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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

ε = - A (dB)/(dt)

$B = μ₀ n I$ $B = \mu_0 n I$

irms = εrmsR

Core Logic

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 $B = μ₀ n I$ $B = \mu_0 n I$. Here $n = 500$ $n = 500$ turns/m. The sliding velocity $v$ $v$ 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 $B$ $B$ field) creates the induced current.

Step 1: Calculate Induced EMF

I = I₀ ω t (dI)/(dt) = I₀ ω ω t

ε = A (dB)/(dt) = A μ₀ n (dI)/(dt) = (π r²) μ₀ n (I₀ ω ω t)

Step 2: Calculate RMS Current

Peak EMF $ε₀ = π r² μ₀ n I₀ ω$ .

i₀ = (ε₀)/(R) = (π r² μ₀ n I₀ ω)/(R)

irms = i₀\sqrt(2) = π r² μ₀ n I₀ ω\sqrt(2) R

Step 3: Numerical Substitution

irms = π \times (10⁻²)² \times (4π \times 10⁻⁷) \times 500 \times 10 \times 1000\sqrt(2) \times 10

irms = π \times 10⁻⁴ \times 4π \times 10⁻⁷ \times 5 \times 10⁶\sqrt(2) \times 10

irms = 20π² \times 10⁻⁶ \times 10\sqrt(2) \times 10 = 20π²\sqrt(2) \times 10⁻⁶ A

Using $π² \approx 9.87$ $\pi^2 \approx 9.87$, $20π² \approx 197.39$ $20\pi^2 \approx 197.39$. Thus, $irms \approx 197\sqrt(2)$ . $α = 197$ $\alpha = 197$.

Pattern Recognition

Motion parallel to uniform field lines inside an ideal solenoid induces no motional EMF. Only the time derivative of the AC current $(dI)/(dt)$ $\frac{dI}{dt}$ contributes to Faraday induction.

Chapter Mix

Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

More Electromagnetic Induction Previous-Year Questions — Page 6

Q53 jee_main_2024_31_jan_evening Faraday's Law

The magnetic flux

φ

(in weber) linked with a closed circuit of resistance

8 Ω

varies with time (in seconds) as

φ = 5t² - 36t + 1

. The induced current in the circuit at

t = 2 s

is ________ A.

Numerical Answer. Answer: 2 to 2

Solution

Related Formula

ε = -(dφ)/(dt)

I = (|ε|)/(R)

Core Logic

Calculate the time derivative of the magnetic flux to find the induced EMF. Evaluate it at the requested time, and then apply Ohm's law to find the current magnitude.

Step 1: Calculate Induced EMF

ε = -(d)/(dt) (5t² - 36t + 1)

ε = -(10t - 36)

At $t = 2 s$ $t = 2 \text{ s}$:

ε = - (10 \times 2 - 36)

ε = -(20 - 36) = 16 V

Step 2: Calculate Induced Current

I = (ε)/(R) = (16)/(8) = 2 A

Pattern Recognition

Flux polynomials ( $At² - Bt + C$ $At^2 - Bt + C$) instantly trigger a simple derivative test. Remember to drop the negative sign for final current magnitude unless direction is specifically asked.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q jee_main_2024_31_jan_morning Mutual Inductance

A small square loop of wire of side

is placed inside a large square loop of wire of side

L

(

L = ²

). The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is

\sqrt(x) \times 10⁻⁷ H

, where

x =

Numerical Answer. Answer: 128 to 128

Solution

Related Formula

M = (φ₂)/(i₁)

Bstraight wire segment = (μ₀ i)/(4π d) ( θ₁ + θ₂)

Core Logic

Mutual Inductance diagram for Q57 - JEE Main 2024 Morning

Assume a current $i$ $i$ flows through the larger square loop of side $L$ $L$. The magnetic field generated by it at its center acts as a uniform field across the very small inner loop of side $\ell$.

The magnetic field at the center of the large square loop (distance $d = L/2$ $d = L/2$ from each side, angles $45^°$ $45^\circ$):

B = 4 \times [ (μ₀ i)/(4π (L/2)) ( 45^° + 45^°) ]

B = (μ₀ i)/(π (L/2)) ( 2\sqrt(2) )

B = 2\sqrt(2) μ₀ iπ L

Step 2: Mutual Inductance Calculation

Flux linkage for the inner loop:

φ = B \cdot ²

φ = 2\sqrt(2) μ₀ iπ L ²

Given $L = ²$ $L = \ell^2$:

φ = 2\sqrt(2) μ₀ iπ ( ²) ² = 2\sqrt(2) μ₀ iπ

Mutual inductance $M$ $M$:

M = (φ)/(i) = 2\sqrt(2) μ₀π

Using $μ₀ = 4π \times 10⁻⁷$ $\mu_0 = 4\pi \times 10^{-7}$:

M = 2\sqrt(2) \times 4π \times 10⁻⁷π

M = 8\sqrt(2) \times 10⁻⁷ H

M = \sqrt(128) \times 10⁻⁷ H

Comparing with $\sqrt(x) \times 10⁻⁷$ $\sqrt{x} \times 10^{-7}$, we get $x = 128$ $x = 128$.

Chapter Mix

Class 12 Physics: Electromagnetic Induction

Q44 jee_main_2024_31_jan_morning Faraday's Law

A coil is placed perpendicular to a magnetic field of

5000 ~T

. When the field is changed to

3000 ~T

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:

A. $7$ $7$
B. $70$ $70$
C. $35$ $35$
D. $140$ $140$

Solution

Related Formula

ε = N | (Δφ)/(Δ t) |

Δφ = (Δ B) A θ

Core Logic

Given data: Initial Magnetic Field, $Bᵢ = 5000 T$ $B_i = 5000\mathrm{\,T}$ Final Magnetic Field, $Bf = 3000 T$ $B_f = 3000\mathrm{\,T}$ Time interval, $Δ t = 2 s$ $\Delta t = 2\mathrm{\,s}$ Diameter, $d = 0.02 m \Rightarrow r = 0.01 m$ Induced emf, $ε = 22 V$ $\varepsilon = 22\mathrm{\,V}$

Change in magnetic field magnitude $|Δ B| = 5000 - 3000 = 2000 T$ . Area of the coil $A = π r² = π (0.01)² = 10⁻⁴π m²$ .

Step 2: Equation Evaluation

Δφ = |Δ B| A = (2000) π (0.01)² = 0.2π

Using Faraday's Law:

22 = N ( (0.2π)/(2) )

$22 = N (0.1π)$ $22 = N (0.1\pi)$ Taking $π \approx 22/7$ $\pi \approx 22/7$:

22 = N ( 0.1 \times (22)/(7) )

1 = (N)/(70) \Rightarrow N = 70

Chapter Mix

Class 12 Physics: Electromagnetic Induction

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 6

Solution

Related Formula

Core Logic

Step 1: Calculate Induced EMF

Step 2: Calculate Induced Current

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Mutual Inductance Calculation

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Equation Evaluation

Chapter Mix

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More Electromagnetic Induction Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Calculate Induced EMF

Step 2: Calculate Induced Current

Pattern Recognition

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Related Formula

Core Logic

Step 2: Mutual Inductance Calculation

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Related Formula

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