A coil is places perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2 s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is:

Solution

Physics sir · 4 steps

Find the rate of change of B, multiply by the area of one turn, and the emf tells you N.

1. Faraday's law. The coil is perpendicular to the field, so the flux through each turn is BA. With N turns, the size of the induced emf is the number of turns times the rate of change of flux through one turn.

   ε = N A (Δ B)/(Δ t)

2. Rate and area. (Δ B)/(Δ t) = (5000 - 3000)/2 = 1000 T s⁻¹. The radius is half the diameter, r = 0.01 m, so A = π r² = π × 10⁻⁴ m².
3. Solve for N. N = 22/(π × 10⁻⁴ × 1000) = 22/0.1π. With π = 22/7, this is N = (22 × 7)/2.2 = 70.
4. So. The coil has 70 turns, which is option (b).

The idea to keep. Induced emf is N times the rate of change of flux. Divide the change in B by the time it takes, and use the radius, not the diameter, when you find the area.

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