Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 (omega t) A, where omega = 1000 rad/s . A circular conducting loop (B) of radius 1 cm coaxially slid 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 alpha/sqrt(2) muA . The value of alpha is ____. [Resistance of the loop = 10Omega ]

B = mu0 n I Since the loop B is well inside the long solenoid (10 cm deep into a 100 cm length), the magnetic field is uniform across its cross-section and approximately equal to the ideal solenoid field B = mu0 n I. Here n = 500 turns/m. The sliding velocity v just means it's moving through a uniform field region, so motional EMF does not contribute to the flux change; only the transformer EMF (time-varying B field) creates the induced current. Peak EMF epsilon0 = pi r2 mu0 n I0 omega. Using pi2 approx 9.87, 20pi2 approx 197.39. Thus, irms approx (197/sqrt(2)) , muA. alpha = 197. Motion parallel to uniform field lines inside an ideal solenoid induces no motional EMF. Only the time derivative of the AC current (dI/dt) contributes to Faraday induction. Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 (omega t) A, where omega = 1000 rad/s . A circular conducting loop (B) of radius 1 cm c

B = mu0 n I Since the loop B is well inside the long solenoid (10 cm deep into a 100 cm length), the magnetic field is uniform across its cross-section and approximately equal to the ideal solenoid field B = mu0 n I. Here n = 500 turns/m. The sliding velocity v just means it's moving through a uniform field region, so motional EMF does not contribute to the flux change; only the transformer EMF (time-varying B field) creates the induced current. Peak EMF epsilon0 = pi r2 mu0 n I0 omega. Using pi

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 (omega t) A, where omega = 1000 rad/s . A circular conducting loop (B) of radius 1 cm coaxially slid 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 alpha/sqrt(2) muA . The value of alpha is ____. [Resistance of the loop = 10Omega ]

B = mu0 n I Since the loop B is well inside the long solenoid (10 cm deep into a 100 cm length), the magnetic field is uniform across its cross-section and approximately equal to the ideal solenoid field B = mu0 n I. Here n = 500 turns/m. The sliding velocity v just means it's moving through a uniform field region, so motional EMF does not contribute to the flux change; only the transformer EMF (time-varying B field) creates the induced current. Peak EMF epsilon0 = pi r2 mu0 n I0 omega. Using pi2 approx 9.87, 20pi2 approx 197.39. Thus, irms approx (197/sqrt(2)) , muA. alpha = 197. Motion parallel to uniform field lines inside an ideal solenoid induces no motional EMF. Only the time derivative of the AC current (dI/dt) contributes to Faraday induction. Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 (omega t) A, where omega = 1000 rad/s . A circular conducting loop (B) of radius 1 cm c

B = mu0 n I Since the loop B is well inside the long solenoid (10 cm deep into a 100 cm length), the magnetic field is uniform across its cross-section and approximately equal to the ideal solenoid field B = mu0 n I. Here n = 500 turns/m. The sliding velocity v just means it's moving through a uniform field region, so motional EMF does not contribute to the flux change; only the transformer EMF (time-varying B field) creates the induced current. Peak EMF epsilon0 = pi r2 mu0 n I0 omega. Using pi

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current I = 10 (omega…

Year	2026	2025	2024	Total
Questions	12	6	10	28

Q50 jee_main_2026_22_january_morning Energy Density in Inductor

Inductance of a coil with

10⁴

turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of

10~Ω

. The energy density in the inductor when the current reaches

( 1e)

of its maximum value is

α π \times 1e² ~J / m³

. The value of

α

is \_\_\_\_. (

μ₀ = 4π \times 10⁻⁷ Tm / A

)

Numerical Answer. Answer: 20 to 20

Solution

Related Formula

Ed = (B²)/(2μ₀), B = μ₀ n I

Core Logic

Solution inductor energy density diagram for Q50 - JEE Main 2026 Morning

Maximum current:

I₀ = (10)/(10) = 1 A

Current at given instant:

I = (I₀)/(e) = (1)/(e)

Energy density expression:

Ed = (μ₀ n² I²)/(2) = 4π \times 10⁻⁷ \times (10⁴)² \times (1/e)²2 = (20π)/(e²) J/m³

Therefore, $α = 20$ $\alpha = 20$.

Pattern Recognition

Sees: Energy density in inductor at exponential current growth stage. Shortcut: Express magnetic field $B$ $B$ in terms of turns density $n$ $n$ and current $I$ $I$, substitute into energy density formula $Ed = (B²)/(2μ₀)$ $E_d = \frac{B^2}{2\mu_0}$. Check: Numerical answer is 20. ✓

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q38 jee_main_2026_22_january_evening Transient Behavior in LR Circuits

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.

LR circuit diagram with ammeter for Q38 - JEE Main 2026 Evening — The figure shows a circuit powered by a 9V battery with three 9-ohm resistors and two 4mH inductors connected in parallel branches.

A. $1$ $1$
B. zero
C. $3$ $3$
D. $2$ $2$

Solution

Related Formula

I(t=0^+) for an ideal inductor = 0 (Open Circuit)

Core Logic

At $t = 0$ $t = 0$ (immediately after closing switch K), inductors oppose any instant change in current and behave as open circuits ( $IL = 0$ $I_L = 0$).

Removing the branches containing the $4 ~mH$ $4 \mathrm{~mH}$ inductors leaves only the middle branch containing a single $9 Ω$ $9 \Omega$ resistor connected to the $9 ~V$ $9 \mathrm{~V}$ battery.

Calculating total initial current $I$ $I$ measured by the ammeter:

I = (V)/(R) = 9 ~V9 Ω = 1 ~A

Equivalent circuit diagram at t=0 for Q38 - JEE Main 2026 Evening — The figure shows a circuit powered by a 9V battery with three 9-ohm resistors and two 4mH inductors connected in parallel branches.

Step 1: Final Conclusion

The reading of the ammeter at the moment switch K is turned ON is $1 ~A$ $1 \mathrm{~A}$.

Pattern Recognition

LR Circuit Transient Rule: At $t=0$ $t=0$, Replace Inductor $arrow$ $\rightarrow$ Open Circuit. At $t=\infty$ $t=\infty$, Replace Inductor $arrow$ $\rightarrow$ Short Circuit wire.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q46 jee_main_2026_22_january_evening Induced EMF in Rotating Loop

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 = (22)/(7)

)

Numerical Answer. Answer: 14 to 14

Solution

Related Formula

E = B A ω (ω t)

Core Logic

Given $B = 0.5 ~T, ω = 100 ~rad/s, θ = ω t = 30^°$ and $E = 15.4 \times 10⁻³ ~V$ :

15.4 \times 10⁻³ = B (π r²) ω (30^°)

15.4 \times 10⁻³ = 0.5 \times ((22)/(7) r²) \times 100 \times (1)/(2)

15.4 \times 10⁻³ = (550)/(7) r²

r² = 15.4 \times 10⁻³ \times 7550 = 107.8 \times 10⁻³550 = 1.96 \times 10⁻⁴ ~m²

r = 1.96 \times 10⁻⁴ = 1.4 \times 10⁻² ~m = 14 ~mm

Step 1: Final Conclusion

The radius of the circular loop is $14 ~mm$ $14 \mathrm{~mm}$.

Pattern Recognition

AC Generator induced EMF formula: $E = B A ω θ$ $\mathcal{E} = B A \omega \sin\theta$. Substitute $30^° = 1/2$ $\sin 30^\circ = 1/2$ and solve directly for radius $r$ $r$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q41 jee_main_2026_23_january_morning Motional EMF

A 20 m long uniform copper wire held horizontally is allowed to fall under the gravity (g = 10 m/s

²

) through a uniform horizontal magnetic field of 0.5 Gauss perpendicular to the length of the wire. The induced EMF across the wire it travels a vertical distance of 200 m is ____ mV.

A. $0.2\sqrt(10)$ $0.2\sqrt{10}$
B. $20\sqrt(10)$ $20\sqrt{10}$
C. $2 \sqrt(10)$ $2 \sqrt{10}$
D. $200\sqrt(10)$ $200\sqrt{10}$

Solution

Related Formula

$v = \sqrt(2gh)$ $v = \sqrt{2gh}$

ε = Bvl

Step 1: Calculate Velocity

Falling freely under gravity for $200 m$ $200\text{ m}$: $v = \sqrt(2gh)$ $v = \sqrt{2gh}$

v = \sqrt(2 \times 10 \times 200) = \sqrt(4000) = 20\sqrt(10) m/s

Step 2: Calculate Motional EMF

Convert magnetic field to Tesla: $0.5 Gauss = 0.5 \times 10⁻⁴ T$ . The velocity, magnetic field, and length are mutually perpendicular.

ε = Bvl

ε = (0.5 \times 10⁻⁴) \times (20\sqrt(10)) \times 20

ε = (10 \times 10⁻⁴) \times 20\sqrt(10)

ε = 200\sqrt(10) \times 10⁻⁴ V

ε = 20\sqrt(10) \times 10⁻³ V = 20\sqrt(10) mV

Pattern Recognition

Sees: "wire falling in horizontal magnetic field" → find velocity using kinematics $v = \sqrt(2gh)$ $v = \sqrt{2gh}$, then just plug into standard motional EMF $ε = Bvl$ $\varepsilon = Bvl$. Remember to convert Gauss to Tesla.

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Class 12 Physics: Electromagnetic Induction Class 11 Physics: Motion in a Straight Line

Q46 jee_main_2026_23_january_morning Motional EMF

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

)

Numerical Answer. Answer: 100 to 100

Solution

Related Formula

ε = (1)/(2) Bω l²

mgl(1 - θ) = (1)/(2)(ml²)ω²

Core Logic

The metallic string of the pendulum cuts the perpendicular magnetic field lines as it swings. The induced EMF along the length of a rotating rod is maximum when its angular velocity $ω$ $\omega$ is maximum. This maximum $ω$ $\omega$ occurs at the lowest point of the swing.

Step 1: Find Maximum Angular Velocity

Using conservation of mechanical energy from extreme to mean position:

mgl(1 - 60°) = (1)/(2) I ω²

mgl(1 - (1)/(2)) = (1)/(2)(ml²)ω²

(1)/(2)mgl = (1)/(2)ml²ω²

ω = \sqrt((g)/(l))

Step 2: Substitute Values

Given $l = 10 cm = 0.1 m$ $l = 10 \text{ cm} = 0.1 \text{ m}$ and $g = 10 m/s²$ $g = 10 \text{ m/s}^{2}$.

ω = \sqrt((10)/(0.1)) = \sqrt(100) = 10 rad/s

Step 3: Calculate Maximum EMF

ε = (1)/(2) B ω l²

ε = (1)/(2) (2 T)(10 rad/s)(0.1 m)²

ε = 10 \times 0.01 = 0.1 V

Convert to millivolts:

ε = 100 mV

Pattern Recognition

Sees: "pendulum wire in magnetic field" → It's just a rotating rod! EMF is $(1)/(2) B ω l²$ $\frac{1}{2} B \omega l^2$. Find max $ω$ $\omega$ via energy conservation at the lowest point.

Chapter Mix

Class 12 Physics: Electromagnetic Induction Class 11 Physics: Work, Energy and Power

Solution & Explanation

Related Formula

Core Logic

Step 1: Calculate Induced EMF

Step 2: Calculate RMS Current

Step 3: Numerical Substitution

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electromagnetic Induction Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate Velocity

Step 2: Calculate Motional EMF

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Maximum Angular Velocity

Step 2: Substitute Values

Step 3: Calculate Maximum EMF

Pattern Recognition

Chapter Mix

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