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NVAEasyJEE 2023Faraday's Laws of EMI

JEE Physics 2023 Question with Solution

A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field B=0.8 TB = 0.8 \, \text{T}. When released, the radius of the loop starts shrinking at a constant rate of 2 cm/s2 \, \text{cm/s}. The induced emf in the loop at an instant when the radius of the loop is 10 cm10 \, \text{cm} will be — mV.

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: magnetic field B=0.8 TB = 0.8 \, \text{T}, radius r=10 cm=0.1 mr = 10 \, \text{cm} = 0.1 \, \text{m}, and shrinking rate drdt=−2 cm/s=−0.02 m/s\frac{dr}{dt} = -2 \, \text{cm/s} = -0.02 \, \text{m/s}.

Find: the induced emf in the loop.

Magnetic flux through the circular loop is

Φ=B⋅A=Bπr2\Phi = B \cdot A = B\pi r^2

Using Faraday's law,

E=−dΦdtE = -\frac{d\Phi}{dt}

So,

E=−ddt(Bπr2)=−B⋅2πr(drdt)E = -\frac{d}{dt}(B\pi r^2) = -B \cdot 2\pi r \left(\frac{dr}{dt}\right)

Substituting the values,

E=−0.8⋅2π⋅0.1⋅(−0.02)E = -0.8 \cdot 2\pi \cdot 0.1 \cdot (-0.02) E≈0.010 VE \approx 0.010 \, \text{V}

Converting to millivolt,

E=10 mVE = 10 \, \text{mV}

Therefore, the induced emf is 10 mV10 \, \text{mV}.

Flux Differentiation Details

Given: the area of the loop changes because the radius decreases uniformly.

Find: emf from the rate of change of magnetic flux.

Since the plane of the loop is perpendicular to the magnetic field, the angle between area vector and field is 0∘0^\circ. Hence flux is maximum and equals

Φ=BA=Bπr2\Phi = BA = B\pi r^2

Differentiate carefully with respect to time:

dΦdt=Bπddt(r2)=Bπ⋅2rdrdt\frac{d\Phi}{dt} = B\pi \frac{d}{dt}(r^2) = B\pi \cdot 2r\frac{dr}{dt}

Therefore,

E=−2πBrdrdtE = -2\pi Br\frac{dr}{dt}

Because the radius is shrinking, drdt<0\frac{dr}{dt} < 0. This makes the emf magnitude positive:

∣E∣=2πBr∣drdt∣|E| = 2\pi B r \left|\frac{dr}{dt}\right| =2π⋅0.8⋅0.1⋅0.02≈0.010 V= 2\pi \cdot 0.8 \cdot 0.1 \cdot 0.02 \approx 0.010 \, \text{V}

Thus the numerical value asked is 10.

Common mistakes

  • Using diameter instead of radius in the area formula. The area of a circle is πr2\pi r^2, not πd2\pi d^2. Always substitute the given radius directly.

  • Forgetting to convert 10 cm10 \, \text{cm} and 2 cm/s2 \, \text{cm/s} into SI units. The formula is easiest to use in metres and metres per second, so take 0.1 m0.1 \, \text{m} and 0.02 m/s0.02 \, \text{m/s}.

  • Ignoring the chain rule while differentiating r2r^2. Since rr changes with time, ddt(r2)=2rdrdt\frac{d}{dt}(r^2) = 2r\frac{dr}{dt}, not just 2r2r.

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