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MCQEasyJEE 2023Faraday's Laws of EMI

JEE Physics 2023 Question with Solution

A conducting loop of radius π1/2∗(10) cm\pi^{1/2} * (10) \, \text{cm} is placed perpendicular to a uniform magnetic field of 0.5 T0.5 \, \text{T}. The magnetic field is decreased to zero in 0.5 s0.5 \, \text{s} at a steady rate. The induced emf in the circular loop at 0.25 s0.25 \, \text{s} is:

  • A

    1 mV1 \, \text{mV}

  • B

    10 mV10 \, \text{mV}

  • C

    100 mV100 \, \text{mV}

  • D

    5 mV5 \, \text{mV}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Radius of the loop is written as r=10π cmr = \frac{10}{\sqrt{\pi}} \, \text{cm}, magnetic field changes uniformly from 0.5 T0.5 \, \text{T} to 00 in 0.5 s0.5 \, \text{s}.

Find: The induced emf at 0.25 s0.25 \, \text{s}.

Since the field decreases at a steady rate, the induced emf remains constant throughout the interval.

The area of the loop is

A=πr2=π(10π)2=100 cm2=10−2 m2A = \pi r^2 = \pi \left(\frac{10}{\sqrt{\pi}}\right)^2 = 100 \, \text{cm}^2 = 10^{-2} \, \text{m}^2

The magnetic flux through the loop is

Φ=B⋅A=0.5⋅10−2=5⋅10−3 Wb\Phi = B \cdot A = 0.5 \cdot 10^{-2} = 5 \cdot 10^{-3} \, \text{Wb}

The induced emf is

E=∣ΔΦΔt∣=5⋅10−30.5=10 mV\mathcal{E} = \left| \frac{\Delta \Phi}{\Delta t} \right| = \frac{5 \cdot 10^{-3}}{0.5} = 10 \, \text{mV}

Therefore, the induced emf is 10 mV10 \, \text{mV}. The solution working gives option B, although the solution states D.

Common mistakes

  • Using the radius as π×10 cm\sqrt{\pi} \times 10 \, \text{cm} instead of 10π cm\frac{10}{\sqrt{\pi}} \, \text{cm}. This gives the wrong area. Evaluate the radius carefully before substituting into A=πr2A = \pi r^2.

  • Assuming the emf depends on the instant 0.25 s0.25 \, \text{s} separately. Because the magnetic field decreases at a steady rate, dBdt\frac{dB}{dt} is constant, so the induced emf is the same at every instant during the change.

  • Forgetting to convert 100 cm2100 \, \text{cm}^2 into 10−2 m210^{-2} \, \text{m}^2. Area must be in SI units before calculating flux in weber.

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