Question

A 500-turn coil with an average radius of 0.06 m is placed in a uniform magnetic field so that o=40 degrees. The field increases at a rate of 0.160 T/s. What is the magnitude of the resulting emf?O0.693VO0.0014VO0.905VO0.116V

Answer 1

Given:

no. of turn in coil is

$n=500$

radius of the coil is

$r=0.06\text{ m}$

rate of change of magnetic field is

$\frac{dB}{dt}=0.160\text{ T/s}$

coil makes an angle with the magnetic field is

$\theta=40^{\circ}$

Required: magnitude of the resulting emf in the coil

Explanation:

emf in the any coil is given by

$\epsilon=-nA\frac{dB}{dt}\cos40^{\circ}$

plugging all the values in the above relation, we get

$\begin{gathered} \epsilon=500\times3.14\times(0.06\text{ m})^2\times0.160\text{ }\times0.766 \\ \epsilon=0.693\text{ V} \end{gathered}$

Thus, the emf in the coil is

$0.693\text{ V}$