Jarvis OS
PYQ papersPricingSign inGet started
  1. Home
  2. /JEE Physics PYQs
  3. /Electromagnetic Induction

Electromagnetic Induction — JEE Previous Year Questions

Every Electromagnetic Induction question asked in JEE Main and JEE Advanced across the last 186 papers — 155 questions, each with its correct answer. Free to read, no account needed.

Questions

155

Papers it appeared in

120/186

Appearance rate

65%

All 155 Electromagnetic Induction questions

Most recent papers first.

Q1·PhysicsSingle correctJEE Advanced 2026
List-I contains four conducting loops lying in the XYXYXY plane, as shown in the figures. The loops are rotating about ZZZ axis passing through the point OOO with time period TTT in clockwise direction. The region x>0x > 0x>0 contains a uniform magnetic field BBB in the +z+z+z direction. List-II contains the qualitative variation of the induced current i(t)i(t)i(t) for each of these loops. Choose the option which describes the correct match between the entries in List-I to those in List-II.
  1. (A)P → 5, Q → 4, R → 1, S → 3
  2. (B)P → 3, Q → 2, R → 5, S → 4
  3. (C)P → 3, Q → 2, R → 1, S → 4
  4. (D)P → 5, Q → 1, R → 2, S → 3

Correct answer: (C)

Step-by-step solution →
Q2·PhysicsSingle correctJEE Advanced 2026
Consider a circuit consisting of a capacitor of capacitance CCC and a coil with NNN turns per unit length, cross sectional area SSS and length ddd, where d2≫Sd^2 \gg Sd2≫S. There is another coil of length d/2d/2d/2, cross sectional area S/2S/2S/2 and 2N2N2N turns per unit length completely inside the larger coil, as shown in the figure. The ends of this smaller coil are connected with each other by an insulated conducting wire. The self-inductance of the larger coil is LLL. Neglecting edge effects and all the Ohmic resistances, the resonant frequency of the circuit is:
  1. (A)415 LC\frac{4}{\sqrt{15\,LC}}15LC​4​
  2. (B)65 LC\frac{6}{\sqrt{5\,LC}}5LC​6​
  3. (C)23 LC\frac{2}{\sqrt{3\,LC}}3LC​2​
  4. (D)23 LC\sqrt{\frac{2}{3\,LC}}3LC2​​

Correct answer: (C)

Step-by-step solution →
Q3·PhysicsSingle correctJEE Main 2026
A 30 cm long solenoid has 10 turns per cm and area of 5 cm2^22. The current through the solenoid coil varies from 2 A to 4 A in 3.14 s. The e.m.f. induced in the coil is α×10−5\alpha \times 10^{-5}α×10−5 V. The value of α\alphaα is ________.
  1. (A)60
  2. (B)12
  3. (C)120
  4. (D)34

Correct answer: (B)

Step-by-step solution →
Q4·PhysicsSingle correctJEE Main 2026
An inductor of inductance 10 mH having resistance of 100 Ω is connected to battery of E.M.F. 1.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 2 mA and 4 mA is _______.
  1. (A)4/3
  2. (B)3/4
  3. (C)5/3
  4. (D)3/5

Correct answer: (A)

Step-by-step solution →
Q5·PhysicsSingle correctJEE Main 2026
A square loop of side 2 cm is placed in a time varying magnetic field with magnitude as B=0.4sin⁡(300t)B = 0.4\sin(300t)B=0.4sin(300t) Tesla. The normal to the plane of loop makes an angle of 60° with the field. The maximum induced emf produced in the loop is __________ mV.
  1. (A)12
  2. (B)18
  3. (C)21
  4. (D)24

Correct answer: (D)

Step-by-step solution →
Q6·PhysicsNumericalJEE Main 2026
In the given circuit below inductance values of L1L_1L1​, L2L_2L2​ and L3L_3L3​ are same. The magnetic energy stored in the entire circuit is (Ut)(U_t)(Ut​) and that stored in the L2L_2L2​ inductor is (Ul)(U_l)(Ul​). Ut/UlU_t/U_lUt​/Ul​ is _____. (Ignore the mutual inductance if any)

Correct answer: 6

Step-by-step solution →
Q7·PhysicsSingle correctJEE Main 2026
A metal rod of length L rotates about one end at origin with a uniform angular velocity ω. The magnetic field radially falls off as B(r) = B₀e−λr^{-\lambda r}−λr; λ being a positive constant. The emf induced (neglecting the centripetal force on electrons in the rod) is :
  1. (A)B0ω[1λ2−e−λL(1λ2+Lλ)]B_{0}\omega\left[\dfrac{1}{\lambda^{2}} - e^{-\lambda L}\left(\dfrac{1}{\lambda^{2}} + \dfrac{L}{\lambda}\right)\right]B0​ω[λ21​−e−λL(λ21​+λL​)]
  2. (B)B0ω[1λ2+e−λL(1λ2+Lλ)]B_{0}\omega\left[\dfrac{1}{\lambda^{2}} + e^{-\lambda L}\left(\dfrac{1}{\lambda^{2}} + \dfrac{L}{\lambda}\right)\right]B0​ω[λ21​+e−λL(λ21​+λL​)]
  3. (C)B0ω[4λ2−e−2λL(1λ2+2Lλ)]B_{0}\omega\left[\dfrac{4}{\lambda^{2}} - e^{-2\lambda L}\left(\dfrac{1}{\lambda^{2}} + \dfrac{2L}{\lambda}\right)\right]B0​ω[λ24​−e−2λL(λ21​+λ2L​)]
  4. (D)B0ω[3λ2−e−3λL(3λ2+Lλ)]B_{0}\omega\left[\dfrac{3}{\lambda^{2}} - e^{-3\lambda L}\left(\dfrac{3}{\lambda^{2}} + \dfrac{L}{\lambda}\right)\right]B0​ω[λ23​−e−3λL(λ23​+λL​)]

Correct answer: (A)

Step-by-step solution →
Q8·PhysicsNumericalJEE Main 2026
A circular loop of radius 20 cm and resistance 2 Ω is placed in a time varying magnetic field B⃗=(2t2+2t+3)\vec{B} = (2t^2 + 2t + 3)B=(2t2+2t+3) T. At t=0t = 0t=0, for the plane of the loop being perpendicular to the magnetic field and, the induced current in the loop at t=3t = 3t=3 s is α50\frac{\alpha}{50}50α​ A. The value of α\alphaα is ________. (Take π=22/7\pi = 22/7π=22/7)

Correct answer: 44

Step-by-step solution →
Q9·PhysicsSingle correctJEE Main 2026
A circular current loop of radius RRR is placed inside square loop of side length LLL (L>>R)(L >> R)(L>>R) such that they are co-planar and their centers coincide. The permeability of free space is μ0\mu_0μ0​. The mutual inductance between circular loop and square loop is ________.
  1. (A)22 μ0L2R2\sqrt{2}\,\frac{\mu_0 L^2}{R}22​Rμ0​L2​
  2. (B)2 μ0L2R\sqrt{2}\,\frac{\mu_0 L^2}{R}2​Rμ0​L2​
  3. (C)2 μ0R2L\sqrt{2}\,\frac{\mu_0 R^2}{L}2​Lμ0​R2​
  4. (D)22 μ0R2L2\sqrt{2}\,\frac{\mu_0 R^2}{L}22​Lμ0​R2​

Correct answer: (D)

Step-by-step solution →
Q10·PhysicsSingle correctJEE Main 2026
When a coil is placed in a time dependent magnetic field the power dissipated in it is PPP. The number of turns, area of the coil and radius of the coil wire are NNN, AAA and rrr respectively. For a second coils number of turns, area of the coil and radius of the coil wire are 2N2N2N, 2A2A2A and 3r3r3r respectively. When the first coil is replaced with second coil the power dissipated in it is 2 αP\sqrt{2}\,\alpha P2​αP. The value of α\alphaα is ______.
  1. (A)36
  2. (B)1282128\sqrt{2}1282​
  3. (C)16
  4. (D)64

Correct answer: (A)

Step-by-step solution →
Q11·PhysicsSingle correctJEE Main 2026
A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s2^22) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travells a vertical distance of 200 m is ______ mV.
  1. (A)0.2100.2\sqrt{10}0.210​
  2. (B)201020\sqrt{10}2010​
  3. (C)2102\sqrt{10}210​
  4. (D)20010200\sqrt{10}20010​

Correct answer: (B)

Step-by-step solution →
Q12·PhysicsSingle correctJEE Main 2026
Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current III = 10 sin (ωt) A, where ω = 1000 rad./s. A circular conducting loop (B) of radius 1 cm coaxially slided through the solenoid at a speed v = 1 cm/s. The r.m.s. current through the loop when the coil B is inserted 10 cm inside the solenoid is α/2\alpha/\sqrt{2}α/2​ μA. The value of α is __________. [Resistance of the loop = 10 Ω]
  1. (A)197
  2. (B)80
  3. (C)280
  4. (D)100

Correct answer: (A)

Step-by-step solution →
Q13·PhysicsNumericalJEE Main 2026
A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of 60° with vertical, then maximum induced EMF between the point of suspension and point of oscillation is ______ mV. (Take g = 10 m/s2^22)

Correct answer: 100

Step-by-step solution →
Q14·PhysicsSingle correctJEE Main 2026
A circular loop of radius 7 cm is placed in uniform magnetic field of 0.2 T directed perpendicular to plane of loop. The loop is converted into a square loop in 0.5 s. The EMF induced in the loop is ___________ mV.
  1. (A)6.6
  2. (B)13.2
  3. (C)8.25
  4. (D)1.32

Correct answer: (D)

Step-by-step solution →
Q15·PhysicsNumericalJEE Main 2026
Inductance of a coil with 10410^4104 turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10Ω. The energy density in the inductor when the current reaches (1e)\left(\frac{1}{e}\right)(e1​) of its maximum value is απ×1e2\alpha\pi \times \frac{1}{e^2}απ×e21​ J/m3^33. The value of α is ____. (μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7 Tm/A).

Correct answer: 20

Step-by-step solution →
Q16·PhysicsNumericalJEE Main 2026
A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is _______ mm. (Take π=227)\left(\text{Take } \pi = \dfrac{22}{7}\right)(Take π=722​)

Correct answer: 14

Step-by-step solution →
Q17·PhysicsSingle correctJEE Main 2026
Three identical coils C1C_1C1​, C2C_2C2​ and C3C_3C3​ are closely placed such that they share a common axis. C2C_2C2​ is exactly midway. C1C_1C1​ carries current III in anti-clockwise direction while C3C_3C3​ carries current III in clockwise direction. An induced current flows through C2C_2C2​ will be in clockwise direction when
  1. (A)C1C_1C1​ and C3C_3C3​ move with equal speeds away from C2C_2C2​
  2. (B)C1C_1C1​ moves towards C2C_2C2​ and C3C_3C3​ moves away from C2C_2C2​
  3. (C)C1C_1C1​ moves away from C2C_2C2​ and C3C_3C3​ moves towards C2C_2C2​
  4. (D)C1C_1C1​ and C3C_3C3​ move with equal speeds towards C2C_2C2​

Correct answer: (B)

Step-by-step solution →
Q18·PhysicsSingle correctJEE Main 2026
Figure shows the circuit that contains three resistances (9 Ω each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is _________A.
  1. (A)1
  2. (B)zero
  3. (C)3
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q19·PhysicsSingle correctJEE Main 2026
XPQY is a vertical smooth long loop having a total resistance R where PX is parallel to QY and separation between them is lll. A constant magnetic field B perpendicular to the plane of the loop exists in the entire space. A rod CD of length L (L > lll) and mass m is made to slide down from rest under the gravity as shown in figure. The terminal speed acquired by the rod is_________m/s. (g = acceleration due to gravity)
  1. (A)2mgRB2l2\frac{2mgR}{B^{2}l^{2}}B2l22mgR​
  2. (B)8mgRB2l2\frac{8mgR}{B^{2}l^{2}}B2l28mgR​
  3. (C)2mgRB2L2\frac{2mgR}{B^{2}L^{2}}B2L22mgR​
  4. (D)mgRB2l2\frac{mgR}{B^{2}l^{2}}B2l2mgR​

Correct answer: (D)

Step-by-step solution →
Q20·PhysicsSingle correctJEE Main 2026
A conducting circular loop of area 1.0 m2\mathrm{m}^{2}m2 is placed perpendicular to a magnetic field which varies as B = sin(100 t) Tesla. If the resistance of the loop is 100 Ω , then the average thermal energy dissipated in the loop in one period is _______J.
  1. (A)π2\frac{\pi}{2}2π​
  2. (B)2π2\pi2π
  3. (C)π\piπ
  4. (D)π2\pi^{2}π2

Correct answer: (C)

Step-by-step solution →
Q21·PhysicsSingle correctJEE Main 2026
A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2Ω then the force needed to move the rod towards right with constant speed (v) of 1.5 m/s is _________ N.
  1. (A)7.5×10−27.5 \times 10^{-2}7.5×10−2
  2. (B)5.7×10−35.7 \times 10^{-3}5.7×10−3
  3. (C)5.7×10−25.7 \times 10^{-2}5.7×10−2
  4. (D)7.5×10−37.5 \times 10^{-3}7.5×10−3

Correct answer: (D)

Step-by-step solution →
Q22·PhysicsSingle correctJEE Advanced 2025
A conducting square loop initially lies in the XZ plane with its lower edge hinged along the X-axis. Only in the region y ≥ 0, there is a time dependent magnetic field pointing along the z-direction, B⃗(t)=B0(cos⁡ωt)k^\vec{B}(t) = B_{0}(\cos\omega t)\hat{k}B(t)=B0​(cosωt)k^, where B0B_{0}B0​ is a constant. The magnetic field is zero everywhere else. At time t = 0, the loop starts rotating with constant angular speed ω\omegaω about the X axis in the clockwise direction as viewed from the +X axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. (V) in the loop as a function of time:
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q23·PhysicsMultiple correctJEE Advanced 2025
A conducting square loop of side L, mass M and resistance R is moving in the XY plane with its edges parallel to the X and Y axes. The region y ≥ 0 has a uniform magnetic field, B⃗=B0k^\vec{B} = B_{0}\hat{k}B=B0​k^. The magnetic field is zero everywhere else. At time t = 0, the loop starts to enter the magnetic field with an initial velocity v0j^v_{0}\hat{j}v0​j^​ m/s, as shown in the figure. Considering the quantity K=B02L2RMK = \frac{B_{0}^{2}L^{2}}{RM}K=RMB02​L2​ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:
  1. (A)If v0v_{0}v0​ = 1.5 KL, the loop will stop before it enters completely inside the region of magnetic field.
  2. (B)When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.
  3. (C)If v0=KL10v_{0} = \frac{KL}{10}v0​=10KL​, the loop comes to rest at t=(1K)ln⁡(52)t = \left(\frac{1}{K}\right)\ln\left(\frac{5}{2}\right)t=(K1​)ln(25​).
  4. (D)If v0v_{0}v0​ = 3KL, the complete loop enters inside the region of magnetic field at time t=(1K)ln⁡(32)t = \left(\frac{1}{K}\right)\ln\left(\frac{3}{2}\right)t=(K1​)ln(23​).

Correct answer: (B), (D)

Step-by-step solution →
Q24·PhysicsIntegerJEE Main 2025
An inductor of self inductance 1 H connected in series with a resistor of 100π Ω100\pi\ \Omega100π Ω and an AC supply of 100π100\pi100π volt, 50 Hz. Maximum current flowing in the circuit is ______ A.

Correct answer: 1

Step-by-step solution →
Q25·PhysicsIntegerJEE Main 2025
Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of 12\dfrac{1}{\sqrt{2}}2​1​ Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of 10 cm/s, induced emf between points A and E is ______ mV.

Correct answer: 10

Step-by-step solution →
Q26·PhysicsSingle correctJEE Main 2025
A solenoid having area A and length lll is filled with a material having relative permeability 2. The magnetic energy stored in the solenoid is:
  1. (A)B2Alμ0\dfrac{B^2Al}{\mu_0}μ0​B2Al​
  2. (B)B2Al2μ0\dfrac{B^2Al}{2\mu_0}2μ0​B2Al​
  3. (C)B2AlB^2AlB2Al
  4. (D)B2Al4μ0\dfrac{B^2Al}{4\mu_0}4μ0​B2Al​

Correct answer: (D)

Step-by-step solution →
Q27·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): Choke coil is simply a coil having a large inductance but a small resistance. Choke coils are used with fluorescent mercury-tube fittings. If household electric power is directly connected to a mercury tube, the tube will be damaged. Reason (R): By using the choke coil, the voltage across the tube is reduced by a factor RR2+ω2L2\dfrac{R}{\sqrt{R^2+\omega^2 L^2}}R2+ω2L2​R​, where ω\omegaω is frequency of the supply across resistor R and inductor L. If the choke coil were not used, the voltage across the resistor would be the same as the applied voltage. In the light of the above statements, choose the most appropriate answer from the options given below:
  1. (A)Both (A) and (R) are true but (R) is not the correct explanation of (A).
  2. (B)(A) is false but (R) is true.
  3. (C)Both (A) and (R) are true and (R) is the correct explanation of (A).
  4. (D)(A) is true but (R) is false.

Correct answer: (C)

Step-by-step solution →
Q28·PhysicsSingle correctJEE Main 2025
Consider I1I_1I1​ and I2I_2I2​ are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L1L_1L1​ = self inductance of coil 1, M12M_{12}M12​ = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be
  1. (A)ε1=−L1dI1dt+M12dI2dt\varepsilon_1=-L_1\dfrac{dI_1}{dt}+M_{12}\dfrac{dI_2}{dt}ε1​=−L1​dtdI1​​+M12​dtdI2​​
  2. (B)ε1=−L1dI1dt−M12dI1dt\varepsilon_1=-L_1\dfrac{dI_1}{dt}-M_{12}\dfrac{dI_1}{dt}ε1​=−L1​dtdI1​​−M12​dtdI1​​
  3. (C)ε1=−L1dI1dt−M12dI2dt\varepsilon_1=-L_1\dfrac{dI_1}{dt}-M_{12}\dfrac{dI_2}{dt}ε1​=−L1​dtdI1​​−M12​dtdI2​​
  4. (D)ε1=−L1dI2dt−M12dI1dt\varepsilon_1=-L_1\dfrac{dI_2}{dt}-M_{12}\dfrac{dI_1}{dt}ε1​=−L1​dtdI2​​−M12​dtdI1​​

Correct answer: (C)

Step-by-step solution →
Q29·PhysicsSingle correctJEE Main 2025
A coil of area A and N turns is rotating with angular velocity ω\omegaω in a uniform magnetic field B⃗\vec{B}B about an axis perpendicular to B⃗\vec{B}B. Magnetic flux ϕ\phiϕ and induced emf ε\varepsilonε across it, at an instant when B⃗\vec{B}B is parallel to the plane of coil, are:
  1. (A)ϕ=AB, ε=0\phi=AB,\ \varepsilon=0ϕ=AB, ε=0
  2. (B)ϕ=0, ε=NABω\phi=0,\ \varepsilon=NAB\omegaϕ=0, ε=NABω
  3. (C)ϕ=0, ε=0\phi=0,\ \varepsilon=0ϕ=0, ε=0
  4. (D)ϕ=AB, ε=NABω\phi=AB,\ \varepsilon=NAB\omegaϕ=AB, ε=NABω

Correct answer: (B)

Step-by-step solution →
Q30·PhysicsSingle correctJEE Main 2025
A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10π10\pi10π rad s−1^{-1}−1 about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim? (π=3.14\pi=3.14π=3.14)
  1. (A)0.0628 V
  2. (B)0.5024 V
  3. (C)0.2512 V
  4. (D)0.1256 V

Correct answer: (C)

Step-by-step solution →
Q31·PhysicsIntegerJEE Main 2025
A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field B exists into the page. The bar starts to move from the vertex at time t = 0 with a constant velocity. If the induced EMF is E∝tnE\propto t^nE∝tn, then value of n is ______.

Correct answer: 1

Step-by-step solution →
Q32·PhysicsSingle correctJEE Main 2025
Regarding self-inductance : A : The self-inductance of the coil depends on its geometry. B : Self-inductance does not depend on the permeability of the medium. C : Self-induced e.m.f. opposes any change in the current in a circuit. D : Self-inductance is electromagnetic analogue of mass in mechanics. E : Work needs to be done against self-induced e.m.f. in establishing the current. Choose the correct answer from the options given below:
  1. (A)A, B, C, D only
  2. (B)A, C, D, E only
  3. (C)A, B, C, E only
  4. (D)B, C, D, E only

Correct answer: (B)

Step-by-step solution →
Q33·PhysicsSingle correctJEE Main 2025
A rectangular metallic loop is moving out of a uniform magnetic field region to a field free region with a constant speed. When the loop is partially inside the magnetic field, the plot of magnitude of induced emf (ε\varepsilonε) with time (t) is given by
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q34·PhysicsSingle correctJEE Advanced 2024
A region in the form of an equilateral triangle (in x−yx-yx−y plane) of height L has a uniform magnetic field B⃗\vec{B}B pointing in the +z+z+z -direction. A conducting loop PQR , in the form of an equilateral triangle of the same height L , is placed in the x−yx-yx−y plane with its vertex P at x=0x = 0x=0 in the orientation shown in the figure. At t=0t = 0t=0 , the loop starts entering the region of the magnetic field with a uniform velocity v⃗\vec{v}v along the +x+x+x-direction. The plane of the loop and its orientation remain unchanged throughout its motion.
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q35·PhysicsSingle correctJEE Main 2024
A square loop of side 15 cm is being moved towards right at a constant speed of 2 cm/s as shown in the figure. The front edge enters the 50 cm wide magnetic field at t=0t=0t=0. The value of induced emf in the loop at t=10t=10t=10 s will be:
  1. (A)0.3 mV
  2. (B)4.5 mV
  3. (C)0 mV
  4. (D)3 mV

Correct answer: (C)

Step-by-step solution →
Q36·PhysicsNumericalJEE Main 2024
A square loop PQRS having 101010 turns, area 3.6×10−3 m23.6\times10^{-3}\,m^23.6×10−3m2 and resistance 100 Ω100\,\Omega100Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B=0.5B=0.5B=0.5 T as shown. Work done in pulling the loop out of the field in 1.01.01.0 s is ___ ×10−6\times10^{-6}×10−6 J.

Correct answer: 3

Step-by-step solution →
Q37·PhysicsSingle correctJEE Main 2024
In a coil, the current changes form −2-2−2 A to +2+2+2A in 0.2 s and induces an emf of 0.1 V. The self-inductance of the coil is :
  1. (A)5 mH
  2. (B)1 mH
  3. (C)2.5 mH
  4. (D)4 mH

Correct answer: (A)

Step-by-step solution →
Q38·PhysicsSingle correctJEE Main 2024
Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in the figure (radius of A is aaa, radius of B is bbb, with b≫ab \gg ab≫a). The mutual inductance between them is:
  1. (A)μ0πa22b\dfrac{\mu_0\pi a^2}{2b}2bμ0​πa2​
  2. (B)μ02πb2a\dfrac{\mu_0}{2\pi}\dfrac{b^2}{a}2πμ0​​ab2​
  3. (C)μ0πb22a\dfrac{\mu_0\pi b^2}{2a}2aμ0​πb2​
  4. (D)μ02πa2b\dfrac{\mu_0}{2\pi}\dfrac{a^2}{b}2πμ0​​ba2​

Correct answer: (A)

Step-by-step solution →
Q39·PhysicsNumericalJEE Main 2024
The current in an inductor is given by I=(3t+8)I=(3t+8)I=(3t+8), where ttt is in seconds. The magnitude of the induced emf produced in the inductor is 12 mV. The self-inductance of the inductor is __________ mH.

Correct answer: 4

Step-by-step solution →
Q40·PhysicsNumericalJEE Main 2024
A rod of length 60 cm rotates with a uniform angular velocity 20 rad s−1^{-1}−1 about its perpendicular bisector, in a uniform magnetic field 0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is ______ V.

Correct answer: 0

Step-by-step solution →
Q41·PhysicsNumericalJEE Main 2024
A rectangular loop of sides 12 cm12\,cm12cm and 5 cm5\,cm5cm, with its sides parallel to the x-axis and y-axis respectively moves with a velocity of 5 cm/s5\,cm/s5cm/s in the positive x axis direction, in a space containing a variable magnetic field in the positive z direction. The field has a gradient of 10−3 T/cm10^{-3}\,T/cm10−3T/cm along the negative x direction and it is decreasing with time at the rate of 10−3 T/s10^{-3}\,T/s10−3T/s. If the resistance of the loop is 6 mΩ6\,m\Omega6mΩ, the power dissipated by the loop as heat is ___ ×10−9 W\times10^{-9}\,W×10−9W.

Correct answer: 216

Step-by-step solution →
Q42·PhysicsNumericalJEE Main 2024
A coil of 200 turns and area 0.20 m2^22 is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is 2πβ\dfrac{2\pi}{\beta}β2π​ volt. The value of β\betaβ is __________.

Correct answer: 5

Step-by-step solution →
Q43·PhysicsSingle correctJEE Main 2024
A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is :
  1. (A)7
  2. (B)70
  3. (C)35
  4. (D)140

Correct answer: (B)

Step-by-step solution →
Q44·PhysicsNumericalJEE Main 2024
A small square loop of wire of side ℓ\ellℓ is placed inside a large square loop of wire of side L (L=ℓ2)(L=\ell^2)(L=ℓ2). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is x×10−7\sqrt{x}\times10^{-7}x​×10−7 H, where x=x=x= ______.

Correct answer: 128

Step-by-step solution →
Q45·PhysicsNumericalJEE Main 2024
The magnetic flux ϕ\phiϕ (in weber) linked with a closed circuit of resistance 8 Ω8\,\Omega8Ω varies with time (in seconds) as ϕ=5t2−36t+1\phi=5t^2-36t+1ϕ=5t2−36t+1. The induced current in the circuit at t=2t=2t=2 s is ______ A.

Correct answer: 2

Step-by-step solution →
Q46·PhysicsNumericalJEE Main 2024
A ceiling fan having 333 blades of length 80 cm80\,cm80cm each is rotating with an angular velocity of 1200 rpm1200\,rpm1200rpm. The magnetic field of earth in that region is 0.5 G0.5\,G0.5G and angle of dip is 30∘30^\circ30∘. The emf induced across the blades is Nπ×10−5 VN\pi\times10^{-5}\,VNπ×10−5V. The value of NNN is ___

Correct answer: 32

Step-by-step solution →
Q47·PhysicsSingle correctJEE Main 2024
A capacitor of capacitance 100 μ100\,\mu100μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be :
  1. (A)3.23.23.2 A
  2. (B)1.51.51.5 A
  3. (C)2.02.02.0 A
  4. (D)1.21.21.2 A

Correct answer: (B)

Step-by-step solution →
Q48·PhysicsNumericalJEE Main 2024
A horizontal straight wire 5 m long extending from east to west is falling freely at right angle to horizontal component of earth's magnetic field 0.60×10−40.60\times10^{-4}0.60×10−4 Wbm−2^{-2}−2. The instantaneous value of emf induced in the wire when its velocity is 10 ms−1^{-1}−1 is ___ ×10−3\times10^{-3}×10−3 V.

Correct answer: 3

Step-by-step solution →
Q49·PhysicsNumericalJEE Main 2024
A square loop of side 10 cm and resistance 0.7 Ω0.7\,\Omega0.7Ω is placed vertically in east-west plane. A uniform magnetic field of 0.20 T is set up across the plane in north east direction. The magnetic field is decreased to zero in 1 s at a steady rate. Then, magnitude of induced emf is x×10−3\sqrt{x}\times10^{-3}x​×10−3 V. The value of x is ______ .

Correct answer: 2

Step-by-step solution →
Q50·PhysicsSingle correctJEE Main 2024
In an a.c. circuit, voltage and current are given by: V=100sin⁡(100 t)V=100\sin(100\,t)V=100sin(100t) V and I=100sin⁡(100 t+π3)I=100\sin\left(100\,t+\dfrac{\pi}{3}\right)I=100sin(100t+3π​) mA respectively. The average power dissipated in one cycle is:
  1. (A)5 W
  2. (B)10 W
  3. (C)2.5 W
  4. (D)25 W

Correct answer: (C)

Step-by-step solution →
Q51·PhysicsNumericalJEE Main 2024
Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation i=i0sin⁡ωti=i_0\sin\omega ti=i0​sinωt, where i0=5i_0=5i0​=5 A and ω=50π\omega=50\piω=50π rad/s. The maximum value of emf in the second coil is πα\dfrac{\pi}{\alpha}απ​ V. The value of α\alphaα is __________.

Correct answer: 2

Step-by-step solution →
Q52·PhysicsSingle correctJEE Main 2024
A rectangular loop of length 2.5 m and width 2 m is placed at 60∘60^{\circ}60∘ to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is:
  1. (A)−2-2−2 V
  2. (B)+2+2+2 V
  3. (C)+1+1+1 V
  4. (D)−1-1−1 V

Correct answer: (C)

Step-by-step solution →
Q53·PhysicsSingle correctJEE Advanced 2023
A thin conducting rod MN of mass 20 gm, length 25 cm and resistance 10 Ω\OmegaΩ is held on frictionless, long, perfectly conducting vertical rails as shown in the figure. There is a uniform magnetic field B0=4B_0 = 4B0​=4 T directed perpendicular to the plane of the rod-rail arrangement. The rod is released from rest at time t = 0 and it moves down along the rails. Assume air drag is negligible. Match each quantity in List-I with an appropriate value from List-II, and choose the correct option. [Given: The acceleration due to gravity g = 10 m s−2s^{-2}s−2 and e−1=0.4e^{-1} = 0.4e−1=0.4]
List-IList-II
P.At t=0.2t = 0.2t=0.2 s, the magnitude of the induced emf in Volt1.0.07
Q.At t=0.2t = 0.2t=0.2 s, the magnitude of the magnetic force in Newton2.0.14
R.At t=0.2t = 0.2t=0.2 s, the power dissipated as heat in Watt3.1.20
S.The magnitude of terminal velocity of the rod in m s−1s^{-1}s−14.0.12
5.2.00
  1. (A)P →\rightarrow→ 5, Q →\rightarrow→ 2, R →\rightarrow→ 3, S →\rightarrow→ 1
  2. (B)P →\rightarrow→ 3, Q →\rightarrow→ 1, R →\rightarrow→ 4, S →\rightarrow→ 5
  3. (C)P →\rightarrow→ 4, Q →\rightarrow→ 3, R →\rightarrow→ 1, S →\rightarrow→ 2
  4. (D)P →\rightarrow→ 3, Q →\rightarrow→ 4, R →\rightarrow→ 2, S →\rightarrow→ 5

Correct answer: (D)

Step-by-step solution →
Q54·PhysicsIntegerJEE Advanced 2023
A rectangular conducting loop of length 4 cm and width 2 cm is in the xy-plane, as shown in the figure. It is being moved away from a thin and long conducting wire along the direction 32x^+12y^\frac{\sqrt{3}}{2}\hat{x} + \frac{1}{2}\hat{y}23​​x^+21​y^​ with a constant speed v. The wire is carrying a steady current III = 10 A in the positive x-direction. A current of 10 μ\muμA flows through the loop when it is at a distance d = 4 cm from the wire. If the resistance of the loop is 0.1 Ω\OmegaΩ, then the value of v is ________ m s−1^{-1}−1. [Given: The permeability of free space μ0=4π×10−7\mu_0 = 4\pi\times10^{-7}μ0​=4π×10−7 N A−2^{-2}−2]

Correct answer: 4

Step-by-step solution →
Q55·PhysicsNumericalJEE Main 2023
A 202020 cm long metallic rod is rotated with 210210210 rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.20.20.2 T parallel to the axis exists everywhere. The emf developed between the centre and the ring is __________ mV. Take π=227\pi = \frac{22}{7}π=722​.

Correct answer: 88

Step-by-step solution →
Q56·PhysicsSingle correctJEE Main 2023
A 12 V battery connected to a coil of resistance 6 Ω\OmegaΩ through a switch, drives a constant current in the circuit. The switch is opened in 1 ms. The emf induced across the coil is 20 V. The inductance of the coil is :
  1. (A)5 mH
  2. (B)12 mH
  3. (C)8 mH
  4. (D)10 mH

Correct answer: (D)

Step-by-step solution →
Q57·PhysicsSingle correctJEE Main 2023
Which of the following Maxwell's equations is valid for time varying conditions but not valid for static conditions:
  1. (A)∮B⃗⋅dl⃗=μ0I\displaystyle\oint\vec B\cdot d\vec l=\mu_{0}I∮B⋅dl=μ0​I
  2. (B)∮E⃗⋅dl⃗=0\displaystyle\oint\vec E\cdot d\vec l=0∮E⋅dl=0
  3. (C)∮E⃗⋅dl⃗=−∂ΦB∂t\displaystyle\oint\vec E\cdot d\vec l=-\dfrac{\partial\Phi_{B}}{\partial t}∮E⋅dl=−∂t∂ΦB​​
  4. (D)∮D⃗⋅dA⃗=Q\displaystyle\oint\vec D\cdot d\vec A=Q∮D⋅dA=Q

Correct answer: (C)

Step-by-step solution →
Q58·PhysicsNumericalJEE Main 2023
In the given figure, an inductor and a resistor are connected in series with a battery of emf EEE volt. E22b J/s\dfrac{E^{2}}{2b}\,J/s2bE2​J/s represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of ba\dfrac{b}{a}ab​ will be _____.

Correct answer: 25

Step-by-step solution →
Q59·PhysicsNumericalJEE Main 2023
An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area 24 cm2^22. The two ends of the wire are connected to a resistor. The total resistance in the circuit is 12 Ω\OmegaΩ. If an externally applied uniform magnetic field in the core along its axis changes from 1.5 T in one direction to 1.5 T in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ____ mC.

Correct answer: 60

Step-by-step solution →
Q60·PhysicsNumericalJEE Main 2023
A conducting circular loop is placed in a uniform magnetic field of 0.4 T with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm/s. The magnitude of induced emf in the loop at an instant when the radius of the loop is 1 cm will be _________ μ\muμV.

Correct answer: 50

Step-by-step solution →
Q61·PhysicsSingle correctJEE Main 2023
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: A bar magnet dropped through a metallic cylindrical pipe takes more time to come down compared to a non-magnetic bar with the same geometry and mass. Reason R: For the magnetic bar, eddy currents are produced in the metallic pipe which oppose the motion of the magnetic bar. In the light of the above statements, choose the correct answer from the options given below.
  1. (A)Both A and R are true but R is NOT the correct explanation of A
  2. (B)A is true but R is false
  3. (C)Both A and R are true and R is the correct explanation of A
  4. (D)A is false but R is true

Correct answer: (C)

Step-by-step solution →
Q62·PhysicsNumericalJEE Main 2023
A metallic cube of side 151515 cm is moving along the y-axis at a uniform velocity of 222 ms−1^{-1}−1 in a region of uniform magnetic field of magnitude 0.50.50.5 T directed along the z-axis. In equilibrium the potential difference between the faces of higher and lower potential developed because of the motion through the field will be ____ mV.

Correct answer: 150

Step-by-step solution →
Q63·PhysicsNumericalJEE Main 2023
The magnetic field B crossing normally a square metallic plate of area 4 m2\text{m}^2m2 is changing with time as shown in figure. The magnitude of induced emf in the plate during t=2t = 2t=2s to t=4t = 4t=4s, is ________ mV

Correct answer: 8

Step-by-step solution →
Q64·PhysicsNumericalJEE Main 2023
A coil has an inductance of 222 H and resistance of 4 Ω4\ \Omega4 Ω. A 101010 V supply is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ____ ×10−3\times10^{-3}×10−3 J.

Correct answer: 625

Step-by-step solution →
Q65·PhysicsNumericalJEE Main 2023
A square loop of side 2.02.02.0 cm is placed inside a long solenoid that has 505050 turns per centimetre and carries a sinusoidally varying current of amplitude 2.52.52.5 A and angular frequency 700700700 rad s−1^{-1}−1. The central axes of the loop and the solenoid coincide. The amplitude of the emf induced in the loop is x×10−4x\times10^{-4}x×10−4 V. The value of x is _______ (take π=227\pi=\tfrac{22}{7}π=722​).

Correct answer: 44

Step-by-step solution →
Q66·PhysicsNumericalJEE Main 2023
A 1m long metal rod XY completes the circuit as shown in figure. The plane of the circuit is perpendicular to the magnetic field of flux density 0.15 T. If the resistance of the circuit is 5 Ω5\,\Omega5Ω, the force needed to move the rod in direction, as indicated, with a constant speed of 4 m/s will be _________ ×10−3\times10^{-3}×10−3 N.

Correct answer: 18

Step-by-step solution →
Q67·PhysicsSingle correctJEE Main 2023
A bar magnet is released from rest along the axis of a very long vertical copper tube. After some time the magnet will:
  1. (A)Move down with almost constant speed
  2. (B)Oscillate inside the tube
  3. (C)Move down with an acceleration greater than g
  4. (D)Move down with an acceleration equal to g

Correct answer: (A)

Step-by-step solution →
Q68·PhysicsSingle correctJEE Main 2023
An emf of 0.08 V is induced in a metal rod of length 10 cm held normal to a uniform magnetic field of 0.4 T, when it moves with a velocity of:
  1. (A)2 ms−1^{-1}−1
  2. (B)3.2 ms−1^{-1}−1
  3. (C)0.5 ms−1^{-1}−1
  4. (D)20 ms−1^{-1}−1

Correct answer: (A)

Step-by-step solution →
Q69·PhysicsSingle correctJEE Main 2023
The induced emf can be produced in a coil by A. moving the coil with uniform speed inside uniform magnetic field B. moving the coil with non-uniform speed inside uniform magnetic field C. rotating the coil inside the uniform magnetic field D. changing the area of the coil inside the uniform magnetic field Choose the correct answer from the options given below:
  1. (A)B and D only
  2. (B)B and C only
  3. (C)A and C only
  4. (D)C and D only

Correct answer: (D)

Step-by-step solution →
Q70·PhysicsNumericalJEE Main 2023
Two concentric circular coils with radii 1 cm and 1000 cm, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ____ ×10−8\times 10^{-8}×10−8 H. (Take π2=10\pi^2=10π2=10)

Correct answer: 4

Step-by-step solution →
Q71·PhysicsSingle correctJEE Main 2023
Match List I (electrical machines and resonance phenomena) with List II (the underlying principle or condition). Choose the correct answer from the options given below:
List IList II
A.AC generatorI.Presence of both L and C
B.TransformerII.Electromagnetic Induction
C.Resonance phenomenon to occurIII.Quality factor
D.Sharpness of resonanceIV.Mutual Induction
  1. (A)A-IV, B-III, C-I, D-II
  2. (B)A-IV, B-II, C-I, D-III
  3. (C)A-II, B-IV, C-I, D-III
  4. (D)A-II, B-I, C-III, D-IV

Correct answer: (C)

Step-by-step solution →
Q72·PhysicsNumericalJEE Main 2023
A square shaped coil of area 707070 cm2^22 having 600600600 turns rotates in a magnetic field of 0.40.40.4 wbm−2^{-2}−2, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500500500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at 60°60°60° with the field, will be _________ V. (Take π=227\pi=\dfrac{22}{7}π=722​)

Correct answer: 44

Step-by-step solution →
Q73·PhysicsSingle correctJEE Main 2023
A coil is wrapped in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed: A. By changing the magnitude of the magnetic field within the coil. B. By changing the area of coil within the magnetic field. C. By changing the angle between the direction of magnetic field and the plane of the coil. D. By reversing the magnetic field direction abruptly without changing its magnitude. Choose the most appropriate answer from the options given below:
  1. (A)A and B only
  2. (B)A, B and D only
  3. (C)B, C and D only
  4. (D)A and C only

Correct answer: (C)

Step-by-step solution →
Q74·PhysicsSingle correctJEE Main 2023
Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following Assume negligible air friction)
  1. (A)Insulating ball will reach the earth's surface earlier than the metal ball
  2. (B)Metal ball will reach the earth's surface earlier than the insulating ball
  3. (C)Both will reach the earth's surface simultaneously
  4. (D)Time taken by them to reach the earth's surface will be independent of the properties of their materials

Correct answer: (A)

Step-by-step solution →
Q75·PhysicsNumericalJEE Main 2023
In an ac generator, a rectangular coil of 100 turns each having area 14×10−214\times10^{-2}14×10−2 m2^22 is rotated at 360 rev/min about an axis perpendicular to a uniform magnetic field of magnitude 3.0 T. The maximum value of the emf produced will be _______ V. (Take π=227\pi=\dfrac{22}{7}π=722​)

Correct answer: 1584

Step-by-step solution →
Q76·PhysicsNumericalJEE Main 2023
As per the given figure, if dIdt=−1\dfrac{dI}{dt}=-1dtdI​=−1 A/s then the value of VABV_{AB}VAB​ at this instant will be _________ V.

Correct answer: 30

Step-by-step solution →
Q77·PhysicsSingle correctJEE Main 2023
A square loop of area 25 cm225\,cm^225cm2 has a resistance of 10 Ω10\,\Omega10Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T40.0\,T40.0T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec1.0\,sec1.0sec, will be:
  1. (A)1.0×10−3 J1.0\times10^{-3}\,J1.0×10−3J
  2. (B)2.5×10−3 J2.5\times10^{-3}\,J2.5×10−3J
  3. (C)5×10−3 J5\times10^{-3}\,J5×10−3J
  4. (D)1.0×10−4 J1.0\times10^{-4}\,J1.0×10−4J

Correct answer: (A)

Step-by-step solution →
Q78·PhysicsSingle correctJEE Main 2023
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius RRR is placed inside a large square loop of wire of side LLL (L≫RL \gg RL≫R). The loops are coplanar and their centers coincide:
  1. (A)M=2 μ0R2LM=\dfrac{\sqrt{2}\,\mu_0 R^2}{L}M=L2​μ0​R2​
  2. (B)M=22 μ0RL2M=\dfrac{2\sqrt{2}\,\mu_0 R}{L^2}M=L222​μ0​R​
  3. (C)M=2 μ0RL2M=\dfrac{\sqrt{2}\,\mu_0 R}{L^2}M=L22​μ0​R​
  4. (D)M=22 μ0R2LM=\dfrac{2\sqrt{2}\,\mu_0 R^2}{L}M=L22​μ0​R2​

Correct answer: (D)

Step-by-step solution →
Q79·PhysicsNumericalJEE Main 2023
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B=0.8B = 0.8B=0.8 T. When released the radius of the loop starts shrinking at a constant rate of 2 cm s−1^{-1}−1. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be ________ mV.

Correct answer: 10

Step-by-step solution →
Q80·PhysicsSingle correctJEE Main 2023
A wire of length 111 m moving with velocity 888 m/s at right angles to a magnetic field of 222 T. The magnitude of induced emf, between the ends of wire will be
  1. (A)202020 V
  2. (B)888 V
  3. (C)121212 V
  4. (D)161616 V

Correct answer: (D)

Step-by-step solution →
Q81·PhysicsSingle correctJEE Main 2023
A metallic rod of length 'LLL' is rotated with an angular speed of 'ω\omegaω' normal to a uniform magnetic field 'BBB' about an axis passing through one end of rod as shown in figure. The induced emf will be:
  1. (A)14BL2ω\dfrac14 BL^2\omega41​BL2ω
  2. (B)12B2L2ω\dfrac12 B^2 L^2\omega21​B2L2ω
  3. (C)14B2Lω\dfrac14 B^2 L\omega41​B2Lω
  4. (D)12BL2ω\dfrac12 BL^2\omega21​BL2ω

Correct answer: (D)

Step-by-step solution →
Q82·PhysicsSingle correctJEE Main 2023
A conducting circular loop of radius 10π\tfrac{10}{\sqrt{\pi}}π​10​ cm is placed perpendicular to a uniform magnetic field of 0.5 T. The magnetic field is decreased to zero in 0.5 s at a steady rate. The induced emf in the circular loop at 0.25 s is:
  1. (A)emf = 1 mV
  2. (B)emf = 5 mV
  3. (C)emf = 100 mV
  4. (D)emf = 10 mV

Correct answer: (D)

Step-by-step solution →
Q83·PhysicsNumericalJEE Main 2023
Three identical resistors with resistance R=12 ΩR=12\,\OmegaR=12Ω and two identical inductors with self inductance L=5L=5L=5 mH are connected to an ideal battery with emf of 121212 V as shown in figure. The current through the battery long after the switch has been closed will be _________ A.

Correct answer: 3

Step-by-step solution →
Q84·PhysicsSingle correctJEE Main 2022
A coil of inductance 1 H and resistance 100 Ω is connected to a battery of 6 V. Determine approximately : (a) The time elapsed before the current acquires half of its steady – state value (b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circuit is switched on. (Given In2 = 0.693, e−3/2=0.25e^{-3/2} = 0.25e−3/2=0.25)
  1. (A)t = 10 ms; U = 2 mJ
  2. (B)t = 10 ms; U = 1 mJ
  3. (C)t = 7 ms; U = 1 mJ
  4. (D)t = 7 ms; U = 2 mJ

Correct answer: (C)

Step-by-step solution →
Q85·PhysicsNumericalJEE Main 2022
For the given circuit the current through battery of 6 V just after closing the switch 'S' will be ......... A.

Correct answer: 1

Step-by-step solution →
Q86·PhysicsNumericalJEE Main 2022
A conducting circular loop is placed in X − Y plane in presence of magnetic field B⃗=(3t3j^+3t2k^)\vec{B} = \left(3t^{3}\hat{j} + 3t^{2}\hat{k}\right)B=(3t3j^​+3t2k^) in SI unit. If the radius of the loop is 1m, the induced emf in the loop, at time, t = 2s is nπV. The value of n is……..

Correct answer: 12

Step-by-step solution →
Q87·PhysicsNumericalJEE Main 2022
In a coil of resistance 8Ω8\Omega8Ω, the magnetic flux due to an external magnetic field varies with time as ϕ=23(9−t2)\phi = \frac{2}{3}\left(9 - t^2\right)ϕ=32​(9−t2). The value of total heat produced in the coil, till the flux becomes zero, will be________J.

Correct answer: 2

Step-by-step solution →
Q88·PhysicsSingle correctJEE Main 2022
A small square loop of wire of side lll is placed inside a large square loop of wire L (L >> lll). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is :
  1. (A)22μ0L2πℓ\frac{2\sqrt{2}\mu_0 L^{2}}{\pi \ell}πℓ22​μ0​L2​
  2. (B)μ0ℓ222πL\frac{\mu_0 \ell^{2}}{2\sqrt{2}\pi L}22​πLμ0​ℓ2​
  3. (C)22μ0ℓ2πL\frac{2\sqrt{2}\mu_0 \ell^{2}}{\pi L}πL22​μ0​ℓ2​
  4. (D)μ0L222πℓ\frac{\mu_0 L^{2}}{2\sqrt{2}\pi \ell}22​πℓμ0​L2​

Correct answer: (C)

Step-by-step solution →
Q89·PhysicsSingle correctJEE Main 2022
A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be: (Assume the coil to be short circuited.)
  1. (A)Halved
  2. (B)Quadrupled
  3. (C)The same
  4. (D)Doubled

Correct answer: (D)

Step-by-step solution →
Q90·PhysicsNumericalJEE Main 2022
A metallic rod of length 20 cm is palced in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4×10−34 \times 10^{-3}4×10−3 T and the angle of dip is 45°. The emf induced in the rod is ________ mV.

Correct answer: 16

Step-by-step solution →
Q91·PhysicsSingle correctJEE Main 2022
A metallic conductor of length 1m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad/s. If the horizontal component of earth's magnetic field is 0.2×10−40.2 \times 10^{-4}0.2×10−4 T, then emf induced between the two ends of the conductor is :
  1. (A)5μV
  2. (B)50μV
  3. (C)5mV
  4. (D)50mV

Correct answer: (B)

Step-by-step solution →
Q92·PhysicsNumericalJEE Main 2022
A 10Ω, 20 mH coil carrying constant current is connected to a battery of 20 V through a switch is opened current becomes zero in 100μs. The average emf induced in the coil is ............ V.

Correct answer: 400

Step-by-step solution →
Q93·PhysicsSingle correctJEE Main 2022
Two coils of self inductance L1L_1L1​ and L2L_2L2​ are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be :
  1. (A)1L1+1L2+1M\frac{1}{L_1} + \frac{1}{L_2} + \frac{1}{M}L1​1​+L2​1​+M1​
  2. (B)L1+L2+ML_1 + L_2 + ML1​+L2​+M
  3. (C)L1+L2+2ML_1 + L_2 + 2ML1​+L2​+2M
  4. (D)L1+L2−2ML_1 + L_2 - 2ML1​+L2​−2M

Correct answer: (D)

Step-by-step solution →
Q94·PhysicsNumericalJEE Main 2022
The current in a coil of self inductance 2.0 H is increasing according to I=2sin⁡(t2)I = 2\sin(t^{2})I=2sin(t2) A. The amount of energy spent during the period when current changes from 0 to 2A is ______ J.

Correct answer: 4

Step-by-step solution →
Q95·PhysicsSingle correctJEE Main 2022
Match List-I with List-II List-I List-II (A) AC generator (I) Detects the presence of current in the circuit (B) Galvanometer (II) Converts mechanical energy into electrical energy (C) Transformer (III) Works on the principle of resonance in AC circuit (D) Metal detector (IV) Changes an alternating voltage for smaller or greater value Choose the correct answer from the options given below :-
  1. (A)(A)–(II), B–(I), (C)–(IV), (D)–(III)
  2. (B)(A)–(II), B–(I), (C)–(III), (D)–(IV)
  3. (C)(A)–(III), B–(IV), (C)–(II), (D)–(I)
  4. (D)(A)–(III), B–(I), (C)–(II), (D)–(IV)

Correct answer: (A)

Step-by-step solution →
Q96·PhysicsNumericalJEE Main 2022
A circular coil of 1000 turns each with area 1m2^22 is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T. The maximum voltage generation will be ______V.

Correct answer: 440

Step-by-step solution →
Q97·PhysicsSingle correctJEE Advanced 2021
A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius aaa, with its center at the origin. A magnetic dipole of moment mmm is brought along the axis of this loop from infinity to a point at distance rrr (>> aaa) from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole mmm, at a point on its axis at distance rrr, is μ02πmr3\frac{\mu_0}{2\pi}\frac{m}{r^3}2πμ0​​r3m​, where μ0\mu_0μ0​ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_1m1​ and m2m_2m2​, separated by a distance r on the common axis, with their north poles facing each other, is km1m2r4\frac{km_1m_2}{r^4}r4km1​m2​​, where kkk is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles. The work done in bringing the dipole from infinity to a distance rrr from the center of the loop by the given process is proportional to
  1. (A)mr5\frac{m}{r^5}r5m​
  2. (B)m2r5\frac{m^2}{r^5}r5m2​
  3. (C)m2r6\frac{m^2}{r^6}r6m2​
  4. (D)m2r7\frac{m^2}{r^7}r7m2​

Correct answer: (C)

Step-by-step solution →
Q98·PhysicsMultiple correctJEE Advanced 2021
A long straight wire carries a current, III = 2 ampere. A semi-circular conducting rod is placed beside it on two conducting parallel rails of negligible resistance. Both the rails are parallel to the wire. The wire, the rod and the rails lie in the same horizontal plane, as shown in the figure. Two ends of the semi-circular rod are at distances 1cm and 4 cm from the wire. At time t = 0, the rod starts moving on the rails with a speed v = 3.0 m/s (see the figure). A resistor R = 1.4 Ω\OmegaΩ and a capacitor C0C_0C0​ = 5.0 μ\muμF are connected in series between the rails. At time t = 0, C0C_0C0​ is uncharged. Which of the following statement(s) is(are) correct ? [μ0=4π×10−7\mu_0 = 4\pi \times 10^{-7}μ0​=4π×10−7 SI units. Take ln 2 = 0.7]
  1. (A)Maximum current through RRR is 1.2×10−61.2 \times 10^{-6}1.2×10−6 ampere
  2. (B)Maximum current through RRR is 3.8×10−63.8 \times 10^{-6}3.8×10−6 ampere
  3. (C)Maximum charge on capacitor C0C_0C0​ is 8.4×10−128.4 \times 10^{-12}8.4×10−12 coulomb
  4. (D)Maximum charge on capacitor C0C_0C0​ is 2.4×10−122.4 \times 10^{-12}2.4×10−12 coulomb

Correct answer: (A), (C)

Step-by-step solution →
Q99·PhysicsSingle correctJEE Advanced 2021
A special metal S conducts electricity without any resistance. A closed wire loop, made of S, does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius aaa, with its center at the origin. A magnetic dipole of moment mmm is brought along the axis of this loop from infinity to a point at distance rrr (>> aaa) from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole mmm, at a point on its axis at distance rrr, is μ02πmr3\frac{\mu_0}{2\pi}\frac{m}{r^3}2πμ0​​r3m​, where μ0\mu_0μ0​ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, m1m_1m1​ and m2m_2m2​, separated by a distance r on the common axis, with their north poles facing each other, is km1m2r4\frac{km_1m_2}{r^4}r4km1​m2​​, where kkk is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles. When the dipole mmm is placed at a distance rrr from the center of the loop (as shown in the figure), the current induced in the loop will be proportional to
  1. (A)mr3\frac{m}{r^3}r3m​
  2. (B)m2r2\frac{m^2}{r^2}r2m2​
  3. (C)mr2\frac{m}{r^2}r2m​
  4. (D)m2r\frac{m^2}{r}rm2​

Correct answer: (A)

Step-by-step solution →
Q100·PhysicsSingle correctJEE Main 2021
A square loop of side 20 cm and resistance 1Ω is moved towards right with a constant speed v0_{0}0​. The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4Ω. What should be the value of v0_{0}0​ so that a steady current of 2 mA flows in the loop ?
  1. (A)1 m/s
  2. (B)1 cm/s
  3. (C)102^{2}2 m/s
  4. (D)10−2^{-2}−2 cm/s

Correct answer: (B)

Step-by-step solution →
Q101·PhysicsSingle correctJEE Main 2021
For the given circuit the current iii through the battery when the key in closed and the steady state has been reached is__________.
  1. (A)6 A
  2. (B)25 A
  3. (C)10 A
  4. (D)0 A

Correct answer: (C)

Step-by-step solution →
Q102·PhysicsSingle correctJEE Main 2021
A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :
  1. (A)μ04π82 a2b\frac{\mu_{0}}{4\pi}8\sqrt{2}\,\frac{a^{2}}{b}4πμ0​​82​ba2​
  2. (B)μ04π82a\frac{\mu_{0}}{4\pi}\frac{8\sqrt{2}}{a}4πμ0​​a82​​
  3. (C)μ04π82 b2a\frac{\mu_{0}}{4\pi}8\sqrt{2}\,\frac{b^{2}}{a}4πμ0​​82​ab2​
  4. (D)μ04π82b\frac{\mu_{0}}{4\pi}\frac{8\sqrt{2}}{b}4πμ0​​b82​​

Correct answer: (A)

Step-by-step solution →
Q103·PhysicsSingle correctJEE Main 2021
A coil is placed in a magnetic field B⃗\vec{B}B as shown below : A current is induced in the coil because B⃗\vec{B}B is :
  1. (A)Outward and decreasing with time
  2. (B)Parallel to the plane of coil and decreasing with time
  3. (C)Outward and increasing with time
  4. (D)Parallel to the plane of coil and increasing with time

Correct answer: (A)

Step-by-step solution →
Q104·PhysicsSingle correctJEE Main 2021
A constant magnetic field of 1 T is applied in the x > 0 region. A metallic circular ring of radius 1m is moving with a constant velocity of 1 m/s along the x-axis. At t = 0s, the centre of O of the ring is at x = − 1m. What will be the value of the induced emf in the ring at t = 1s? (Assume the velocity of the ring does not change.)
  1. (A)1 V
  2. (B)2π V
  3. (C)2 V
  4. (D)0 V

Correct answer: (C)

Step-by-step solution →
Q105·PhysicsSingle correctJEE Main 2021
An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :
  1. (A)0.4
  2. (B)0.8
  3. (C)0.125
  4. (D)0.2

Correct answer: (D)

Step-by-step solution →
Q106·PhysicsNumericalJEE Main 2021
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s−1^{-1}−1 in a uniform horizontal magnetic field of 3.0×10−23.0 \times 10^{-2}3.0×10−2 T. The maximum emf induced the coil will be .......... ×10−2\times 10^{-2}×10−2 volt (rounded off to the nearest integer)

Correct answer: 60

Step-by-step solution →
Q107·PhysicsNumericalJEE Main 2021
In the given figure the magnetic flux through the loop increases according to the relation φβ_\betaβ​(t)10t2^22 + 20t, where φβ_\betaβ​ is in milliwebers and t is in seconds. The magnitude of current throughR = 2Ω resistor at t = 5s is ________ mA.

Correct answer: 60

Step-by-step solution →
Q108·PhysicsNumericalJEE Main 2021
Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T1T_1T1​ is connected to T2T_2T2​. As the current in R = 6Ω attains a maximum value of steady state level, T1T_1T1​ is disconnected from T2T_2T2​ and immediately connected to T3T_3T3​. Potential drop across r = 3Ω resistor immediately after T1T_1T1​ is connected to T3T_3T3​ is ____ V. (Round off to the Nearest Integer)

Correct answer: 3

Step-by-step solution →
Q109·PhysicsNumericalJEE Main 2021
A circular conducting coil of radius 1 m being heated by the change of magnetic field B⃗\vec{B}B passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2μΩ .The magnetic field is slowly switched off such that its magnitude changes in time as B=4π×10−3 T(1−t100)B = \frac{4}{\pi} \times 10^{-3}\ \text{T}\left(1 - \frac{t}{100}\right)B=π4​×10−3 T(1−100t​) The energy dissipated by the coil before the magnetic field is switched off completely is E = _________ mJ

Correct answer: 80

Step-by-step solution →
Q110·PhysicsNumericalJEE Main 2021
An inductor of 10 mH is connected to a 20 V battery through a resistor of 10kΩ and a switch . After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after 1 μs is x100\frac{x}{100}100x​ mA. Then x is equal to ________. (Take e−1e^{-1}e−1 = 0.37 )

Correct answer: 74

Step-by-step solution →
Q111·PhysicsSingle correctJEE Main 2021
The arm PQ of a rectangular conductor is moving from x = 0 to x = 2b outwards and then inwards from x = 2b to x = 0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph the variation of different quantities with distance.
  1. (A)A − Flux, B − EMF, C − Power dissipated
  2. (B)A − EMF, B − Power dissipated, C − Flux
  3. (C)A − Power dissipated, B − Flux, C − EMF
  4. (D)A − Flux, B − Power dissipated, C − EMF

Correct answer: (A)

Step-by-step solution →
Q112·PhysicsSingle correctJEE Main 2021
The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is :
  1. (A)LRℓn5\frac{L}{R}\ell n5RL​ℓn5
  2. (B)infinite
  3. (C)LRℓn2\frac{L}{R}\ell n2RL​ℓn2
  4. (D)LRℓn10\frac{L}{R}\ell n10RL​ℓn10

Correct answer: (C)

Step-by-step solution →
Q113·PhysicsSingle correctJEE Main 2021
A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure Two resistors R1_11​ and R2_22​ are connected across the ends of the rails. There is a uniform magnetic field B⃗\vec{B}B pointing into the page. An external agent pulls the bar to the left at a constant speed v. The correct statement about the directions of induced currents I1_11​ and I2_22​ flowing through R1_11​ and R2_22​ respectively is :
  1. (A)Both I1_11​ and I2_22​ are in anticlockwise direction
  2. (B)Both I1_11​ and I2_22​ are in clockwise direction
  3. (C)I1_11​ is in clockwise direction and I2_22​ is in anticlockwise direction
  4. (D)I1_11​ is in anticlockwise direction and I2_22​ is in clockwise direction

Correct answer: (C)

Step-by-step solution →
Q114·PhysicsSingle correctJEE Main 2021
The magnetic field in a region is given by B⃗=B0(xa)k^\vec{B} = B_{0}\left(\frac{x}{a}\right)\hat{k}B=B0​(ax​)k^. A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity v⃗=v0i^\vec{v} = v_{0}\hat{i}v=v0​i^. The emf induced in the loop is :
  1. (A)B0v02d2a\frac{B_{0}v_{0}^{2}d}{2a}2aB0​v02​d​
  2. (B)B0v0d2a\frac{B_{0}v_{0}d}{2a}2aB0​v0​d​
  3. (C)B0v0d2a\frac{B_{0}v_{0}d^{2}}{a}aB0​v0​d2​
  4. (D)B0v0d22a\frac{B_{0}v_{0}d^{2}}{2a}2aB0​v0​d2​

Correct answer: (C)

Step-by-step solution →
Q115·PhysicsSingle correctJEE Main 2021
An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5×10−42.5 \times 10^{-4}2.5×10−4 Wb/m2^22 and the angle of dip is 60°. The emf induced between the tips of the plane wings will be __________.
  1. (A)88.37 mV
  2. (B)62.50 mV
  3. (C)54.125 mV
  4. (D)108.25 mV

Correct answer: (D)

Step-by-step solution →
Q116·PhysicsNumericalJEE Main 2021
A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V =3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is ________ J.

Correct answer: 144

Step-by-step solution →
Q117·PhysicsSingle correctJEE Main 2021
The current (i) at time t=0 and t=∞t = \inftyt=∞ respectively for the given circuit is :
  1. (A)18E55,5E18\dfrac{18E}{55}, \dfrac{5E}{18}5518E​,185E​
  2. (B)5E18,18E55\dfrac{5E}{18}, \dfrac{18E}{55}185E​,5518E​
  3. (C)5E18,10E33\dfrac{5E}{18}, \dfrac{10E}{33}185E​,3310E​
  4. (D)10E33,5E18\dfrac{10E}{33}, \dfrac{5E}{18}3310E​,185E​

Correct answer: (C)

Step-by-step solution →
Q118·PhysicsSingle correctJEE Main 2021
Figure shows a circuit that contains four identical resistors with resistance R = 2.0 Ω\OmegaΩ. Two identical inductors with inductance L = 2.0 mH and an ideal battery with emf E = 9.V. The current 'i' just after the switch 's' is closed will be :
  1. (A)9A
  2. (B)3.0 A
  3. (C)2.25 A
  4. (D)3.37 A

Correct answer: (C)

Step-by-step solution →
Q119·PhysicsSingle correctJEE Advanced 2020
A light disc made of aluminium (a nonmagnetic material) is kept horizontally and is free to rotate about its axis as shown in the figure. A strong magnet is held vertically at a point above the disc away from its axis. On revolving the magnet about the axis of the disc, the disc will (figure is schematic and not drawn to scale)-
  1. (A)rotate in the direction opposite to the direction of magnet's motion
  2. (B)rotate in the same direction as the direction of magnet's motion
  3. (C)not rotate and its temperature will remain unchanged
  4. (D)not rotate but its temperature will slowly rise

Correct answer: (B)

Step-by-step solution →
Q120·PhysicsNumericalJEE Advanced 2020
The inductors of two LRLRLR circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced EMFEMFEMF in the inductors by the time the currents reach their steady state values is_________ mJ.

Correct answer: 55.00

Step-by-step solution →
Q121·PhysicsNumericalJEE Main 2020
A part of complete circuit is shown in the figure. At some instant, the value of current I is 1 A and it is decreasing at a rate of 102As−110^2\text{As}^{-1}102As−1. The value of the potential difference VP−VQV_P-V_QVP​−VQ​, (in volts) at that instant, is __________.

Correct answer: 33.00

Step-by-step solution →
Q122·PhysicsNumericalJEE Main 2020
Two concentric circular coils, C1_11​ and C2_22​ are placed in the XY plane. C1_11​ has 500 turns, and a radius of 1 cm. C2_22​ has 200 turns and radius of 20 cm. C2_22​ carries a time dependent current I(t)=(5t2−2t+3)I(t) = (5t^2 - 2t + 3)I(t)=(5t2−2t+3) A where t is in s. The emf induced in C1_11​ (in mV), at the instant t = 1 s is 4x\dfrac{4}{x}x4​. The value of x is ________.

Correct answer: 5.00

Step-by-step solution →
Q123·PhysicsSingle correctJEE Main 2020
A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then the time at which the energy stored in the inductor reaches (1/n) times of its maximum value, is :
  1. (A)LRln⁡(nn+1)\frac{L}{R}\ln\left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)RL​ln(n​+1n​​)
  2. (B)LRln⁡(n+1n−1)\frac{L}{R}\ln\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)RL​ln(n​−1n​+1​)
  3. (C)LRln⁡(n−1n)\frac{L}{R}\ln\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right)RL​ln(n​n​−1​)
  4. (D)LRln⁡(nn−1)\frac{L}{R}\ln\left(\frac{\sqrt{n}}{\sqrt{n}-1}\right)RL​ln(n​−1n​​)

Correct answer: (D)

Step-by-step solution →
Q124·PhysicsSingle correctJEE Main 2020
A small bar magnet is moved through a coil at constant speed from one end to the other. Which of the following series of observations will be seen on the galvanometer G attached across the coil? Three positions shown describe: (a) the magnet's entry (b) magnet is completely inside and (c) magnet's exit
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q125·PhysicsSingle correctJEE Main 2020
An elliptical loop having resistance R, of semi major axis a and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency ω, the average power loss in the loop due to joule heating is:
  1. (A)π2a2b2B2ω2R\frac{\pi^{2}a^{2}b^{2}B^{2}\omega^{2}}{R}Rπ2a2b2B2ω2​
  2. (B)zero
  3. (C)πabBωR\frac{\pi abB\omega}{R}RπabBω​
  4. (D)π2a2b2B2ω22R\frac{\pi^{2}a^{2}b^{2}B^{2}\omega^{2}}{2R}2Rπ2a2b2B2ω2​

Correct answer: (D)

Step-by-step solution →
Q126·PhysicsSingle correctJEE Main 2020
A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dBdt=0.032 Ts−1\frac{dB}{dt}=0.032\,Ts^{-1}dtdB​=0.032Ts−1. The induced current in the loop is close to (Resistivity of the metal wire is 1.23×10−8 Ωm1.23\times 10^{-8}\,\Omega m1.23×10−8Ωm )
  1. (A)0.61A0.61A0.61A
  2. (B)0.43 A0.43\,A0.43A
  3. (C)0.53 A0.53\,A0.53A
  4. (D)0.34 A0.34\,A0.34A

Correct answer: (A)

Step-by-step solution →
Q127·PhysicsNumericalJEE Main 2020
In a fluorescent lamp choke (a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be ____.

Correct answer: 10

Step-by-step solution →
Q128·PhysicsSingle correctJEE Main 2020
As shown in the figure, a battery of emf ε\varepsilonε is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc_{c}c​ (tc_{c}c​ is the time constant of the circuit) is
  1. (A)εLeR2\dfrac{\varepsilon L}{eR^{2}}eR2εL​
  2. (B)εLR2(1−1e)\dfrac{\varepsilon L}{R^{2}}\left(1-\dfrac{1}{e}\right)R2εL​(1−e1​)
  3. (C)εReL2\dfrac{\varepsilon R}{eL^{2}}eL2εR​
  4. (D)εLR2\dfrac{\varepsilon L}{R^{2}}R2εL​

Correct answer: (A)

Step-by-step solution →
Q129·PhysicsSingle correctJEE Main 2020
At time t = 0 magnetic field of 1000 Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to 500 Gauss, in the next 5s, then induced EMF in the loops is:
  1. (A)28 μ\muμV
  2. (B)56 μ\muμV
  3. (C)48 μ\muμV
  4. (D)36 μ\muμV

Correct answer: (B)

Step-by-step solution →
Q130·PhysicsSingle correctJEE Main 2020
A planar loop of wire rotates in a uniform magnetic field. Initially, at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:
  1. (A)2.5 s and 5.0 s
  2. (B)5.0 s and 10.0 s
  3. (C)2.5 s and 7.5 s
  4. (D)5.0 and 7.5 s

Correct answer: (A)

Step-by-step solution →
Q131·PhysicsSingle correctJEE Main 2020
A long solenoid of radius R carries a time (t) dependent current I (t) = I0t(1−t)I_0 t (1-t)I0​t(1−t). A ring of radius 2R is placed cordially near its middle. During the time internal 0≤t≤10 \le t \le 10≤t≤1, the induced current (IRI_RIR​) and the induced EMF (VRV_RVR​) in the ring changes as:
  1. (A)At t = 0.25 direction of IRI_RIR​ reverses and VRV_RVR​ is maximum.
  2. (B)Direction of IRI_RIR​ remains unchanged and VRV_RVR​ is zero at t = 0.25
  3. (C)Direction of IRI_RIR​ remains unchanged and VRV_RVR​ is maximum at t = 0.5
  4. (D)At t = 0.5 direction of IRI_RIR​ reverses and VRV_RVR​ is zero.

Correct answer: (D)

Step-by-step solution →
Q132·PhysicsNumericalJEE Main 2020
A loop ABCDEFA of straight edges has six corner points A (0, 0, 0), B (5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and F(0, 0, 5). The magnetic field in this region in B⃗=(3i^+4k^)\vec{B} = (3\hat{i} + 4\hat{k})B=(3i^+4k^)T. The quantity of flux through the loop ABCDEFA (in Wb) is _________.

Correct answer: 175

Step-by-step solution →
Q133·PhysicsNumericalJEE Advanced 2019
A 10 cm long perfectly conducting wire PQ is moving with a velocity 1 cm/s on a pair of horizontal rails of zero resistance. One side of the rails is connected to an inductor L = 1 mH and a resistance R = 1 Ω\OmegaΩ as shown in the figure. The horizontal rails, L and R lie in the same plane with a uniform magnetic field B = 1 T perpendicular to the plane. If the key S is closed at certain instant, the current in the circuit after 1 millisecond is x×10−3x \times 10^{-3}x×10−3 A, where the value of x is ________. [Assume the velocity of wire PQ remains constant (1 cm/s) after key S is closed. Given: e−1=0.37e^{-1} = 0.37e−1=0.37, where e is base of the natural logarithm]

Correct answer: 0.63

Step-by-step solution →
Q134·PhysicsMultiple correctJEE Advanced 2019
A conducting wire of parabolic shape, initially y=x2y = x^2y=x2, is moving with velocity V⃗=V0i^\vec{V} = V_0\hat{i}V=V0​i^ in a non uniform magnetic field B⃗=B0(1+(yL)β)k^\vec{B} = B_0\left(1+\left(\frac{y}{L}\right)^{\beta}\right)\hat{k}B=B0​(1+(Ly​)β)k^, as shown in figure. If V0V_0V0​, B0B_0B0​, L and β are positive constants and Δφ is the potential difference developed between the ends of the wire, then the correct statement(s) is/are:
  1. (A)∣Δϕ∣|\Delta\phi|∣Δϕ∣ is proportional to the length of the wire projected on the y-axis.
  2. (B)∣Δϕ∣|\Delta\phi|∣Δϕ∣ remains the same if the parabolic wire is replaced by a straight wire, y = x initially, of length 2\sqrt{2}2​ L
  3. (C)∣Δϕ∣=12B0V0L|\Delta\phi| = \frac{1}{2}B_0V_0L∣Δϕ∣=21​B0​V0​L for β = 0
  4. (D)∣Δϕ∣=43B0V0L|\Delta\phi| = \frac{4}{3}B_0V_0L∣Δϕ∣=34​B0​V0​L for β = 2

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

Step-by-step solution →
Q135·PhysicsSingle correctJEE Main 2019
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of 1 cms−1^{-1}−1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 Ω, the current in the loop at that instant will be close to :
  1. (A)115 μA
  2. (B)170 μA
  3. (C)60 μA
  4. (D)150 μA

Correct answer: (B)

Step-by-step solution →
Q136·PhysicsSingle correctJEE Main 2019
Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between t = 0 and t=LRt = \dfrac{L}{R}t=RL​ is :
  1. (A)EL7.3R2\dfrac{EL}{7.3R^{2}}7.3R2EL​
  2. (B)EL2.7R2\dfrac{EL}{2.7R^{2}}2.7R2EL​
  3. (C)7.3ELR2\dfrac{7.3EL}{R^{2}}R27.3EL​
  4. (D)2.7ELR2\dfrac{2.7EL}{R^{2}}R22.7EL​

Correct answer: (B)

Step-by-step solution →
Q137·PhysicsSingle correctJEE Main 2019
A coil of self inductance 10 mH and resistance 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω. After the switch is closed the time taken for the current to attain 80% of the saturation values is: [take ln 5 = 1.6]
  1. (A)0.016 s
  2. (B)0.324 s
  3. (C)0.002 s
  4. (D)0.103 s

Correct answer: (A)

Step-by-step solution →
Q138·PhysicsSingle correctJEE Main 2019
Two coils P and Q are separated by some distance. When a current of 3 A flows through coil P a magnetic flux of 10−310^{-3}10−3 Wb passes through Q. No current is passed through Q. When no current passes through P and a current of 2 A passes through Q, the flux through P is:
  1. (A)6.67×10−36.67 \times 10^{-3}6.67×10−3 Wb
  2. (B)6.67×10−46.67 \times 10^{-4}6.67×10−4 Wb
  3. (C)3.67×10−43.67 \times 10^{-4}3.67×10−4 Wb
  4. (D)3.67×10−33.67 \times 10^{-3}3.67×10−3 Wb

Correct answer: (B)

Step-by-step solution →
Q139·PhysicsSingle correctJEE Main 2019
The total number of turns and cross-section area in solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to:
  1. (A)1/L
  2. (B)L
  3. (C)1/L2^22
  4. (D)L2^22

Correct answer: (A)

Step-by-step solution →
Q140·PhysicsSingle correctJEE Main 2019
A very long solenoid of radius R is carrying current I(t)=kte−αtI(t) = kte^{-\alpha t}I(t)=kte−αt (k>0)(k > 0)(k>0), as a function of time (t≥0)(t \geq 0)(t≥0). counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:-
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (D)

Step-by-step solution →
Q141·PhysicsSingle correctJEE Main 2019
A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is
  1. (A)2ln⁡2\dfrac{2}{\ln 2}ln22​
  2. (B)ln⁡2\ln 2ln2
  3. (C)12ln⁡2\dfrac{1}{2}\ln 221​ln2
  4. (D)2ln⁡22\ln 22ln2

Correct answer: (D)

Step-by-step solution →
Q142·PhysicsSingle correctJEE Main 2019
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5 N m−1^{-1}−1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 Ω10\,\Omega10Ω and air drag negligible, N will be close to:
  1. (A)50000
  2. (B)10000
  3. (C)1000
  4. (D)5000

Correct answer: (D)

Step-by-step solution →
Q143·PhysicsSingle correctJEE Main 2019
A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms−1^{-1}−1, at right angles to the horizontal component of the earth's magnetic field, of 0.3×10−40.3\times10^{-4}0.3×10−4 Wb/m2^{2}2. The value of the induced emf in wire is :
  1. (A)1.5×10−31.5\times10^{-3}1.5×10−3 V
  2. (B)1.1×10−31.1\times10^{-3}1.1×10−3 V
  3. (C)2.5×10−32.5\times10^{-3}2.5×10−3 V
  4. (D)0.3×10−30.3\times10^{-3}0.3×10−3 V

Correct answer: (B)

Step-by-step solution →
Q144·PhysicsSingle correctJEE Main 2019
In the figure shown, a circuit contains two identical resistors with resistance R = 5 Ω\OmegaΩ and an inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed?
  1. (A)5.5 A
  2. (B)7.5 A
  3. (C)3 A
  4. (D)6 A

Correct answer: (D)

Step-by-step solution →
Q145·PhysicsSingle correctJEE Main 2019
There are two long co – axial solenoids of same length l. The inner and outer coils have radii r1_11​ and r2_22​ and number of turns per unit length n1_11​ and n2_22​ respectively. The ratio of mutual inductance to the self – inductance of the inner – coil is:
  1. (A)n1n2\frac{n_1}{n_2}n2​n1​​
  2. (B)n2n1⋅r1r2\frac{n_2}{n_1}\cdot\frac{r_1}{r_2}n1​n2​​⋅r2​r1​​
  3. (C)n2n1⋅r22r12\frac{n_2}{n_1}\cdot\frac{r_2^2}{r_1^2}n1​n2​​⋅r12​r22​​
  4. (D)n2n1\frac{n_2}{n_1}n1​n2​​

Correct answer: (D)

Step-by-step solution →
Q146·PhysicsSingle correctJEE Main 2019
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then self inductance of the coil:
  1. (A)decreases by a factor of 9
  2. (B)increases by a factor of 27
  3. (C)increases by a factor of 3
  4. (D)decreases by a factor of 939\sqrt{3}93​

Correct answer: (C)

Step-by-step solution →
Q147·PhysicsSingle correctJEE Main 2019
A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is:
  1. (A)12 mV
  2. (B)6 mV
  3. (C)1 mV
  4. (D)2 mV

Correct answer: (A)

Step-by-step solution →
Q148·PhysicsMultiple correctJEE Advanced 2018
In the figure below, the switches S1S_{1}S1​ and S2S_{2}S2​ are closed simultaneously at t=0t = 0t=0 and a current starts to flow in the circuit. Both the batteries have the same magnitude of the electromotive force (emf) and the polarities are as indicated in the figure. Ignore mutual inductance between the inductors. The current III in the middle wire reaches its maximum magnitude ImaxI_{max}Imax​ at time t=τt = \taut=τ. Which of the following statements is (are) true?
  1. (A)Imax=V2RI_{max} = \frac{V}{2R}Imax​=2RV​
  2. (B)Imax=V4RI_{max} = \frac{V}{4R}Imax​=4RV​
  3. (C)τ=LRln⁡2\tau = \frac{L}{R}\ln 2τ=RL​ln2
  4. (D)τ=2LRln⁡2\tau = \frac{2L}{R}\ln 2τ=R2L​ln2

Correct answer: (B), (D)

Step-by-step solution →
Q149·PhysicsMultiple correctJEE Advanced 2017
A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform magnetic field B⃗\vec{B}B points into the plane of the paper. A uniform magnetic field B⃗\vec{B}B points into the plane of the paper. At t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity ω\omegaω in the magnetic field. Which of the following options is/are correct?
  1. (A)The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper.
  2. (B)The net emf induced due to both the loops is proportional to cos ωt\omega tωt.
  3. (C)The emf induced in the loop is proportional to the sun of the areas of the two loops.
  4. (D)The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone.

Correct answer: (A), (D)

Step-by-step solution →
Q150·PhysicsMultiple correctJEE Advanced 2017
A source of constant voltage V is connected to a resistance R and two ideal inductors L1L_{1}L1​ and L2L_{2}L2​ through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t=0t = 0t=0, the switch is closed and current begins to flow. Which of the following options is/are correct?
  1. (A)After a long time, the current through L1L_{1}L1​ will be VRL2L1+L2\dfrac{V}{R}\dfrac{L_{2}}{L_{1}+L_{2}}RV​L1​+L2​L2​​
  2. (B)After a long time, the current through L2L_{2}L2​ will be VRL1L1+L2\dfrac{V}{R}\dfrac{L_{1}}{L_{1}+L_{2}}RV​L1​+L2​L1​​
  3. (C)The ratio of the currents through L1L_{1}L1​ and L2L_{2}L2​ is fixed at all times (t>0)(t > 0)(t>0)
  4. (D)At t=0t = 0t=0, the current through the resistance R is VR\dfrac{V}{R}RV​

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

Step-by-step solution →
Q151·PhysicsMultiple correctJEE Advanced 2016
A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the 90°90°90° vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 A s−110\ \text{A s}^{-1}10 A s−1. Which of the following statement(s) is(are) true?
  1. (A)The magnitude of induced emfemfemf in the wire is (μ0π)\left(\frac{\mu_0}{\pi}\right)(πμ0​​) volt
  2. (B)If the loop is rotated at a constant angular speed about the wire, an additional emfemfemf of (μ0π)\left(\frac{\mu_0}{\pi}\right)(πμ0​​) volt is induced in the wire
  3. (C)The induced current in the wire is in opposite direction to the current along the hypotenuse
  4. (D)There is a repulsive force between the wire and the loop

Correct answer: (A), (D)

Step-by-step solution →
Q152·PhysicsMultiple correctJEE Advanced 2016
A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0v_{0}v0​ in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0B_{0}B0​ into the plane of the paper, as shown in the figure. For sufficiently large v0v_{0}v0​, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive. Which of the following schematic plot(s) is(are) correct? (Ignore gravity)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (C), (D)

Step-by-step solution →
Q153·PhysicsIntegerJEE Advanced 2016
Two inductors L1L_1L1​ (inductance 1 mH, internal resistance 3Ω) and L2L_2L2​ (inductance 2 mH, internal resistance 4Ω), and a resistor R (resistance 12Ω) are all connected in parallel across a 5V battery. The circuit is switched on at time t=0t = 0t=0. The ratio of the maximum to the minimum current (Imax/IminI_{max} / I_{min}Imax​/Imin​) drawn from the battery is

Correct answer: 8

Step-by-step solution →
Q154·PhysicsSingle correctJEE Advanced 2013
A point Q is moving in a circular orbit of radius RRR in the x-y plane with an angular velocity ω\omegaω. This can be considered as equivalent to a loop carrying a steady current Qω2π\frac{Q\omega}{2\pi}2πQω​. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 000 to BBB in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ\gammaγ. The change in the magnetic dipole moment associated with the orbit, at the end of time interval of the magnetic field change, is
  1. (A)−γBQR2-\gamma BQR^{2}−γBQR2
  2. (B)−γBQR22-\gamma\frac{BQR^{2}}{2}−γ2BQR2​
  3. (C)γBQR22\gamma\frac{BQR^{2}}{2}γ2BQR2​
  4. (D)γBQR2\gamma BQR^{2}γBQR2

Correct answer: (B)

Step-by-step solution →
Q155·PhysicsSingle correctJEE Advanced 2013
A point Q is moving in a circular orbit of radius RRR in the x-y plane with an angular velocity ω\omegaω. This can be considered as equivalent to a loop carrying a steady current Qω2π\frac{Q\omega}{2\pi}2πQω​. A uniform magnetic field along the positive z-axis is now switched on, which increases at a constant rate from 000 to BBB in one second. Assume that the radius of the orbit remains constant. The application of the magnetic field induces an emf in the orbit. The induced emf is defined as the work done by an induced electric field in moving a unit positive charge around closed loop. It is known that, for an orbiting charge, the magnetic dipole moment is proportional to the angular momentum with a proportionality constant γ\gammaγ. The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change, is
  1. (A)BR4\frac{BR}{4}4BR​
  2. (B)BR2\frac{BR}{2}2BR​
  3. (C)BRBRBR
  4. (D)2BR2BR2BR

Correct answer: (B)

Step-by-step solution →

Electromagnetic Induction — frequently asked

How many questions from Electromagnetic Induction appear in JEE?

Electromagnetic Induction has appeared in 120 of the last 186 JEE Main and JEE Advanced papers — about 65% of them — contributing 155 questions in total across those papers.

Is Electromagnetic Induction an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 65% 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 Electromagnetic Induction 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.

Other Physics chapters

  • Properties of Solids and Liquids 344
  • Current Electricity 309
  • Rotational Motion 266
  • Geometrical Optics 259
  • Kinematics 255
  • Magnetic Field of Current 211
  • Electromagnetic Waves 208
  • Thermodynamics 201

All 28 Physics chapters →

Practise Electromagnetic Induction until it stops costing you marks.

Build a timed test from these 155 questions in one click. Jarvis marks it, names the specific misconception behind each wrong answer, and brings the ones you failed back at the right interval.

Practise Electromagnetic Induction freeSee pricing

Free plan, no time limit · No credit card needed

Jarvis OS

AI-powered JEE preparation.

Question papers

JEE Main PYQsJEE Advanced PYQsPhysics PYQsChemistry PYQsMaths PYQs

Product

TourPricingSign inCreate account

Legal

Privacy PolicyTerms of ServiceRefund & CancellationShipping & DeliveryContact Us

Operated by

Venyou Craft Private Limited

info@jarvisos.net

© 2026 Jarvis OS