Question

A 40.0 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s. What is the magnitude of the induced emf in the coil if (a) the field is perpendicular to the plane of the coil

Answer 1

Answer:

Induced emf, $V=3.76\times 10^{-4}\ V$

Explanation:

We have,

Number of turns in the coil, N = 40

Radius of coil, r = 3 cm = 0.03 m

The field increases from 0 to 0.75 T at a constant rate in a time interval of 225 s.

It is required to find the magnitude of the induced emf in the coil if the field is perpendicular to the plane of the coil. The induced emf is given by :

$\epsilon=\dfrac{d\phi}{dt}$

$\phi=NBA\cos\theta$, is magnetic flux

$\epsilon=NA\dfrac{dB}{dt}\\\\\epsilon=\dfrac{40\times \pi (0.03)^2\times (0.75-0)}{225}\\\\\epsilon=3.76\times 10^{-4}\ V$

So, the magnitude of induced emf is $3.76\times 10^{-4}\ V$.