Question

A coil consists of 200 turns of wire. Each turn is a square of side d = 18 cm, and a uniform magnetic field directed perpendicular to the plane of the coil is turned on. If the field changes linearly from 0 to 0.50 T in 0.80 s, what is the magnitude of the induced emf in the coil while the field is changing?

Answer 1

Answer:

Induced emf in the coil, $\epsilon=4.05\ volts$

Explanation:

Given that.

Number of turns in the coil, N = 200

Side of square, d = 18 cm = 0.18 m

The field changes linearly from 0 to 0.50 T in 0.80 s.

To find,

The magnitude of the induced emf in the coil while the field is changing.

Solution,

We know that due to change in the magnetic field, an emf gets induced in the coil. The formula of induced emf is given by :

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

$\phi$ = magnetic flux

$\epsilon=-\dfrac{d(NBA)}{dt}$

A is the ares of square

$\epsilon=AN\dfrac{d(B)}{dt}$

$\epsilon=AN\dfrac{B_f-B_i}{t}$

$\epsilon=(0.18)^2\times 200 \times \dfrac{0.5-0}{0.8}$

$\epsilon=4.05\ volts$

So, the induced emf in the coil is 4.05 volts. Hence, this is the required solution.