A conducting circular loop of area 1.0 , m2 is placed perpendicular to a magnetic field which varies as B = (100 , t) Tesla. If the resistance of the loop is 100 , Omega, then the average thermal energy dissipated in the loop in one period is ____ J.

Given area of the loop, A = 1 m2 and magnetic field B = (100t). The magnetic flux passing through the loop is: Induced EMF E = | (dphi/dt) | = 100(100t) Instantaneous power P = (E2/R) = (1002 2(100t)/100) = 1002(100t). Thermal energy dissipated in one time period T: The angular frequency omega = 100 rad/s, so time period T = (2pi/omega) = (2pi/100) = (pi/50) sec. For a sinusoidal signal, the integral of 2(omega t) over one full period T is always T/2. Thus, P , dt = Pmax x (T/2). Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

A conducting circular loop of area 1.0 , m2 is placed perpendicular to a magnetic field which varies as B = (100 , t) Tesla. If the resistance of the loop is 100 , Omega, then the average thermal ener

Given area of the loop, A = 1 m2 and magnetic field B = (100t). The magnetic flux passing through the loop is: Induced EMF E = | (dphi/dt) | = 100(100t) Instantaneous power P = (E2/R) = (1002 2(100t)/100) = 1002(100t). Thermal energy dissipated in one time period T: The angular frequency omega = 100 rad/s, so time period T = (2pi/omega) = (2pi/100) = (pi/50) sec. For a sinusoidal signal, the integral of 2(omega t) over one full period T is always T/2. Thus, P , dt = Pmax x (T/2). Class 12 Physic

A conducting circular loop of area 1.0 , m2 is placed perpendicular to a magnetic field which varies as B = (100 , t) Tesla. If the resistance of the loop is 100 , Omega, then the average thermal energy dissipated in the loop in one period is ____ J.

Given area of the loop, A = 1 m2 and magnetic field B = (100t). The magnetic flux passing through the loop is: Induced EMF E = | (dphi/dt) | = 100(100t) Instantaneous power P = (E2/R) = (1002 2(100t)/100) = 1002(100t). Thermal energy dissipated in one time period T: The angular frequency omega = 100 rad/s, so time period T = (2pi/omega) = (2pi/100) = (pi/50) sec. For a sinusoidal signal, the integral of 2(omega t) over one full period T is always T/2. Thus, P , dt = Pmax x (T/2). Class 12 Physics: Electromagnetic Induction Class 12 Physics: Alternating Current

A conducting circular loop of area 1.0 , m2 is placed perpendicular to a magnetic field which varies as B = (100 , t) Tesla. If the resistance of the loop is 100 , Omega, then the average thermal ener

Given area of the loop, A = 1 m2 and magnetic field B = (100t). The magnetic flux passing through the loop is: Induced EMF E = | (dphi/dt) | = 100(100t) Instantaneous power P = (E2/R) = (1002 2(100t)/100) = 1002(100t). Thermal energy dissipated in one time period T: The angular frequency omega = 100 rad/s, so time period T = (2pi/omega) = (2pi/100) = (pi/50) sec. For a sinusoidal signal, the integral of 2(omega t) over one full period T is always T/2. Thus, P , dt = Pmax x (T/2). Class 12 Physic

A conducting circular loop of area 1.0 , m2 is placed perpendicular to a magnetic field which varies as B = (100 , t)…

Year	2026	2025	2024	Total
Questions	12	6	10	28

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

φ = B \cdot A

E = -(dφ)/(dt)

P = (E²)/(R)

Core Logic

Given area of the loop, $A = 1 m²$ $A = 1\text{ m}^2$ and magnetic field $B = (100t)$ $B = \sin(100t)$. The magnetic flux passing through the loop is:

φ = B \cdot A = (100t) \times 1 = (100t)

Induced EMF $E = | (dφ)/(dt) | = 100 (100t)$

Step 1: Calculating Power and Energy

Instantaneous power $P = (E²)/(R) = (100² ²(100t))/(100) = 100 ²(100t)$ .

Thermal energy dissipated in one time period $T$ $T$:

Q = \int₀T P dt = \int₀T 100 ²(100t) dt

The angular frequency $ω = 100 rad/s$ $\omega = 100\text{ rad/s}$, so time period $T = (2π)/(ω) = (2π)/(100) = (π)/(50) sec$ .

Q = 100 \int₀π/50 ²(100t) dt = 100 \int₀π/50 (1 + (200t))/(2) dt

Q = 50 [ t + ( (200t))/(200) ]₀π/50

Q = 50 ( (π)/(50) - 0 ) = π Joules

Pattern Recognition

For a sinusoidal signal, the integral of $²(ω t)$ $\cos^2(\omega t)$ over one full period $T$ $T$ is always $T/2$ $T/2$. Thus, $\int P dt = Pmax \times (T)/(2)$ .

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: Calculating Power and Energy

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

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More Electromagnetic Induction Previous-Year Questions — Page 6

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Related Formula

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