Question

A 40 turn coil of wire of radius 3.0 cm is placed between the poles of an electromagnet. The field increase 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 the field is perpendicular to the plane of the coil? (2 points)

Answer 1

The induced emf in the coil is
`3.76* 10^(-4)\ V`.

Step-by-step explanation:

Given that,

Number of turns in the coil, N = 40

Radius of the wire, r = 3 cm = 0.03 m

The field increase from 0 to 0.75 T at a constant rate in a time interval of 225 s. The field is perpendicular to the plane of the coil. We need to find the magnitude of induced emf in the coil. It is given by :

`\epsilon=(-d\phi)/(dt)\\\\\epsilon=(-d(NBA))/(dt)\\\\\epsilon=NA(-dB)/(dt)\\\\\epsilon=40* \pi (0.03)^2* (0.75-0)/(225)\\\\\epsilon=3.76* 10^(-4)\ V`

So, the induced emf in the coil is
`3.76* 10^(-4)\ V`