Question

A 500-turn coil of wire 2. 0 cm in diameter is in a magnetic field that increases from 0. 00 T to 0. 30 T in 20 ms. The axis of the coil is parallel to the field. What is the emf of the coil?

Answer 1

The magnetic flux through the coil is given by:

Φ = BAcosθ

where B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil. Since the axis of the coil is parallel to the field, θ = 0 and the equation simplifies to:

Φ = BA

The rate of change of magnetic flux is given by:

dΦ/dt = d(BA)/dt = A(dB/dt)

Plugging in the given values, we get:

A = πr^2 = π(1.0 cm)^2 = 3.14 × 10^-4 m^2

dB/dt = (0.30 T - 0.00 T)/(20 ms) = 1.5 × 10^4 T/s

Therefore, the emf induced in the coil is:

emf = -N(dΦ/dt) = -500 × 3.14 × 10^-4 m^2 × 1.5 × 10^4 T/s = -0.235 V

Note that the negative sign indicates that the emf is induced in a direction that opposes the change in magnetic flux.