StreakPeaked· Practice

ExamsJEE MainPhysics

The radius of a single-turn circular coil shrinks steadily at 10⁻² m/s in a constant uniform magnetic field of 10⁻³ Wb/m² directed perpendicular to the coil's plane. Find the radius of the coil at the instant the induced emf is 1 microvolt.

  1. 5/pi cm
  2. 2/pi cm
  3. 3/pi cm
  4. 4/pi cm

Correct answer: 5/pi cm

Solution

Flux phi = B*pi*r². Magnitude of induced emf = B*2*pi*r*|dr/dt|. Setting this equal to 1e-6 V: r = emf/(B*2*pi*|dr/dt|) = 1e-6/(1e-3 * 2*pi * 1e-2) = 1e-6/(2*pi*1e-5) = 0.05/pi m = 5/pi cm.

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →