NVAMediumJEE 2026Faraday's Laws of EMI

JEE Physics 2026 Question with Solution

A conducting circular loop is rotated about its diameter at a constant angular speed of 100rad s1100 \, \text{rad s}^{-1} in a magnetic field of 0.5T0.5 \, \text{T}, perpendicular to the axis of rotation. When the loop is rotated by 3030^\circ from the horizontal position, the induced EMF is 15.4mV15.4 \, \text{mV}. The radius of the loop is _____ mm.

(Take π=227\pi = \dfrac{22}{7})

Answer

Correct answer:100

Step-by-step solution

Standard Method

Given: angular speed ω=100rad s1\omega = 100 \, \text{rad s}^{-1}, magnetic field B=0.5TB = 0.5 \, \text{T}, induced EMF e=15.4×103Ve = 15.4 \times 10^{-3} \, \text{V}, and angle θ=30\theta = 30^\circ.

Find: radius of the loop in millimetres.

The solution uses the rotating-loop relation

e=BAωsinθe = BA\omega \sin\theta

with

A=πr2A = \pi r^2

So,

15.4×103=0.5×πr2×100×sin3015.4 \times 10^{-3} = 0.5 \times \pi r^2 \times 100 \times \sin 30^\circ

Using

sin30=12\sin 30^\circ = \frac{1}{2}

we get

15.4×103=12.5πr215.4 \times 10^{-3} = 12.5\pi r^2

Hence,

r2=15.4×10312.5πr^2 = \frac{15.4 \times 10^{-3}}{12.5\pi}

Using

π=227\pi = \frac{22}{7}

this gives

r2=15.4×10312.5×227=4.9×104r^2 = \frac{15.4 \times 10^{-3}}{12.5 \times \frac{22}{7}} = 4.9 \times 10^{-4}

and therefore

r=0.022m=22mmr = 0.022 \, \text{m} = 22 \, \text{mm}

However, the solution finally states:

r=0.10mr = 0.10 \, \text{m}

and concludes Final Answer: 100.

So, taking the solution, the recorded answer is 100100. The working shown earlier in the same HTML leads to 22mm22 \, \text{mm}, so there is a discrepancy in the source solution.

Common mistakes

  • Using e=BAωcosθe = BA\omega \cos\theta instead of e=BAωsinθe = BA\omega \sin\theta. This is wrong because EMF comes from the time derivative of flux Φ=BAcosθ\Phi = BA\cos\theta. Differentiate first, then substitute the angle.

  • Forgetting to convert 15.4mV15.4 \, \text{mV} into volts. This is wrong because the SI form required in the equation is 15.4×103V15.4 \times 10^{-3} \, \text{V}. Always convert milli to 10310^{-3} before calculation.

  • Substituting the area as A=2πrA = 2\pi r instead of A=πr2A = \pi r^2. This is wrong because magnetic flux depends on enclosed area, not circumference. Use the area of the circular loop.

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