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MCQMediumJEE 2026Faraday's Laws of EMI

JEE Physics 2026 Question with Solution

A conducting circular loop of area 1.0 m21.0 \, \text{m}^2 is placed perpendicular to a magnetic field which varies as B=sin⁡(100t)B = \sin(100t) Tesla. If the resistance of the loop is 100 Ω100 \, \Omega, then the average thermal energy dissipated in the loop in one period is _____ J.

  • A

    2π2\pi

  • B

    π\pi

  • C

    π2\pi^2

  • D

    π/2\pi/2

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Area of loop = 1.0 m21.0 \, \text{m}^2, resistance = 100 Ω100 \, \Omega, magnetic field B(t)=sin⁡(100t)B(t) = \sin(100t). The loop is perpendicular to the field, so cos⁡θ=1\cos\theta = 1.

Find: Thermal energy dissipated in one full time period.

Using Faraday's law, magnetic flux is

Φ(t)=B(t)⋅A=sin⁡(100t)⋅1=sin⁡(100t)\Phi(t) = B(t) \cdot A = \sin(100t) \cdot 1 = \sin(100t)

Therefore, induced emf is

E(t)=−dΦdt=−ddt(sin⁡(100t))=−100cos⁡(100t)\mathcal{E}(t) = -\frac{d\Phi}{dt} = -\frac{d}{dt}(\sin(100t)) = -100\cos(100t)

Current in the loop is

I(t)=E(t)R=−100cos⁡(100t)100=−cos⁡(100t)I(t) = \frac{\mathcal{E}(t)}{R} = \frac{-100\cos(100t)}{100} = -\cos(100t)

So instantaneous power dissipated is

P(t)=I2R=cos⁡2(100t)⋅100=100cos⁡2(100t)P(t) = I^2R = \cos^2(100t) \cdot 100 = 100\cos^2(100t)

The angular frequency is ω=100 rad/s\omega = 100 \, \text{rad/s}, hence the time period is

T=2πω=2π100=π50T = \frac{2\pi}{\omega} = \frac{2\pi}{100} = \frac{\pi}{50}

Energy dissipated in one period is

E=∫0TP(t) dt=∫0π/50100cos⁡2(100t) dtE = \int_0^T P(t) \, dt = \int_0^{\pi/50} 100\cos^2(100t) \, dt

Using cos⁡2θ=1+cos⁡2θ2\cos^2\theta = \frac{1+\cos 2\theta}{2},

E=100∫0π/501+cos⁡(200t)2 dtE = 100\int_0^{\pi/50} \frac{1+\cos(200t)}{2} \, dt E=50[t+sin⁡(200t)200]0π/50E = 50\left[t + \frac{\sin(200t)}{200}\right]_0^{\pi/50} E=50[π50+sin⁡(4π)200]E = 50\left[\frac{\pi}{50} + \frac{\sin(4\pi)}{200}\right]

Since sin⁡(4π)=0\sin(4\pi)=0,

E=50(π50)=πE = 50\left(\frac{\pi}{50}\right) = \pi

Therefore, the thermal energy dissipated in one period is π J\pi \, \text{J}. The correct option is B.

RMS Power Method

Given: E(t)=−100cos⁡(100t)\mathcal{E}(t) = -100\cos(100t), so the emf amplitude is 100 V100 \, \text{V} and resistance is 100 Ω100 \, \Omega.

Find: Energy dissipated in one period.

For a sinusoidal emf, the average value of cos⁡2(100t)\cos^2(100t) over one full cycle is 12\frac{1}{2}. Hence average power is

⟨P⟩=E022R=10022⋅100=50 W\langle P \rangle = \frac{\mathcal{E}_0^2}{2R} = \frac{100^2}{2 \cdot 100} = 50 \, \text{W}

The time period is

T=2π100=π50 sT = \frac{2\pi}{100} = \frac{\pi}{50} \, \text{s}

So energy in one period is

E=⟨P⟩T=50⋅π50=πE = \langle P \rangle T = 50 \cdot \frac{\pi}{50} = \pi

Therefore, the thermal energy dissipated in one period is π J\pi \, \text{J}. The correct option is B.

Common mistakes

  • Using B(t)B(t) directly in the power formula is incorrect because power depends on the induced current, and current comes from the induced emf E=−dΦdt\mathcal{E} = -\frac{d\Phi}{dt}. First differentiate the flux, then find current and power.

  • Forgetting that the loop is perpendicular to the magnetic field leads to an incorrect flux factor. Here Φ=BAcos⁡θ\Phi = BA\cos\theta with θ=0\theta = 0, so cos⁡θ=1\cos\theta = 1. Do not introduce an extra sine or cosine factor.

  • Taking the average of cos⁡(100t)\cos(100t) as 12\frac{1}{2} is wrong. Over one full period, the average of cos⁡(100t)\cos(100t) is 00, but the average of cos⁡2(100t)\cos^2(100t) is 12\frac{1}{2}. Power depends on the square.

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