A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field T. When released the radius of the loop starts shrinking at a constant rate of cms. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be _______ mV.
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Answer: 10
Solution
EMF = |dΦ/dt| = B · d(πr²)/dt = B · 2πr · (dr/dt)
= 0.8 × 2π × 0.10 × 0.02 = 0.8 × 2π × 0.002 = 0.8 × 0.004π = 0.003217... ≈ 3.2π mV
Wait, let me recalculate: 0.8 × 2π × 0.10 × 0.02 = 0.8 × 0.004π = 0.0032π ≈ 10.05 mV ≈ 10 mV.
The induced EMF is |dΦ/dt| = B·2πr·(dr/dt). Substituting B = 0.8 T, r = 0.10 m, dr/dt = 0.02 m/s gives EMF = 0.8 × 2π × 0.10 × 0.02 = 0.0032π ≈ 10.05 mV ≈ 10 mV.
= 0.8 × 2π × 0.10 × 0.02 = 0.8 × 2π × 0.002 = 0.8 × 0.004π = 0.003217... ≈ 3.2π mV
Wait, let me recalculate: 0.8 × 2π × 0.10 × 0.02 = 0.8 × 0.004π = 0.0032π ≈ 10.05 mV ≈ 10 mV.
The induced EMF is |dΦ/dt| = B·2πr·(dr/dt). Substituting B = 0.8 T, r = 0.10 m, dr/dt = 0.02 m/s gives EMF = 0.8 × 2π × 0.10 × 0.02 = 0.0032π ≈ 10.05 mV ≈ 10 mV.
[JEE Main 2023 · 29 Jan · Shift 1]