Question

A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of 90°, so that the normal becomes perpendicular to the magnetic field. The coil has an area of 1.5 10-3 m2, 50 turns, and a resistance of 166 Ω. During the time while it is rotating, a charge of 7.3 10-5 C flows in the coil. What is the magnitude of the magnetic field?

Answer 1

Step-by-step explanation:

Expression for magnitude of the induced emf is as follows.

`\epsilon = N (BA)/(t)`

`(Q)/(t)R = (NBA)/(t)`

So, magnitude of the magnetic field is as follows.

B =
`(RQ)/(A * N)`

It is given that,

A =
`1.5 * 10^(-3) m^(2)`

Q =
`7.3 * 10^(-5) C`

N = 50

R = 166
`\ohm`

Putting the given values into the above formula as follows.

B =
`(RQ)/(A * N)`

=
`(166 * 7.3 * 10^(-5))/(1.5 *10^(-3) * 50)`

=
`(1211.8 * 10^(-5))/(75 * 10^(-3))`

=
`16.157 * 10^(-2)`

= 0.1615 T

Thus, we can conclude that magnitude of the magnetic field is 0.1615 T.