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NVAEasyJEE 2023Faraday's Laws of EMI

JEE Physics 2023 Question with Solution

A conducting circular loop is placed in a uniform magnetic field of 0.4 T0.4 \, \text{T} with its plane perpendicular to the field. Somehow, the radius of the loop starts expanding at a constant rate of 1 mm/s1 \, \text{mm/s}. The magnitude of induced emf in the loop at an instant when the radius of the loop is 2 cm2 \, \text{cm} will be _____, μV\mu V.

Answer

Correct answer:50.265

Step-by-step solution

Standard Method

Given:

  • Magnetic field, B=0.4 TB = 0.4 \, \text{T}
  • Rate of expansion, drdt=1 mm/s=1×10−3 m/s\frac{dr}{dt} = 1 \, \text{mm/s} = 1 \times 10^{-3} \, \text{m/s}
  • Instantaneous radius, r=2 cm=2×10−2 mr = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m}
  • The plane of the loop is perpendicular to the magnetic field, so θ=0∘\theta = 0^\circ

Find: The magnitude of induced emf.

The area of the loop is A=πr2A = \pi r^2. The magnetic flux is

Φ=Bπr2cos⁡0∘=Bπr2\Phi = B \pi r^2 \cos 0^\circ = B \pi r^2

Using Faraday's law,

ε=−d(Bπr2)dt=−Bπd(r2)dt=−Bπ(2rdrdt)=−2πBrdrdt\varepsilon = -\frac{d(B \pi r^2)}{dt} = -B \pi \frac{d(r^2)}{dt} = -B \pi \left(2r \frac{dr}{dt}\right) = -2 \pi B r \frac{dr}{dt}

Substituting the given values,

ε=−2π(0.4)(2×10−2)(1×10−3)=−1.6π×10−5 V\varepsilon = -2 \pi (0.4) (2 \times 10^{-2}) (1 \times 10^{-3}) = -1.6 \pi \times 10^{-5} \, \text{V} ε=−16π×10−6 V=−16π μV\varepsilon = -16 \pi \times 10^{-6} \, \text{V} = -16 \pi \, \mu \text{V}

Magnitude of induced emf:

∣ε∣=16π μV|\varepsilon| = 16 \pi \, \mu \text{V} ∣ε∣=16×3.14159 μV≈50.265 μV|\varepsilon| = 16 \times 3.14159 \, \mu \text{V} \approx 50.265 \, \mu \text{V}

Therefore, the magnitude of the induced emf is 50.26550.265.

Flux Differentiation Detail

Given: Φ=BAcos⁡θ\Phi = BA \cos\theta with constant BB and θ=0∘\theta = 0^\circ.

Find: How the changing radius changes the flux.

Since the radius is changing with time, the area changes with time:

A=πr2A = \pi r^2

Hence,

Φ=Bπr2\Phi = B\pi r^2

Differentiate with respect to time:

dΦdt=Bπd(r2)dt\frac{d\Phi}{dt} = B\pi \frac{d(r^2)}{dt}

Using

d(r2)dt=2rdrdt\frac{d(r^2)}{dt} = 2r\frac{dr}{dt}

we get

dΦdt=2πBrdrdt\frac{d\Phi}{dt} = 2\pi B r \frac{dr}{dt}

Therefore,

∣ε∣=∣−dΦdt∣=2πBrdrdt|\varepsilon| = \left| -\frac{d\Phi}{dt} \right| = 2\pi B r \frac{dr}{dt}

This gives the numerical value 50.26550.265 in μV\mu \text{V}.

Common mistakes

  • Using Φ=BAsin⁡θ\Phi = BA \sin\theta instead of Φ=BAcos⁡θ\Phi = BA \cos\theta. Here the plane is perpendicular to the field, so the area vector is parallel to the field and θ=0∘\theta = 0^\circ. Use cos⁡0∘=1\cos 0^\circ = 1.

  • Forgetting to convert 1 mm/s1 \, \text{mm/s} and 2 cm2 \, \text{cm} into SI units. This gives the wrong emf by factors of 1010 or 10001000. Always use meters and seconds before substitution.

  • Differentiating πr2\pi r^2 incorrectly and writing ddt(r2)=2r\frac{d}{dt}(r^2) = 2r. Since rr changes with time, the correct derivative is ddt(r2)=2rdrdt\frac{d}{dt}(r^2) = 2r\frac{dr}{dt}.

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