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 T. When released the radius of the loop starts shrinking at a constant rate of 2 cms⁻¹. The induced emf in the loop at an instant when the radius of the loop is 10 cm will be __________ mV.

Solution

Physics sir · 3 steps

Field is steady, area is shrinking: emf is B times the rate at which the loop's area changes.

1. Flux through loop. The field is perpendicular to the loop, so the flux is φ = Bπ r². Only r changes with time, because B stays constant.

   |ε| = dφ/dt = B · 2π r dr/dt

2. Put in numbers. Use SI units: r = 0.1 m and dr/dt = 0.02 m s⁻¹. Then |ε| = 0.8 × 2π × 0.1 × 0.02 = 1.005 × 10⁻² V.
3. So. |ε| ≈ 10.05 mV, which rounds to 10.

The common slip: 5. Dropping the 2 when differentiating r² gives about 5 mV, not 10. The rate of change of π r² is 2π r dr/dt.

The idea to keep. When a loop changes size in a steady field, write φ = BA and differentiate the area. For a circle, dA/dt = 2π r dr/dt, the circumference times the rate of change of radius.

Answer from the NTA final answer key. Solution by PrepWiser.