the direction of field. The charge particle completes semicircle of radius 333 cm inside magnetic field. The mass of the charge particle is _______ ×10−18\times10^{-18}×10−18 kg.

Correct answer: 144

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2023
Find the magnetic field at the point PPP in the figure. The curved portion is a semicircle of radius rrr connected to two long straight wires.
  1. (A)μ0i2r(1+2π)\tfrac{\mu_0 i}{2r}\left(1+\tfrac{2}{\pi}\right)2rμ0​i​(1+π2​)
  2. (B)μ0i2r(12+12π)\tfrac{\mu_0 i}{2r}\left(\tfrac{1}{2}+\tfrac{1}{2\pi}\right)2rμ0​i​(21​+2π1​)
  3. (C)μ0i2r(1+1π)\tfrac{\mu_0 i}{2r}\left(1+\tfrac{1}{\pi}\right)2rμ0​i​(1+π1​)
  4. (D)μ0i2r(12+1π)\tfrac{\mu_0 i}{2r}\left(\tfrac{1}{2}+\tfrac{1}{\pi}\right)2rμ0​i​(21​+π1​)

Correct answer: (B)

Step-by-step solution →
Q94·PhysicsSingle correctJEE Main 2023
As shown in the figure, a long straight conductor with semicircular arc of radius π10\dfrac{\pi}{10}10π​ m is carrying current I=3I=3I=3 A. The magnitude of the magnetic field at the center O of the arc is: (The permeability of the vacuum =4π×10−7 NA−2=4\pi\times10^{-7}\,NA^{-2}=4π×10−7NA−2)
  1. (A)1 μT1\,\mu T1μT
  2. (B)3 μT3\,\mu T3μT
  3. (C)4 μT4\,\mu T4μT
  4. (D)6 μT6\,\mu T6μT

Correct answer: (B)

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2023
The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50%50\%50%. The percentage change in voltage sensitivity of the galvanometer will be:
  1. (A)0%0\%0%
  2. (B)75%75\%75%
  3. (C)50%50\%50%
  4. (D)100%100\%100%

Correct answer: (A)

Step-by-step solution →
Q96·PhysicsSingle correctJEE Main 2023
A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:
  1. (A)n:Nn:Nn:N
  2. (B)n2:N2n^2:N^2n2:N2
  3. (C)N2:n2N^2:n^2N2:n2
  4. (D)N:nN:nN:n

Correct answer: (C)

Step-by-step solution →
Q97·PhysicsSingle correctJEE Main 2023
A massless square loop, of wire of resistance 10 Ω10\,\Omega10Ω, supporting a mass of 111 g, hangs vertically with one of its sides in a uniform magnetic field of 10310^3103 G, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 111 g? (If sides of the loop =10=10=10 cm, g=10 ms−2g=10\,ms^{-2}g=10ms−2)
  1. (A)110\dfrac{1}{10}101​ V
  2. (B)100100100 V
  3. (C)101010 V
  4. (D)111 V

Correct answer: (C)

Step-by-step solution →
Q98·PhysicsSingle correctJEE Main 2023
A current carrying rectangular loop PQRS is made of uniform wire with PR=QS=5PR=QS=5PR=QS=5 cm and PQ=RS=100PQ=RS=100PQ=RS=100 cm. If the current reading changes from III to 2I2I2I, the ratio of the magnetic force per unit length on the wire PQ due to wire RS in the two cases respectively (fPQI:fPQ2If_{PQ}^{I}:f_{PQ}^{2I}fPQI​:fPQ2I​) is:
  1. (A)1:21:21:2
  2. (B)1:31:31:3
  3. (C)1:41:41:4
  4. (D)1:51:51:5

Correct answer: (C)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Main 2023
The magnetic moments associated with two closely wound circular coils A and B of radius rA=10r_A=10rA​=10 cm and rB=20r_B=20rB​=20 cm respectively are equal if: (Where NA,IAN_A,I_ANA​,IA​ and NB,IBN_B,I_BNB​,IB​ are number of turn and current of A and B respectively)
  1. (A)4NAIA=NBIB4N_A I_A=N_B I_B4NA​IA​=NB​IB​
  2. (B)NA=2NBN_A=2N_BNA​=2NB​
  3. (C)NAIA=4NBIBN_A I_A=4N_B I_BNA​IA​=4NB​IB​
  4. (D)2NAIA=NBIB2N_A I_A=N_B I_B2NA​IA​=NB​IB​

Correct answer: (C)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2023
A current of 2 A is flowing in an equilateral triangle of side 434\sqrt{3}43​ cm. The magnetic field at the centroid O of the triangle is (neglect the effect of earth's magnetic field):
  1. (A)1.43×10−51.4\sqrt{3}\times10^{-5}1.43​×10−5 T
  2. (B)43×10−44\sqrt{3}\times10^{-4}43​×10−4 T
  3. (C)33×10−53\sqrt{3}\times10^{-5}33​×10−5 T
  4. (D)3×10−4\sqrt{3}\times10^{-4}3​×10−4 T

Correct answer: (C)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2023
The magnitude of magnetic induction at mid point OOO due to the current arrangement as shown in the figure will be:
  1. (A)μ0Iπa\dfrac{\mu_0 I}{\pi a}πaμ0​I​
  2. (B)μ0I2πa\dfrac{\mu_0 I}{2\pi a}2πaμ0​I​
  3. (C)000
  4. (D)μ0I4πa\dfrac{\mu_0 I}{4\pi a}4πaμ0​I​

Correct answer: (A)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2023
A single current carrying loop of wire carrying current III flowing in anticlockwise direction seen from +z+z+z direction and lying in the xyxyxy plane is shown in figure. The plot of j^\hat{j}j^​ component of magnetic field (ByB_yBy​) at a distance 'aaa' (less than radius of the coil) and on the yzyzyz plane vs zzz coordinate looks like:
  1. (A)(graph 1)
  2. (B)(graph 2)
  3. (C)(graph 3)
  4. (D)(graph 4)

Correct answer: (A)

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2023
The electric current in a circular coil of four turns produces a magnetic induction 32 T32\,T32T at its centre. The coil is unwound and is rewound into a circular coil of single turn, the magnetic induction at the centre of the coil by the same current will be:
  1. (A)16 T16\,T16T
  2. (B)2 T2\,T2T
  3. (C)8 T8\,T8T
  4. (D)4 T4\,T4T

Correct answer: (B)

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2023
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:
  1. (A)2.4×103 A m−12.4\times10^{3}\ A\,m^{-1}2.4×103 Am−1
  2. (B)1.2×103 A m−11.2\times10^{3}\ A\,m^{-1}1.2×103 Am−1
  3. (C)2.4×10−3 A m−12.4\times10^{-3}\ A\,m^{-1}2.4×10−3 Am−1
  4. (D)1 A m−11\ A\,m^{-1}1 Am−1

Correct answer: (B)

Step-by-step solution →
Q105·PhysicsSingle correctJEE Main 2023
Match List-I (current configurations shown in the figure) with List-II (magnitude of magnetic field at the point O). Choose the correct answer from the options given below:
List-I (Current configuration)List-II (Magnitude of magnetic field at point O)
A.see figureI.B0=μ0I4πr[π+2]B_0=\dfrac{\mu_0 I}{4\pi r}[\pi+2]B0​=4πrμ0​I​[π+2]
B.see figureII.B0=μ04IrB_0=\dfrac{\mu_0}{4}\dfrac{I}{r}B0​=4μ0​​rI​
C.see figureIII.B0=μ0I2πr[π−1]B_0=\dfrac{\mu_0 I}{2\pi r}[\pi-1]B0​=2πrμ0​I​[π−1]
D.see figureIV.B0=μ0I4πr[π+1]B_0=\dfrac{\mu_0 I}{4\pi r}[\pi+1]B0​=4πrμ0​I​[π+1]
  1. (A)A-III, B-I, C-IV, D-II
  2. (B)A-I, B-III, C-IV, D-II
  3. (C)A-III, B-IV, C-I, D-II
  4. (D)A-II, B-I, C-IV, D-III

Correct answer: (A)

Step-by-step solution →
Q106·PhysicsNumericalJEE Main 2023
Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point PPP is at equidistant from both the wires such that the lines joining the point PPP to the wires are perpendicular to each other. The magnitude of magnetic field at PPP is ______ ×10−6\times 10^{-6}×10−6 T (Given: 2=1.4\sqrt{2} = 1.42​=1.4)

Correct answer: 68

Step-by-step solution →
Q107·PhysicsSingle correctJEE Main 2023
For a moving coil galvanometer, the deflection in the coil is 0.050.050.05 rad when a current of 101010 mA is passed through it. If the torsional constant of suspension wire is 4.0×10−54.0\times10^{-5}4.0×10−5 N m rad−1^{-1}−1, the magnetic field is 0.010.010.01 T and the number of turns in the coil is 200200200, the area of each turn (in cm2^22) is:
  1. (A)1.01.01.0
  2. (B)2.02.02.0
  3. (C)1.51.51.5
  4. (D)0.50.50.5

Correct answer: (A)

Step-by-step solution →
Q108·PhysicsNumericalJEE Main 2023
A single turn current loop in the shape of a right angle triangle with sides 555 cm, 121212 cm, 131313 cm is carrying a current of 222 A. The loop is in a uniform magnetic field of magnitude 0.750.750.75 T whose direction is parallel to the current in the 131313 cm side of the loop. The magnitude of the magnetic force on the 555 cm side will be x130\dfrac{x}{130}130x​ N. The value of xxx is _________

Correct answer: 9

Step-by-step solution →
Q109·PhysicsSingle correctJEE Main 2023
A circular loop of radius rrr is carrying current III A. The ratio of magnetic field at the centre of the circular loop and at a distance rrr from the centre of the loop on its axis is:
  1. (A)22:12\sqrt{2}:122​:1
  2. (B)1:321:3\sqrt{2}1:32​
  3. (C)1:21:\sqrt{2}1:2​
  4. (D)32:23\sqrt{2}:232​:2

Correct answer: (A)

Step-by-step solution →
Q110·PhysicsSingle correctJEE Main 2023
A long solenoid is formed by winding 707070 turns cm−1cm^{-1}cm−1. If 2.02.02.0 A current flows, then the magnetic field produced inside the solenoid is _________ (μ0=4π×10−7 T m A−1)(\mu_0=4\pi\times10^{-7}\,T\,m\,A^{-1})(μ0​=4π×10−7TmA−1)
  1. (A)88×10−488\times10^{-4}88×10−4 T
  2. (B)352×10−4352\times10^{-4}352×10−4 T
  3. (C)176×10−4176\times10^{-4}176×10−4 T
  4. (D)1232×10−41232\times10^{-4}1232×10−4 T

Correct answer: (C)

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2023
Two long straight wires PPP and QQQ carrying equal current 10 A each were kept parallel to each other at 5 cm distance. Magnitude of magnetic force experienced by 10 cm length of wire P is F1F_1F1​. If distance between wires is halved and currents on them are doubled, force F2F_2F2​ on 10 cm length of wire P will be:
  1. (A)F18\tfrac{F_1}{8}8F1​​
  2. (B)8F18F_18F1​
  3. (C)10F110F_110F1​
  4. (D)F110\tfrac{F_1}{10}10F1​​

Correct answer: (B)

Step-by-step solution →
Q112·PhysicsSingle correctJEE Advanced 2022
Which one of the following options represents the magnetic field B⃗\vec{B}B at O due to the current flowing in the given wire segments lying on the xy plane?
  1. (A)B⃗=−μ0IL(32+142π)k^\vec{B} = \frac{-\mu_0 I}{L}\left(\frac{3}{2} + \frac{1}{4\sqrt{2}\pi}\right)\hat{k}B=L−μ0​I​(23​+42​π1​)k^
  2. (B)B⃗=−μ0IL(32+122π)k^\vec{B} = -\frac{\mu_0 I}{L}\left(\frac{3}{2} + \frac{1}{2\sqrt{2}\pi}\right)\hat{k}B=−Lμ0​I​(23​+22​π1​)k^
  3. (C)B⃗=−μ0IL(1+142π)k^\vec{B} = \frac{-\mu_0 I}{L}\left(1 + \frac{1}{4\sqrt{2}\pi}\right)\hat{k}B=L−μ0​I​(1+42​π1​)k^
  4. (D)B⃗=−μ0IL(1+14π)k^\vec{B} = \frac{-\mu_0 I}{L}\left(1 + \frac{1}{4\pi}\right)\hat{k}B=L−μ0​I​(1+4π1​)k^

Correct answer: (C)

Step-by-step solution →
Q113·PhysicsNumericalJEE Main 2022
A closely wounded circular coil of radius 5 cm produces a magnetic field of 37.68 x 10−410^{-4}10−4 T at its center. The current through the coil is __________ A. [Given, number of turns in the coil is 100 and π\piπ = 3.14]

Correct answer: 3

Step-by-step solution →
Q114·PhysicsNumericalJEE Main 2022
A wire of length 314 cm carrying current of 14 A is bent to form a circle. The magnetic moment of the coil is ________ A-m2m^2m2. [Given π\piπ = 3.14]

Correct answer: 11

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2022
A wire X of length 50 cm carrying a current of 2 A is placed parallel to a long wire Y of length 5 m. The wire Y carries a current of 3 A. The distance between two wires is 5 cm and currents flow in the same direction. The force acting on the wire Y is :
  1. (A)1.2×10−51.2 \times 10^{-5}1.2×10−5 N directed towards wire X.
  2. (B)1.2×10−41.2 \times 10^{-4}1.2×10−4 N directed away from wire X.
  3. (C)1.2×10−41.2 \times 10^{-4}1.2×10−4 N directed towards wire X.
  4. (D)2.4×10−52.4 \times 10^{-5}2.4×10−5 N directed towards wire X.

Correct answer: (A)

Step-by-step solution →
Q116·PhysicsSingle correctJEE Main 2022
As shown in the figure, a metallic rod of linear density 0.45 kg m−1^{-1}−1 is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be : {Use g = 10 m/s2^{2}2}
  1. (A)30 A
  2. (B)15 A
  3. (C)10 A
  4. (D)3 A

Correct answer: (A)

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2022
A triangular shaped wire carrying 10A current is placed in a uniform magnetic field of 0.5T, as shown in figure. The magnetic force on segment CD is (Given BC = CD = BD = 5 cm).
  1. (A)0.126 N
  2. (B)0.312 N
  3. (C)0.216 N
  4. (D)0.245 N

Correct answer: (C)

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2022
The current sensitivity of a galvanometer can be increased by : (A) decreasing the number of turns (B) increasing the magnetic field (C) decreasing the area of the coil (D) decreasing the torsional constant of the spring Choose the most appropriate answer from the options given below :
  1. (A)(B) and (C) only
  2. (B)(C) and (D) only
  3. (C)(A) and (C) only
  4. (D)(B) and (D) only

Correct answer: (D)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Main 2022
The magnetic field at the center of current carrying circular loop is B1B_1B1​. The magnetic field at a distance of 3\sqrt{3}3​ times radius of the given circular loop from the center on its axis is B2B_2B2​. The value of B1/B2B_1/B_2B1​/B2​ will be
  1. (A)9:49 : 49:4
  2. (B)12:512 : \sqrt{5}12:5​
  3. (C)8:18 : 18:1
  4. (D)5:35 : \sqrt{3}5:3​

Correct answer: (C)

Step-by-step solution →
Q120·PhysicsSingle correctJEE Main 2022
A cyclotron is used to accelerate protons. If the operating magnetic field is 1.0 T and the radius of the cyclotron 'dees' is 60 cm, the kinetic energy of the accelerated protons in MeV will be : [use mpm_pmp​ = 1.6 × 10−27^{-27}−27 kg, e = 1.6 × 10−19^{-19}−19 C]
  1. (A)12
  2. (B)18
  3. (C)16
  4. (D)32

Correct answer: (B)

Step-by-step solution →
Q121·PhysicsSingle correctJEE Main 2022
BX_{X}X​ and BY_{Y}Y​ are the magnetic field at the centre of two coils of two coils X and Y respectively, each carrying equal current. If coil X has 200 turns and 20 cm radius and coil Y has 400 turns and 20 cm radius, the ratio of BX_{X}X​ and BY_{Y}Y​ is
  1. (A)1 : 1
  2. (B)1 : 2
  3. (C)2 : 1
  4. (D)4 : 1

Correct answer: (B)

Step-by-step solution →
Q122·PhysicsSingle correctJEE Main 2022
A charge particle is moving in a uniform magnetic field (2i^+3j^)T\left(2\hat{i} + 3\hat{j}\right)T(2i^+3j^​)T . If it has an acceleration of (αi^−4j^)m/s2\left(\alpha\hat{i} - 4\hat{j}\right)m/s^{2}(αi^−4j^​)m/s2, then the value of α will be
  1. (A)3
  2. (B)6
  3. (C)12
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q123·PhysicsSingle correctJEE Main 2022
A velocity selector consists of electric field E⃗=Ek^\vec{E} = E\hat{k}E=Ek^ and magnetic field B⃗=Bj^\vec{B} = B\hat{j}B=Bj^​ with B=12 mT. The value E required for an electron of energy 728 eV moving along the positive x-axis to pass undeflected is : (Given, mass of electron = 9.1×10−319.1\times10^{-31}9.1×10−31kg)
  1. (A)192 kVm−1^{-1}−1
  2. (B)192 m Vm−1^{-1}−1
  3. (C)9600 kVm−1^{-1}−1
  4. (D)16 kVm−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2022
Two concentric circular loops of radii r1r_1r1​=30 cm and r2r_2r2​=50 cm are placed in X-Y plane as shown in the figure. A current I = 7A is flowing through them in the direction as shown in figure. The net magnetic moment of this system of two circular loops is approximately :
  1. (A)72k^\frac{7}{2}\hat{k}27​k^ Am2^22
  2. (B)−72k^-\frac{7}{2}\hat{k}−27​k^ Am2^22
  3. (C)7k^7\hat{k}7k^ Am2^22
  4. (D)−7k^-7\hat{k}−7k^ Am2^22

Correct answer: (B)

Step-by-step solution →
Q125·PhysicsSingle correctJEE Main 2022
Two charged particles, having same kinetic energy, are allowed to pass through a uniform magnetic field perpendicular to the direction of motion. If the ratio of radii of their circular paths is 6 : 5 and their respective masses ratio is 9 : 4. Then, the ratio of their charges will be :
  1. (A)8 : 5
  2. (B)5 : 4
  3. (C)5 : 3
  4. (D)8 : 7

Correct answer: (B)

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2022
Given below are two statements : Statement I: The electric force changes the speed of the charged particle and hence changes its kinetic energy: whereas the magnetic force does not change the kinetic energy of the charged particle. Statement II: The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both Statement I and Statement II are correct.
  2. (B)Both Statement I and Statement II are incorrect.
  3. (C)Statement I is correct but Statement II is incorrect.
  4. (D)Statement I is incorrect but Statement II is correct.

Correct answer: (C)

Step-by-step solution →
Q127·PhysicsSingle correctJEE Main 2022
A charged particle moves along circular path in a uniform magnetic field in a cyclotron. The kinetic energy of the charged particle increases to 4 times its initial value. What will be the ratio of new radius to the original radius of circular path of the charged particle :
  1. (A)1 : 1
  2. (B)1 : 2
  3. (C)2 : 1
  4. (D)1 : 4

Correct answer: (C)

Step-by-step solution →
Q128·PhysicsSingle correctJEE Main 2022
Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 μT. The equal current flowing in the two conductors is :
  1. (A)30A in the same direction.
  2. (B)30A in the opposite direction.
  3. (C)60A in the opposite direction.
  4. (D)300A in the opposite direction.

Correct answer: (B)

Step-by-step solution →
Q129·PhysicsNumericalJEE Main 2022
A sing1y ionized magnesium atom (A24) ion is accelerated to kinetic energy 5 keV and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be ____________ cm.

Correct answer: 10

Step-by-step solution →
Q130·PhysicsSingle correctJEE Main 2022
Two parallel, long wires are kept 0.20 m apart in vacuum, each carrying current of x A in the same direction. If the force of attraction per meter of each wire is 2×10−62\times10^{-6}2×10−6 N, then the value of x is approximately:
  1. (A)1
  2. (B)2.4
  3. (C)1.4
  4. (D)2

Correct answer: (C)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2022
An infinitely long hollow conducting cylinder with radius R carries a uniform current along its surface. Choose the correct representation of magnetic field (B) as a function of radial distance (r) from the axis of cylinder.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q132·PhysicsSingle correctJEE Main 2022
Two long parallel conductors S1S_{1}S1​ and S2S_{2}S2​ are separated by a distance 10 cm and carrying currents of 4A and 2A respectively. The conductors are placed along x-axis in X-Y plane. There is a point P located between the conductors (as shown in figure). A charge particle of 3π3\pi3π coulomb is passing through the point P with velocity v⃗=(2i^+3j^) m/s\vec{v} = (2\hat{i} + 3\hat{j})\,\text{m}/\text{s}v=(2i^+3j^​)m/s; where i^ & j^\hat{i}\ \& \ \hat{j}i^ & j^​ represents unit vector along x & y axis respectively. The force acting on the charge particle is 4π×10−5(−xi^+2j^) N4\pi \times 10^{-5}(-x\hat{i} + 2\hat{j})\,\text{N}4π×10−5(−xi^+2j^​)N. The value of x is :
  1. (A)2
  2. (B)1
  3. (C)3
  4. (D)−3-3−3

Correct answer: (C)

Step-by-step solution →
Q133·PhysicsNumericalJEE Main 2022
A deuteron and a proton moving with equal kinetic energy enter into to a uniform magnetic field at right angle to the field. If rd_{d}d​ and rp_{p}p​ are the radii of their circular paths respectively, then the ratio rdrp\dfrac{r_d}{r_p}rp​rd​​ will be x\sqrt{x}x​ : 1 where x is ________ .

Correct answer: 2

Step-by-step solution →
Q134·PhysicsNumericalJEE Main 2022
Two 10 cm long, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10−510^{-5}10−5 N, then separation between the wires is _________ cm.

Correct answer: 5

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2022
A long straight wire with a circular cross-section having radius R, is carrying a steady current I. The current I is uniformly distributed across this cross-section. Then the variation of magnetic field due to current I with distance r (r < R) from its centre will be :-
  1. (A)B∝r2B\propto r^2B∝r2
  2. (B)B∝rB\propto rB∝r
  3. (C)B∝1r2B\propto\frac{1}{r^2}B∝r21​
  4. (D)B∝1rB\propto\frac{1}{r}B∝r1​

Correct answer: (B)

Step-by-step solution →
Q136·PhysicsSingle correctJEE Main 2022
A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to
  1. (A)B
  2. (B)2 B
  3. (C)4 B
  4. (D)B2\frac{B}{2}2B​

Correct answer: (A)

Step-by-step solution →
Q137·PhysicsSingle correctJEE Main 2022
Given below are two statements : One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In an uniform magnetic field, speed and energy remains the same for a moving charged particle. Reason (R) : Moving charged particle experiences magnetic force perpendicular to its direction of motion.
  1. (A)Both (A) and (R) are true and (R) is the correct explanation of (A)
  2. (B)Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  3. (C)(A) is true but (R) is false
  4. (D)(A) is false but (R) is true.

Correct answer: (A)

Step-by-step solution →
Q138·PhysicsSingle correctJEE Main 2022
The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance r2\dfrac{r}{2}2r​ from the centre is :
  1. (A)B/2
  2. (B)2B
  3. (C)(25)3B\left(\dfrac{2}{\sqrt{5}}\right)^{3}B(5​2​)3B
  4. (D)(23)3B\left(\dfrac{2}{\sqrt{3}}\right)^{3}B(3​2​)3B

Correct answer: (C)

Step-by-step solution →
Q139·PhysicsSingle correctJEE Main 2022
A proton, a deuteron and an α\alphaα-particle with same kinetic energy enter into a uniform magnetic field at right angle to magnetic field. The ratio of the radii of their respective circular paths is :
  1. (A)1:2:21:\sqrt{2}:\sqrt{2}1:2​:2​
  2. (B)1:1:21:1:\sqrt{2}1:1:2​
  3. (C)2:1:1\sqrt{2}:1:12​:1:1
  4. (D)1:2:11:\sqrt{2}:11:2​:1

Correct answer: (D)

Step-by-step solution →
Q140·PhysicsIntegerJEE Advanced 2021
An α\alphaα-particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the α\alphaα-particle and the sulfur ion move in circular orbits of radii rαr_\alpharα​ and rsr_srs​, respectively. The ratio (rs/rα)(r_s/r_\alpha)(rs​/rα​) is____.

Correct answer: 4

Step-by-step solution →
Q141·PhysicsMultiple correctJEE Advanced 2021
Two concentric circular loops, one of radius RRR and the other of radius 2R2R2R, lie in the xyxyxy-plane with the origin as their common center, as shown in the figure. The smaller loop carries current I1I_1I1​ in the anti-clockwise direction and the larger loop carries current I2I_2I2​ in the clockwise direction, with I2>2I1I_2 > 2I_1I2​>2I1​. B⃗(x,y)\vec{B}(x, y)B(x,y) denotes the magnetic field at a point (x,y)(x, y)(x,y) in the xyxyxy-plane. Which of the following statement(s) is(are) current?
  1. (A)B⃗(x,y)\vec{B}(x, y)B(x,y) is perpendicular to the xyxyxy-plane at any point in the plane
  2. (B)∣B⃗(x,y)∣\left| \vec{B}(x,y) \right|​B(x,y)​ depends on xxx and yyy only through the radial distance r=x2+y2r = \sqrt{x^2 + y^2}r=x2+y2​
  3. (C)∣B⃗(x,y)∣\left| \vec{B}(x,y) \right|​B(x,y)​ is non-zero at all points for r<Rr < Rr<R
  4. (D)B⃗(x,y)\vec{B}(x, y)B(x,y) points normally outward from the xyxyxy-plane for all the points between the two loops

Correct answer: (A), (B)

Step-by-step solution →
Q142·PhysicsSingle correctJEE Main 2021
There are two infinitely long straight current carrying conductors and they are held at right angles to each other so that their common ends meet at the origin as shown in the figure given below. The ratio of current in both conductor is 1 : 1. The magnetic field at point P is ____.
  1. (A)μ0I4πxy[x2+y2+(x+y)]\frac{\mu_{0}I}{4\pi xy}\left[\sqrt{x^{2}+y^{2}}+(x+y)\right]4πxyμ0​I​[x2+y2​+(x+y)]
  2. (B)μ0I4πxy[x2+y2−(x+y)]\frac{\mu_{0}I}{4\pi xy}\left[\sqrt{x^{2}+y^{2}}-(x+y)\right]4πxyμ0​I​[x2+y2​−(x+y)]
  3. (C)μ0Ixy4π[x2+y2−(x+y)]\frac{\mu_{0}Ixy}{4\pi}\left[\sqrt{x^{2}+y^{2}}-(x+y)\right]4πμ0​Ixy​[x2+y2​−(x+y)]
  4. (D)μ0Ixy4π[x2+y2+(x+y)]\frac{\mu_{0}Ixy}{4\pi}\left[\sqrt{x^{2}+y^{2}}+(x+y)\right]4πμ0​Ixy​[x2+y2​+(x+y)]

Correct answer: (A)

Step-by-step solution →
Q143·PhysicsSingle correctJEE Main 2021
A current of 1.5 A is flowing through a triangle, of side 9 cm each. The magnetic field at the centroid of the triangle is : (Assume that the current is flowing in the clockwise direction.)
  1. (A)3×10−73 \times 10^{-7}3×10−7 T, outside the plane of triangle
  2. (B)23×10−72\sqrt{3} \times 10^{-7}23​×10−7 T, outside the plane of triangle
  3. (C)23×10−52\sqrt{3} \times 10^{-5}23​×10−5 T, inside the plane of triangle
  4. (D)3×10−53 \times 10^{-5}3×10−5 T, inside the plane of triangle

Correct answer: (D)

Step-by-step solution →
Q144·PhysicsSingle correctJEE Main 2021
A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil :
  1. (A)μ0 IN2(b−a)log⁡e(ba)\frac{\mu_{0}\,IN}{2(b-a)}\log_{e}\left(\frac{b}{a}\right)2(b−a)μ0​IN​loge​(ab​)
  2. (B)μ0I8[a+ba−b]\frac{\mu_{0}I}{8}\left[\frac{a+b}{a-b}\right]8μ0​I​[a−ba+b​]
  3. (C)μ0I4(a−b)[1a−1b]\frac{\mu_{0}I}{4(a-b)}\left[\frac{1}{a}-\frac{1}{b}\right]4(a−b)μ0​I​[a1​−b1​]
  4. (D)μ0I8(a−ba+b)\frac{\mu_{0}I}{8}\left(\frac{a-b}{a+b}\right)8μ0​I​(a+ba−b​)

Correct answer: (A)

Step-by-step solution →
Q145·PhysicsNumericalJEE Main 2021
A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then in the form of a square of side 'a'. The coil is connected to a voltage source V0V_0V0​. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1:y1 : \sqrt{y}1:y​ where y is ....... .

Correct answer: 3

Step-by-step solution →
Q146·PhysicsSingle correctJEE Main 2021
Two ions of masses 4 amu and 16 amu have charges +2e and +3e respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then :
  1. (A)lighter ion will be deflected less than heavier ion
  2. (B)lighter ion will be deflected more than heavier ion
  3. (C)both ions will be deflected equally
  4. (D)no ion will be deflected.

Correct answer: (B)

Step-by-step solution →
Q147·PhysicsSingle correctJEE Main 2021
A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current i0i_{0}i0​, which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x < a and (ii) a < x < b ?
  1. (A)x2a2\frac{x^{2}}{a^{2}}a2x2​
  2. (B)a2x2\frac{a^{2}}{x^{2}}x2a2​
  3. (C)x2b2−a2\frac{x^{2}}{b^{2}-a^{2}}b2−a2x2​
  4. (D)b2−a2x2\frac{b^{2}-a^{2}}{x^{2}}x2b2−a2​

Correct answer: (A)

Step-by-step solution →
Q148·PhysicsNumericalJEE Main 2021
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be x×10−5\sqrt{x} \times 10^{-5}x​×10−5 Nm. The value of x is...........

Correct answer: 3

Step-by-step solution →
Q149·PhysicsSingle correctJEE Main 2021
The fractional change in the magnetic field intensity at a distance 'r' from centre on the axis of current carrying coil of radius 'a' to the magnetic field intensity at the centre of the same coil is : (Take r < a)
  1. (A)32a2r2\frac{3}{2}\frac{a^2}{r^2}23​r2a2​
  2. (B)23a2r2\frac{2}{3}\frac{a^2}{r^2}32​r2a2​
  3. (C)23r2a2\frac{2}{3}\frac{r^2}{a^2}32​a2r2​
  4. (D)32r2a2\frac{3}{2}\frac{r^2}{a^2}23​a2r2​

Correct answer: (D)

Step-by-step solution →
Q150·PhysicsNumericalJEE Main 2021
If the maximum value of accelerating potential provided by a ratio frequency oscillator is 12 kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is ............ [mp=1.67×10−27[m_p = 1.67 \times 10^{-27}[mp​=1.67×10−27 kg, e=1.6×10−19e = 1.6 \times 10^{-19}e=1.6×10−19 C, Speed of light =3×108= 3 \times 10^8=3×108 m/s]]]

Correct answer: 543

Step-by-step solution →
Q151·PhysicsSingle correctJEE Main 2021
Figure A and B show long straight wires of circular cross − section (a and b with a < b), carrying current I which is uniformly distributed across the cross − section. The magnitude of magnetic field B varies with radius r and can be represented as :
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q152·PhysicsSingle correctJEE Main 2021
Two ions having same mass have charges in the ratio 1 : 2. They are projected normally in a uniform magnetic field with their speeds in the ratio 2 : 3. The ratio of the radii of their circular trajectories is :
  1. (A)3 : 1
  2. (B)2 : 3
  3. (C)4 : 3
  4. (D)1 : 4

Correct answer: (C)

Step-by-step solution →
Q153·PhysicsSingle correctJEE Main 2021
A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let rdr_drd​ and rdr_drd​ be their respective radii of circular path. The value of rdrα\frac{r_d}{r_\alpha}rα​rd​​ is equal to :
  1. (A)2\sqrt{2}2​
  2. (B)2
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)1

Correct answer: (A)

Step-by-step solution →
Q154·PhysicsSingle correctJEE Main 2021
A loop of flexible wire of irregular shape carrying current is placed in an external magnetic field. Identify the effect of the field on the wire.
  1. (A)Loop assumes circular shape with its plane normal to the field.
  2. (B)Loop assumes circular shape with its plane parallel to the field.
  3. (C)Wire gets stretched to become straight.
  4. (D)Shape of the loop remains unchanged.

Correct answer: (A)

Step-by-step solution →
Q155·PhysicsSingle correctJEE Main 2021
A proton and an α-particle, having kinetic energies KpK_pKp​ and KαK_\alphaKα​, respectively, enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of α-particle is 2 : 1. The ratio of Kp:KαK_p : K_\alphaKp​:Kα​ is :
  1. (A)1 : 8
  2. (B)8 : 1
  3. (C)1 : 4
  4. (D)4 : 1

Correct answer: (D)

Step-by-step solution →
Q156·PhysicsSingle correctJEE Main 2021
Four identical long solenoids A, B, C and D are connected to each other as shown in the figure. If the magnetic field at the center of A is 3T, the field at the center of C would be : (Assume that the magnetic field is confined with in the volume of respective solenoid).
  1. (A)12T
  2. (B)6T
  3. (C)9T
  4. (D)1T

Correct answer: (D)

Step-by-step solution →
Q157·PhysicsSingle correctJEE Main 2021
A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point P which lies on the centre of the semicircle ?
  1. (A)μ0I4πr(2−π)\frac{\mu_{0}I}{4\pi r}\left(2 - \pi\right)4πrμ0​I​(2−π)
  2. (B)μ0I4πr(2+π)\frac{\mu_{0}I}{4\pi r}\left(2 + \pi\right)4πrμ0​I​(2+π)
  3. (C)μ0I2πr(2+π)\frac{\mu_{0}I}{2\pi r}\left(2 + \pi\right)2πrμ0​I​(2+π)
  4. (D)μ0I2πr(2−π)\frac{\mu_{0}I}{2\pi r}\left(2 - \pi\right)2πrμ0​I​(2−π)

Correct answer: (B)

Step-by-step solution →
Q158·PhysicsSingle correctJEE Main 2021
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is : (permeability of free space = 4π×10−74\pi \times 10^{-7}4π×10−7 H/m)
  1. (A)π\piπT
  2. (B)2×10−3 π2 \times 10^{-3}\ \pi2×10−3 πT
  3. (C)π5\frac{\pi}{5}5π​ T
  4. (D)10−4π10^{-4}\pi10−4πT

Correct answer: (A)

Step-by-step solution →
Q159·PhysicsSingle correctJEE Main 2021
A charge Q is moving dI→\overrightarrow{dI}dI distance in the magnetic field B⃗\vec{B}B. Find the value of work done by B⃗\vec{B}B.
  1. (A)1
  2. (B)Infinite
  3. (C)Zero
  4. (D)−1

Correct answer: (C)

Step-by-step solution →
Q160·PhysicsSingle correctJEE Main 2021
A proton, a deuteron and an α particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces action on them is ________ and their speed is ______, in the ratio.
  1. (A)2 : 1 : 1 and 4 : 2 : 1
  2. (B)1 : 2 : 4 and 2 : 1 :1
  3. (C)1 : 2 : 4 and 1 : 1 : 2
  4. (D)4 : 2 : 1 and 2 : 1 : 1

Correct answer: (A)

Step-by-step solution →
Q161·PhysicsSingle correctJEE Main 2021
Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 02 m from the centre are in the rato 8 : 1. The radius of coil is ________
  1. (A)0.15 m
  2. (B)0.2 m
  3. (C)0.1 m
  4. (D)1.0 m

Correct answer: (C)

Step-by-step solution →
Q162·PhysicsSingle correctJEE Advanced 2020
A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field BoB_oBo​ parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)-
  1. (A)π2NQBoR2\frac{\pi}{2}NQB_oR^22π​NQBo​R2
  2. (B)πNQBoR2\pi NQB_oR^2πNQBo​R2
  3. (C)2πNQBoR22\pi NQB_oR^22πNQBo​R2
  4. (D)4πNQBoR24\pi NQB_oR^24πNQBo​R2

Correct answer: (B)

Step-by-step solution →
Q163·PhysicsSingle correctJEE Main 2020
An electron is moving along + x direction with a velocity of 6×106ms−16\times10^6\text{ms}^{-1}6×106ms−1. It enters a region of uniform electric field of 300 V/cm pointing along + y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be:
  1. (A)3×10−43\times10^{-4}3×10−4T, along + z direction
  2. (B)5×10−35\times10^{-3}5×10−3T, along – z direction
  3. (C)5×10−35\times10^{-3}5×10−3T, along + z direction
  4. (D)3×10−43\times10^{-4}3×10−4T, along – z direction

Correct answer: (C)

Step-by-step solution →
Q164·PhysicsSingle correctJEE Main 2020
A square loop of side 2a and carrying current I is kept in xz plane with its centre at origin. A long wire carrying the same current I is placed parallel to z-axis and passing through point (0, b, 0), (b >> a). The magnitude of torque on the loop about z-axis will be
  1. (A)2μ0I2a2πb\frac{2\mu_{0}I^{2}a^{2}}{\pi b}πb2μ0​I2a2​
  2. (B)2μ0I2a2bπ(a2+b2)\frac{2\mu_{0}I^{2}a^{2}b}{\pi(a^{2}+b^{2})}π(a2+b2)2μ0​I2a2b​
  3. (C)μ0I2a2b2π(a2+b2)\frac{\mu_{0}I^{2}a^{2}b}{2\pi(a^{2}+b^{2})}2π(a2+b2)μ0​I2a2b​
  4. (D)μ0I2a22πb\frac{\mu_{0}I^{2}a^{2}}{2\pi b}2πbμ0​I2a2​

Correct answer: (B)

Step-by-step solution →
Q165·PhysicsSingle correctJEE Main 2020
A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane with speed v under the influence of magnetic field B⃗\vec{B}B. The magnetic moment of this moving particle:
  1. (A)mv2B⃗2B2\frac{mv^{2}\vec{B}}{2B^{2}}2B2mv2B​
  2. (B)−mv2B⃗2πB2-\frac{mv^{2}\vec{B}}{2\pi B^{2}}−2πB2mv2B​
  3. (C)−mv2B⃗B2-\frac{mv^{2}\vec{B}}{B^{2}}−B2mv2B​
  4. (D)−mv2B⃗2B2-\frac{mv^{2}\vec{B}}{2B^{2}}−2B2mv2B​

Correct answer: (D)

Step-by-step solution →
Q166·PhysicsSingle correctJEE Main 2020
A square loop of side 2a, and carrying current I, is kept in XZ plane with its centre at origin. A long wire carrying the same current I is placed parallel to the z − axis and passing through the point (0, b, 0), (b >> a). The magnitude of the torque on the loop about z − axis is given by:
  1. (A)2μ0I2a3πb2\dfrac{2\mu_0 I^2 a^3}{\pi b^2}πb22μ0​I2a3​
  2. (B)μ0I2a22πb\dfrac{\mu_0 I^2 a^2}{2\pi b}2πbμ0​I2a2​
  3. (C)μ0I2a32πb2\dfrac{\mu_0 I^2 a^3}{2\pi b^2}2πb2μ0​I2a3​
  4. (D)2μ0I2a2πb\dfrac{2\mu_0 I^2 a^2}{\pi b}πb2μ0​I2a2​

Correct answer: (D)

Step-by-step solution →
Q167·PhysicsSingle correctJEE Main 2020
A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is:
  1. (A)6 : 5
  2. (B)2 : 5
  3. (C)6 : 4
  4. (D)4 : 6

Correct answer: (A)

Step-by-step solution →
Q168·PhysicsNumericalJEE Main 2020
A galvanometer coil has 500 turns and each turn has an average are of 3×10−4m23\times 10^{-4}m^{2}3×10−4m2. If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is __________.

Correct answer: 20.00

Step-by-step solution →
Q169·PhysicsSingle correctJEE Main 2020
Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm. 50 turns and carrying current I (Ampere) in units of μ0Iπ\dfrac{\mu_0 I}{\pi}πμ0​I​ is:
  1. (A)5003500\sqrt{3}5003​
  2. (B)50350\sqrt{3}503​
  3. (C)535\sqrt{3}53​
  4. (D)2503250\sqrt{3}2503​

Correct answer: (A)

Step-by-step solution →
Q170·PhysicsSingle correctJEE Main 2020
A charged particle carrying charge 1 μC is moving with velocity (2i^+3j^+4jk^)\left(2\hat{i} + 3\hat{j} + 4j\hat{k}\right)(2i^+3j^​+4jk^) ms−1^{-1}−1. If an external magnetic field of (5i^+3j^−6jk^)×10−3\left(5\hat{i} + 3\hat{j} - 6j\hat{k}\right) \times 10^{-3}(5i^+3j^​−6jk^)×10−3 T exists in the region where the particle is moving then the force on the particle is F⃗×10−9\vec{F} \times 10^{-9}F×10−9 N. The vector F⃗\vec{F}F is:
  1. (A)−30i^+32j^−9k^-30\hat{i} + 32\hat{j} - 9\hat{k}−30i^+32j^​−9k^
  2. (B)−3.0i^+3.2j^−0.9k^-3.0\hat{i} + 3.2\hat{j} - 0.9\hat{k}−3.0i^+3.2j^​−0.9k^
  3. (C)−300i^+320j^−90k^-300\hat{i} + 320\hat{j} - 90\hat{k}−300i^+320j^​−90k^
  4. (D)−0.30i^+0.32j^−0.09k^-0.30\hat{i} + 0.32\hat{j} - 0.09\hat{k}−0.30i^+0.32j^​−0.09k^

Correct answer: (A)

Step-by-step solution →
Q171·PhysicsSingle correctJEE Main 2020
A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance a3\dfrac{a}{3}3a​ and 2a, respectively from the axis of the wire is:
  1. (A)23\dfrac{2}{3}32​
  2. (B)12\dfrac{1}{2}21​
  3. (C)2
  4. (D)32\dfrac{3}{2}23​

Correct answer: (A)

Step-by-step solution →
Q172·PhysicsSingle correctJEE Main 2020
A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field Ei^E\hat{i}Ei^ and a uniform magnetic field Bk^B\hat{k}Bk^ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi^v\hat{i}vi^ and −2vj^-2v\hat{j}−2vj^​. Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale) (a) E=34(mv2qa)E = \dfrac{3}{4}\left(\dfrac{mv^{2}}{qa}\right)E=43​(qamv2​) (b) Rate of work done by the electric field at P is 34(mv3a)\dfrac{3}{4}\left(\dfrac{mv^{3}}{a}\right)43​(amv3​) (c) Rate of work done by both the fields at Q is zero. (d) The difference between the magnitude of angular momentum of the particle at P and Q is 2 mav.
  1. (A)(b), (c), (d)
  2. (B)(a), (b), (c)
  3. (C)(a), (c), (d)
  4. (D)(a), (b), (c), (d)

Correct answer: (B)

Step-by-step solution →
Q173·PhysicsSingle correctJEE Main 2020
An electron gun is placed inside a long solenoid of radius R on its axis. The solenoid has n turns/length and carries a current I. The electron gun shoots an electron along the radius of the solenoid with speed v. If the electron does not hit the surface of the solenoid, maximum possible value of v is (all symbols have their standard meaning):
  1. (A)2eμ0nIRm\frac{2e\mu_{0}nIR}{m}m2eμ0​nIR​
  2. (B)eμ0nIRm\frac{e\mu_{0}nIR}{m}meμ0​nIR​
  3. (C)eμ0nIR4m\frac{e\mu_{0}nIR}{4m}4meμ0​nIR​
  4. (D)eμ0nIR2m\frac{e\mu_{0}nIR}{2m}2meμ0​nIR​

Correct answer: (D)

Step-by-step solution →
Q174·PhysicsSingle correctJEE Main 2020
A very long wire ABADMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is:
  1. (A)μ0I2R\dfrac{\mu_{0} I}{2R}2Rμ0​I​
  2. (B)μ0I2πR(π−12)\dfrac{\mu_{0} I}{2\pi R}\left(\pi-\dfrac{1}{\sqrt{2}}\right)2πRμ0​I​(π−2​1​)
  3. (C)μ0I2πR(π+1)\dfrac{\mu_{0} I}{2\pi R}(\pi+1)2πRμ0​I​(π+1)
  4. (D)μ0I2πR(π+12)\dfrac{\mu_{0} I}{2\pi R}\left(\pi+\dfrac{1}{\sqrt{2}}\right)2πRμ0​I​(π+2​1​)

Correct answer: (D)

Step-by-step solution →
Q175·PhysicsSingle correctJEE Main 2020
Proton with kinetic energy of 1 MeV moves from south the north. It gets an acceleration of 101210^{12}1012 m/s2^{2}2 by an applied magnetic field (west to east). The value of magnetic field: (Rest mass of proton is 1.6×10−271.6 \times 10^{-27}1.6×10−27 kg)
  1. (A)0.71 mT
  2. (B)7.1 mT
  3. (C)0.071 mT
  4. (D)71 mT

Correct answer: (A)

Step-by-step solution →
Q176·PhysicsSingle correctJEE Main 2020
Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by ϕ\phiϕ. The magnetic flux through the area is given by ϕ0\phi_0ϕ0​. Which of the following is correct?
  1. (A)ϕi=−ϕ0\phi_i=-\phi_0ϕi​=−ϕ0​
  2. (B)ϕi>ϕ0\phi_i>\phi_0ϕi​>ϕ0​
  3. (C)ϕi=ϕ0\phi_i=\phi_0ϕi​=ϕ0​
  4. (D)ϕi<ϕ0\phi_i<\phi_0ϕi​<ϕ0​

Correct answer: (A)

Step-by-step solution →
Q177·PhysicsSingle correctJEE Main 2020
A particle of mass m and charge q has an initial velocity v⃗=v0j^\vec{v} = v_0\hat{j}v=v0​j^​. If an electric field E⃗=E0i^\vec{E} = E_0\hat{i}E=E0​i^ and magnetic field B⃗=B0i^\vec{B} = B_0\hat{i}B=B0​i^ act on the particle, its speed will double after a time:
  1. (A)2mv0qE0\dfrac{2mv_0}{qE_0}qE0​2mv0​​
  2. (B)2mv0qE0\dfrac{\sqrt{2}mv_0}{qE_0}qE0​2​mv0​​
  3. (C)3mv0qE0\dfrac{\sqrt{3}mv_0}{qE_0}qE0​3​mv0​​
  4. (D)3mv0qE0\dfrac{3mv_0}{qE_0}qE0​3mv0​​

Correct answer: (C)

Step-by-step solution →
Q178·PhysicsSingle correctJEE Main 2019
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) (μ\muμ0 = 4p ×\times× 10–7 N-A–2)
  1. (A)2.0 ×\times× 10−5^{-5}−5 T
  2. (B)3.0 ×\times× 10−5^{-5}−5 T
  3. (C)2.5 ×\times× 10−5^{-5}−5 T
  4. (D)1.5 ×\times× 10−5^{-5}−5 T

Correct answer: (D)

Step-by-step solution →
Q179·PhysicsSingle correctJEE Main 2019
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40π\piπ rad s−1^{-1}−1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 ×\times× 10−9^{-9}−9 T, then the charge carried by the ring is close to (μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7N/A2^{2}2).
  1. (A)2 ×\times× 10−6^{-6}−6 C
  2. (B)7 ×\times× 10−6^{-6}−6 C
  3. (C)4 ×\times× 10−5^{-5}−5 C
  4. (D)3 ×\times× 10−5^{-5}−5 C

Correct answer: (D)

Step-by-step solution →
Q180·PhysicsSingle correctJEE Main 2019
A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig_{g}g​ flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0_{0}0​ (I0_{0}0​ > Ig_{g}g​) by connecting a shunt resistance RA_{A}A​ to it and (ii) into a voltmeter of range 0 to V(V = GI0_{0}0​) by connecting a series resistance RV_{V}V​ to it. Then,
  1. (A)RARV=G2R_AR_V = G^{2}RA​RV​=G2 and RARV=Ig(I0−Ig)\dfrac{R_A}{R_V} = \dfrac{I_g}{(I_0-I_g)}RV​RA​​=(I0​−Ig​)Ig​​
  2. (B)RARV=G2R_AR_V = G^{2}RA​RV​=G2 and RARV=(IgI0−Ig)2\dfrac{R_A}{R_V} = \left(\dfrac{I_g}{I_0-I_g}\right)^{2}RV​RA​​=(I0​−Ig​Ig​​)2
  3. (C)RARV=G2(IgI0−Ig)R_AR_V = G^{2}\left(\dfrac{I_g}{I_0-I_g}\right)RA​RV​=G2(I0​−Ig​Ig​​) and RARV=(I0−IgIg)2\dfrac{R_A}{R_V} = \left(\dfrac{I_0-I_g}{I_g}\right)^{2}RV​RA​​=(Ig​I0​−Ig​​)2
  4. (D)RA−RV=G2(I0−IgIg)R_A - R_V = G^{2}\left(\dfrac{I_0-I_g}{I_g}\right)RA​−RV​=G2(Ig​I0​−Ig​​) and RARV=(Ig(I0−Ig))2\dfrac{R_A}{R_V} = \left(\dfrac{I_g}{(I_0-I_g)}\right)^{2}RV​RA​​=((I0​−Ig​)Ig​​)2

Correct answer: (B)

Step-by-step solution →
Q181·PhysicsSingle correctJEE Main 2019
An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field B⃗=(1.5×10−3T)k^\vec{B} = (1.5 \times 10^{-3} T)\hat{k}B=(1.5×10−3T)k^ at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is : (electron's charge = 1.6×10−191.6 \times 10^{-19}1.6×10−19 C, mass of electron = 9.1×10−319.1 \times 10^{-31}9.1×10−31 kg)
  1. (A)12.87 cm
  2. (B)2.25 cm
  3. (C)1.22 cm
  4. (D)11.65 cm

Correct answer: (A)

Step-by-step solution →
Q182·PhysicsSingle correctJEE Main 2019
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be
  1. (A)mπ\dfrac{m}{π}πm​
  2. (B)3mπ\dfrac{3m}{π}π3m​
  3. (C)4mπ\dfrac{4m}{π}π4m​
  4. (D)2mπ\dfrac{2m}{π}π2m​

Correct answer: (C)

Step-by-step solution →
Q183·PhysicsSingle correctJEE Main 2019
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is: [Take μ0=4π×10−7\mu_{0} = 4\pi \times 10^{-7}μ0​=4π×10−7 NA−2^{-2}−2]
  1. (A)9 μT
  2. (B)1 μT
  3. (C)3 μT
  4. (D)18 μT

Correct answer: (D)

Step-by-step solution →
Q184·PhysicsSingle correctJEE Main 2019
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let rp_{p}p​, re_{e}e​ and rHe_{He}He​ be their respective radii, then,
  1. (A)re_{e}e​ > rp_{p}p​ = rHe_{He}He​
  2. (B)re_{e}e​ > rp_{p}p​ > rHe_{He}He​
  3. (C)re_{e}e​ < rp_{p}p​ < rHe_{He}He​
  4. (D)re_{e}e​ < rp_{p}p​ = rHe_{He}He​

Correct answer: (D)

Step-by-step solution →
Q185·PhysicsSingle correctJEE Main 2019
A rigid square loop of side 'a' and carrying current I2I_2I2​ is laying on a horizontal surface near a long current I1I_1I1​ wire in the same plane as shown in figure. The net force on the loop due to the wire will be:
  1. (A)Repulsive and equal to μ0I1I22π\dfrac{\mu_0 I_1 I_2}{2\pi}2πμ0​I1​I2​​
  2. (B)Attractive and equal to μ0I1I23π\dfrac{\mu_0 I_1 I_2}{3\pi}3πμ0​I1​I2​​
  3. (C)Zero
  4. (D)Repulsive and equal to μ0I1I24π\dfrac{\mu_0 I_1 I_2}{4\pi}4πμ0​I1​I2​​

Correct answer: (D)

Step-by-step solution →
Q186·PhysicsSingle correctJEE Main 2019
A moving coil galvanometer has a coil with 175 turns and area 1 cm2^{2}2 It uses a torsion band of torsion constant 10−610^{-6}10−6 N – m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 101^{0}10 for a current of 1 mA. The value of B (in Tesla) is approximately:-
  1. (A)10−310^{-3}10−3
  2. (B)10−110^{-1}10−1
  3. (C)10−410^{-4}10−4
  4. (D)10−210^{-2}10−2

Correct answer: (A)

Step-by-step solution →
Q187·PhysicsSingle correctJEE Main 2019
A moving coil galvanometer has resistance 50 Ω and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. The maximum voltage, that can be measured using this voltamenter, will be close to:
  1. (A)15 V
  2. (B)20 V
  3. (C)10 V
  4. (D)40 V

Correct answer: (B)

Step-by-step solution →
Q188·PhysicsSingle correctJEE Main 2019
Two very long, straight and insulated wires are kept at 90090^{0}900 angle from each other In xy-plane as shown in the figure. These wires carry currently of equal magnitude I, whose directions are shown in the figure. The met magnetic field at point P will be:
  1. (A)μ0I2πd(x^+y^)\dfrac{\mu_0 I}{2\pi d}(\hat{x}+\hat{y})2πdμ0​I​(x^+y^​)
  2. (B)+μ0Iπd(z^)\dfrac{+\mu_0 I}{\pi d}(\hat{z})πd+μ0​I​(z^)
  3. (C)Zero
  4. (D)−μ0I2πd(x^+y^)-\dfrac{\mu_0 I}{2\pi d}(\hat{x}+\hat{y})−2πdμ0​I​(x^+y^​)

Correct answer: (C)

Step-by-step solution →
Q189·PhysicsSingle correctJEE Main 2019
A circular coil having N turns and radius r carries a current. It is held in the XZ plane in a magnetic field Bi^B\hat{i}Bi^. The torque on the coil due to the magnetic field is:
  1. (A)Br2IπN\dfrac{Br^{2}I}{\pi N}πNBr2I​
  2. (B)zero
  3. (C)Bπr2IN\dfrac{B\pi r^{2}I}{N}NBπr2I​
  4. (D)Bπr2INB\pi r^{2}INBπr2IN

Correct answer: (D)

Step-by-step solution →
Q190·PhysicsSingle correctJEE Main 2019
As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4 cm, and the magnitude of the magnetic field at O is 10−410^{-4}10−4 T, and the two wires carry equal current (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be (μ0=4π×10−7\mu_0=4\pi\times10^{-7}μ0​=4π×10−7 NA−2^{-2}−2):
  1. (A)20 A, perpendicular out of the page
  2. (B)40 A, perpendicular out of the page
  3. (C)20 A, perpendicular into the page
  4. (D)40 A, perpendicular into the page

Correct answer: (C)

Step-by-step solution →
Q191·PhysicsSingle correctJEE Main 2019
A proton and an α\alphaα-particle (with their masses in the ratio of 1:4 and charges in the ratio of 1:2 are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp:rαr_p : r_\alpharp​:rα​ of the circular paths described by them will be:
  1. (A)1:21:\sqrt{2}1:2​
  2. (B)1:21:21:2
  3. (C)1:31:31:3
  4. (D)1:31:\sqrt{3}1:3​

Correct answer: (A)

Step-by-step solution →
Q192·PhysicsSingle correctJEE Main 2019
In an experiment, electrons are accelerated, from rest, by applying, a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6×10−191.6 \times 10^{-19}1.6×10−19 C Mass of the electron = 9.1×10−319.1 \times 10^{-31}9.1×10−31 kg]
  1. (A)7.5×10−37.5 \times 10^{-3}7.5×10−3 m
  2. (B)7.5×10−27.5 \times 10^{-2}7.5×10−2 m
  3. (C)7.5 m
  4. (D)7.5×10−47.5 \times 10^{-4}7.5×10−4 m

Correct answer: (D)

Step-by-step solution →
Q193·PhysicsSingle correctJEE Main 2019
An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. the radius of the loop is a and distance of its centre from the wire is d (d >> a). If the loop applies a force F on the wire then:
  1. (A)F = 0
  2. (B)F∝adF \propto \frac{a}{d}F∝da​
  3. (C)F∝a2d3F \propto \frac{a^2}{d^3}F∝d3a2​
  4. (D)F∝(ad)2F \propto \left(\frac{a}{d}\right)^2F∝(da​)2

Correct answer: (D)

Step-by-step solution →
Q194·PhysicsSingle correctJEE Main 2019
A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to:
  1. (A)1.0×10−71.0 \times 10^{-7}1.0×10−7 T
  2. (B)1.5×10−71.5 \times 10^{-7}1.5×10−7 T
  3. (C)1.5×10−51.5 \times 10^{-5}1.5×10−5 T
  4. (D)1.0×10−51.0 \times 10^{-5}1.0×10−5 T

Correct answer: (D)

Step-by-step solution →
Q195·PhysicsSingle correctJEE Main 2019
One of the two identical conducing wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (BL_LL​) to that at the centre of the coil (BC_CC​), i.e. BLBC\dfrac{B_L}{B_C}BC​BL​​ will be
  1. (A)N
  2. (B)1N\dfrac{1}{N}N1​
  3. (C)N2^{2}2
  4. (D)1N2\dfrac{1}{N^{2}}N21​

Correct answer: (D)

Step-by-step solution →
Q196·PhysicsSingle correctJEE Main 2019
A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (given charge of electron =1.6×10−19= 1.6 \times 10^{-19}=1.6×10−19 C)
  1. (A)9.1×10−319.1 \times 10^{-31}9.1×10−31 kg
  2. (B)1.6×10−271.6 \times 10^{-27}1.6×10−27 kg
  3. (C)1.6×10−191.6 \times 10^{-19}1.6×10−19 kg
  4. (D)2.0×10−242.0 \times 10^{-24}2.0×10−24 kg

Correct answer: (D)

Step-by-step solution →
Q197·PhysicsNumericalJEE Advanced 2018
In the xyxyxy-plane, the region y>0y > 0y>0 has a uniform magnetic field B1k^B_{1}\hat{k}B1​k^ and the region y<0y < 0y<0 has another uniform magnetic field B2k^B_{2}\hat{k}B2​k^. A positively charged particle is projected from the origin along the positive yyy-axis with speed v0=π m s−1v_{0} = \pi\ m\ s^{-1}v0​=π m s−1 at t=0t = 0t=0, as shown in the figure. Neglect gravity in this problem. Let t=Tt = Tt=T be the time when the particle crosses the xxx-axis from below for the first time. If B2=4B1B_{2} = 4B_{1}B2​=4B1​, the average speed of the particle, in ms−1ms^{-1}ms−1, along the xxx-axis in the time interval TTT is __________.

Correct answer: 2.00

Step-by-step solution →
Q198·PhysicsMultiple correctJEE Advanced 2018
Two infinitely long straight wires lie in the xyxyxy-plane along the lines x=±Rx = \pm Rx=±R. The wire located at x=+Rx = +Rx=+R carries a constant current I1I_{1}I1​ and the wire located at x=−Rx = -Rx=−R carries a constant current I2I_{2}I2​. A circular loop of radius RRR is suspended with its centre at (0,0,3R)(0, 0, \sqrt{3}R)(0,0,3​R) and in a plane parallel to the xyxyxy-plane. This loop carries a constant current III in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the +j^+\hat{j}+j^​ direction. Which of the following statements regarding the magnetic field B⃗\vec{B}B is (are) true?
  1. (A)If I1=I2I_{1} = I_{2}I1​=I2​, then B⃗\vec{B}B cannot be equal to zero at the origin (0,0,0)(0, 0, 0)(0,0,0)
  2. (B)If I1>0I_{1} > 0I1​>0, and I2<0I_{2} < 0I2​<0, then B⃗\vec{B}B can be equal to zero at the origin (0,0,0)(0, 0, 0)(0,0,0)
  3. (C)If I1<0I_{1} < 0I1​<0, and I2>0I_{2} > 0I2​>0, then B⃗\vec{B}B can be equal to zero at the origin (0,0,0)(0, 0, 0)(0,0,0)
  4. (D)If I1=I2I_{1} = I_{2}I1​=I2​, then the zzz-component of the magnetic field at the centre of the loop is (−μ0I2R)\left(-\frac{\mu_{0}I}{2R}\right)(−2Rμ0​I​)

Correct answer: (A), (B), (D)

Step-by-step solution →
Q199·PhysicsSingle correctJEE Advanced 2017
A symmetric star conducting wire loop is carrying a steady state current III as shown in figure. The distance between the diametrically opposite vertices of the star is 4a4a4a. The magnitude of the magnetic field at the center of the loop is
  1. (A)μ0I4πa 6[3−1]\dfrac{\mu_{0}I}{4\pi a}\,6[\sqrt{3}-1]4πaμ0​I​6[3​−1]
  2. (B)μ0I4πa 6[3+1]\dfrac{\mu_{0}I}{4\pi a}\,6[\sqrt{3}+1]4πaμ0​I​6[3​+1]
  3. (C)μ0I4πa 3[3−1]\dfrac{\mu_{0}I}{4\pi a}\,3[\sqrt{3}-1]4πaμ0​I​3[3​−1]
  4. (D)μ0I4πa 3[2−3]\dfrac{\mu_{0}I}{4\pi a}\,3[2-\sqrt{3}]4πaμ0​I​3[2−3​]

Correct answer: (A)

Step-by-step solution →
Q200·PhysicsSingle correctJEE Advanced 2017
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity v⃗\vec{v}v. A uniform electric field E⃗\vec{E}E and magnetic field B⃗\vec{B}B are given in columns 1, 2 and 3, respectively. The quantities E0E_{0}E0​, B0B_{0}B0​ are positive in magnitude. In which case will the particle describe a helical path with axis along the positive z direction?
Column IColumn 2Column 3
(I) Electron with v⃗=2E0B0x^\vec{v} = 2\frac{E_{0}}{B_{0}}\hat{x}v=2B0​E0​​x^(i) E⃗=E02z^\vec{E} = E_{0}^{2}\hat{z}E=E02​z^(P) B⃗=−B0x^\vec{B} = -B_{0}\hat{x}B=−B0​x^
(II) Electron with v⃗=E0B0y^\vec{v} = \frac{E_{0}}{B_{0}}\hat{y}v=B0​E0​​y^​(ii) E⃗=−E0y^\vec{E} = -E_{0}\hat{y}E=−E0​y^​(Q) B⃗=B0x^\vec{B} = B_{0}\hat{x}B=B0​x^
(III) Proton with v⃗=0\vec{v} = 0v=0(iii) E⃗=−E0x^\vec{E} = -E_{0}\hat{x}E=−E0​x^(R) B⃗=B0y^\vec{B} = B_{0}\hat{y}B=B0​y^​
(IV) Proton with v⃗=2E0B0x^\vec{v} = 2\frac{E_{0}}{B_{0}}\hat{x}v=2B0​E0​​x^(iv) E⃗=E0x^\vec{E} = E_{0}\hat{x}E=E0​x^(S) B⃗=B0z^\vec{B} = B_{0}\hat{z}B=B0​z^
  1. (A)(II) (ii) (R)
  2. (B)(IV) (ii) (R)
  3. (C)(IV) (i) (S)
  4. (D)(III) (iii) (P)

Correct answer: (C)

Step-by-step solution →
Q201·PhysicsSingle correctJEE Advanced 2017
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity v⃗\vec{v}v. A uniform electric field E⃗\vec{E}E and magnetic field B⃗\vec{B}B are given in columns 1, 2 and 3, respectively. The quantities E0E_{0}E0​, B0B_{0}B0​ are positive in magnitude. In which case will the particle move in a straight line with constant velocity?
Column IColumn 2Column 3
(I) Electron with v⃗=2E0B0x^\vec{v} = 2\frac{E_{0}}{B_{0}}\hat{x}v=2B0​E0​​x^(i) E⃗=E02z^\vec{E} = E_{0}^{2}\hat{z}E=E02​z^(P) B⃗=−B0x^\vec{B} = -B_{0}\hat{x}B=−B0​x^
(II) Electron with v⃗=E0B0y^\vec{v} = \frac{E_{0}}{B_{0}}\hat{y}v=B0​E0​​y^​(ii) E⃗=−E0y^\vec{E} = -E_{0}\hat{y}E=−E0​y^​(Q) B⃗=B0x^\vec{B} = B_{0}\hat{x}B=B0​x^
(III) Proton with v⃗=0\vec{v} = 0v=0(iii) E⃗=−E0x^\vec{E} = -E_{0}\hat{x}E=−E0​x^(R) B⃗=B0y^\vec{B} = B_{0}\hat{y}B=B0​y^​
(IV) Proton with v⃗=2E0B0x^\vec{v} = 2\frac{E_{0}}{B_{0}}\hat{x}v=2B0​E0​​x^(iv) E⃗=E0x^\vec{E} = E_{0}\hat{x}E=E0​x^(S) B⃗=B0z^\vec{B} = B_{0}\hat{z}B=B0​z^
  1. (A)(II) (iii) (S)
  2. (B)(IV) (i) (S)
  3. (C)(III) (ii) (R)
  4. (D)(III) (iii) (P)

Correct answer: (A)

Step-by-step solution →
Q202·PhysicsSingle correctJEE Advanced 2017
A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity v⃗\vec{v}v. A uniform electric field E⃗\vec{E}E and magnetic field B⃗\vec{B}B are given in columns 1, 2 and 3, respectively. The quantities E0E_{0}E0​, B0B_{0}B0​ are positive in magnitude. In which case would be particle move in a straight line along the negative direction of y-axis (i.e., more along −y^-\hat{y}−y^​) ?
Column IColumn 2Column 3
(I) Electron with v⃗=2E0B0x^\vec{v} = 2\frac{E_{0}}{B_{0}}\hat{x}v=2B0​E0​​x^(i) E⃗=E02z^\vec{E} = E_{0}^{2}\hat{z}E=E02​z^(P) B⃗=−B0x^\vec{B} = -B_{0}\hat{x}B=−B0​x^
(II) Electron with v⃗=E0B0y^\vec{v} = \frac{E_{0}}{B_{0}}\hat{y}v=B0​E0​​y^​(ii) E⃗=−E0y^\vec{E} = -E_{0}\hat{y}E=−E0​y^​(Q) B⃗=B0x^\vec{B} = B_{0}\hat{x}B=B0​x^
(III) Proton with v⃗=0\vec{v} = 0v=0(iii) E⃗=−E0x^\vec{E} = -E_{0}\hat{x}E=−E0​x^(R) B⃗=B0y^\vec{B} = B_{0}\hat{y}B=B0​y^​
(IV) Proton with v⃗=2E0B0x^\vec{v} = 2\frac{E_{0}}{B_{0}}\hat{x}v=2B0​E0​​x^(iv) E⃗=E0x^\vec{E} = E_{0}\hat{x}E=E0​x^(S) B⃗=B0z^\vec{B} = B_{0}\hat{z}B=B0​z^
  1. (A)(IV) (ii) (S)
  2. (B)(III) (ii) (P)
  3. (C)(II) (iii) (Q)
  4. (D)(III) (ii) (R)

Correct answer: (D)

Step-by-step solution →
Q203·PhysicsMultiple correctJEE Advanced 2017
A uniform magnetic field BBB exists in the region between x=0x = 0x=0 and x=3R2x = \dfrac{3R}{2}x=23R​ (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +Q+Q+Q and momentum ppp directed along x-axis enters region 2 from region 1 at point P1P_{1}P1​ (y=−R)(y = -R)(y=−R). Which of the following option(s) is/are correct?
  1. (A)For B>23pQRB > \dfrac{2}{3}\dfrac{p}{QR}B>32​QRp​, the particle will re-enter region 1
  2. (B)For B=813pQRB = \dfrac{8}{13}\dfrac{p}{QR}B=138​QRp​, the particle will enter region 3 through the point P2P_{2}P2​ on x-axis
  3. (C)When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point P1P_{1}P1​ and the farthest point from y-axis is p/2p/\sqrt{2}p/2​
  4. (D)For a fixed BBB, particles of same charge QQQ and same velocity vvv, the distance between the point P1P_{1}P1​ and the point of re-entry into region 1 is inversely proportional to the mass of the particle

Correct answer: (A), (B)

Step-by-step solution →
Q204·PhysicsMultiple correctJEE Advanced 2015
A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform magnetic field B⃗\vec{B}B. If F is the magnitude of the total magnetic force acting on the conductor, then the correct statement(s) is(are)
  1. (A)If B⃗\vec{B}B is along z^\hat{z}z^, F∝(L+R)F \propto (L + R)F∝(L+R)
  2. (B)If B⃗\vec{B}B is along x^\hat{x}x^, F=0F = 0F=0
  3. (C)If B⃗\vec{B}B is along y^\hat{y}y^​, F∝(L+R)F \propto (L + R)F∝(L+R)
  4. (D)If B⃗\vec{B}B is along z^\hat{z}z^, F=0F = 0F=0

Correct answer: (A), (B), (C)

Step-by-step solution →
Q205·PhysicsMultiple correctJEE Advanced 2015
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are ℓ\ellℓ, w and d, respectively. A uniform magnetic field B⃗\vec{B}B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1w_{1}w1​ and w2w_{2}w2​ and thicknesses are d1d_{1}d1​ and d2d_{2}d2​, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1V_{1}V1​ and V2V_{2}V2​ are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is(are)
  1. (A)If w1=w2w_{1} = w_{2}w1​=w2​ and d1=2d2d_{1} = 2d_{2}d1​=2d2​, then V2=2V1V_{2} = 2V_{1}V2​=2V1​
  2. (B)If w1=w2w_{1} = w_{2}w1​=w2​ and d1=2d2d_{1} = 2d_{2}d1​=2d2​, then V2=V1V_{2} = V_{1}V2​=V1​
  3. (C)If w1=2w2w_{1} = 2w_{2}w1​=2w2​ and d1=d2d_{1} = d_{2}d1​=d2​, then V2=2V1V_{2} = 2V_{1}V2​=2V1​
  4. (D)If w1=2w2w_{1} = 2w_{2}w1​=2w2​ and d1=d2d_{1} = d_{2}d1​=d2​, then V2=V1V_{2} = V_{1}V2​=V1​

Correct answer: (A), (D)

Step-by-step solution →
Q206·PhysicsMultiple correctJEE Advanced 2015
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are ℓ\ellℓ, w and d, respectively. A uniform magnetic field B⃗\vec{B}B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Consider two different metallic strips (1 and 2) of same dimensions (lengths ℓ\ellℓ, width w and thickness d) with carrier densities n1n_{1}n1​ and n2n_{2}n2​, respectively. Strip 1 is placed in magnetic field B1B_{1}B1​ and strip 2 is placed in magnetic field B2B_{2}B2​, both along positive y-directions. Then V1V_{1}V1​ and V2V_{2}V2​ are the potential differences developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for both the strips, the correct option(s) is(are)
  1. (A)If B1=B2B_{1} = B_{2}B1​=B2​ and n1=2n2n_{1} = 2n_{2}n1​=2n2​, then V2=2V1V_{2} = 2V_{1}V2​=2V1​
  2. (B)If B1=B2B_{1} = B_{2}B1​=B2​ and n1=2n2n_{1} = 2n_{2}n1​=2n2​, then V2=V1V_{2} = V_{1}V2​=V1​
  3. (C)If B1=2B2B_{1} = 2B_{2}B1​=2B2​ and n1=n2n_{1} = n_{2}n1​=n2​, then V2=0.5V1V_{2} = 0.5V_{1}V2​=0.5V1​
  4. (D)If B1=2B2B_{1} = 2B_{2}B1​=2B2​ and n1=n2n_{1} = n_{2}n1​=n2​, then V2=V1V_{2} = V_{1}V2​=V1​

Correct answer: (A), (C)

Step-by-step solution →
Q207·PhysicsIntegerJEE Advanced 2014
Two parallel wires in the plane of the paper are distance X0X_0X0​ apart. A point charge is moving with speed uuu between the wires in the same plane at a distance X1X_1X1​ from one of the wires. When the wires carry current of magnitude III in the same direction, the radius of curvature of the path of the point charge is R1R_1R1​. In contrast, if the currents III in the two wires have directions opposite to each other, the radius of curvature of the path is R2R_2R2​. If X0X1=3\frac{X_0}{X_1} = 3X1​X0​​=3, the value of R1R2\frac{R_1}{R_2}R2​R1​​ is

Correct answer: 3

Step-by-step solution →
Q208·PhysicsSingle correctJEE Advanced 2014
The figure shows a circular loop of radius aaa with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is ddd. The loop and the wires are carrying the same current III. The current in the loop is in the counterclockwise direction if seen from above. When d≈ad \approx ad≈a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height hhh above the loop. In that case
  1. (A)current in wire 1 and wire 2 is the direction PQ and RS, respectively and h≈ah \approx ah≈a
  2. (B)current in wire 1 and wire 2 is the direction PQ and SR, respectively and h≈ah \approx ah≈a
  3. (C)current in wire 1 and wire 2 is the direction PQ and SR, respectively and h≈1.2ah \approx 1.2ah≈1.2a
  4. (D)current in wire 1 and wire 2 is the direction PQ and RS, respectively and h≈1.2ah \approx 1.2ah≈1.2a

Correct answer: (C)

Step-by-step solution →
Q209·PhysicsSingle correctJEE Advanced 2014
The figure shows a circular loop of radius aaa with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is ddd. The loop and the wires are carrying the same current III. The current in the loop is in the counterclockwise direction if seen from above. Consider d≫ad \gg ad≫a, and the loop is rotated about its diameter parallel to the wires by 30∘30^{\circ}30∘ from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)
  1. (A)μ0I2a2d\frac{\mu_0 I^{2} a^{2}}{d}dμ0​I2a2​
  2. (B)μ0I2a22d\frac{\mu_0 I^{2} a^{2}}{2d}2dμ0​I2a2​
  3. (C)3μ0I2a2d\frac{\sqrt{3}\mu_0 I^{2} a^{2}}{d}d3​μ0​I2a2​
  4. (D)3μ0I2a22d\frac{\sqrt{3}\mu_0 I^{2} a^{2}}{2d}2d3​μ0​I2a2​

Correct answer: (B)

Step-by-step solution →
Q210·PhysicsMultiple correctJEE Advanced 2013
A particle of mass M and positive charge Q, moving with a constant velocity u1=4i^u_{1}=4\hat{i}u1​=4i^ ms−1^{-1}−1, enters a region of uniform static magnetic field normal to the x-y plane. The region of the magnetic field extends from x = 0 to x = L for all values of y. After passing through this region, the particle emerges on the other side after 10 milliseconds with a velocity u2=2(3i^+j^)u_{2}=2(\sqrt{3}\hat{i}+\hat{j})u2​=2(3​i^+j^​) m/s. The correct statement(s) is (are)
  1. (A)The direction of the magnetic field is −z-z−z direction.
  2. (B)The direction of the magnetic field is +z+z+z direction.
  3. (C)The magnitude of the magnetic field is 50πM3Q\frac{50\pi M}{3Q}3Q50πM​ units.
  4. (D)The magnitude of the magnetic field is 100πM3Q\frac{100\pi M}{3Q}3Q100πM​ units.

Correct answer: (A), (C)

Step-by-step solution →
Q211·PhysicsMultiple correctJEE Advanced 2013
A steady current III flows along an infinitely long hollow cylindrical conductor of radius RRR. This cylinder is placed coaxially inside an infinite solenoid of radius 2R2R2R. The solenoid has nnn turns per unit length and carries a steady current III. Consider a point PPP at a distance rrr from the common axis. The correct statement(s) is (are)
  1. (A)In the region 0<r<R0 < r < R0<r<R, the magnetic field is non-zero
  2. (B)In the region R<r<2RR < r < 2RR<r<2R, the magnetic field is along the common axis.
  3. (C)In the region R<r<2RR < r < 2RR<r<2R, the magnetic field is tangential to the circle of radius rrr, centered on the axis.
  4. (D)In the region r>2Rr > 2Rr>2R, the magnetic field is non-zero.

Correct answer: (A), (D)

Step-by-step solution →

Magnetic Field of Current — frequently asked

How many questions from Magnetic Field of Current appear in JEE?

Magnetic Field of Current has appeared in 147 of the last 186 JEE Main and JEE Advanced papers — about 79% of them — contributing 211 questions in total across those papers.

Is Magnetic Field of Current an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 79% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Magnetic Field of Current questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

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