Question

A field of 0.0125 T is at right angles to a coil of
area 5×10-3 m2 with 1000 turns. It is removed from the field in 1/20 s. Find the e.m.f. produced.

Answer 1

Answer:

hope the answer will help you ☺❤

Answer 2

Answer:

The emf produced in the coil is 1.25V.

Explanation:

Given the magnetic field, B = 0.0125 T

Area of the coil, $A= 5\times 10^{-3} m^{2}$

Number of turns in the coil, N = 1000

Time, $t=\frac{1}{20} =0.05s$

According to Faraday's law, any change in magnetic flux $\phi$ induces an electromotive force(emf) in a closed conducting loop.

In a loop consisting of N turns, the magnitude of emf induced is equal to the rate of change of flux, i.e.,

$\mathcal{E} = N\dfrac{d\phi_{B}}{dt}$                         ...(1)

And change in magnetic flux depends on the varying magnetic field and the area it exists.

Magnetic flux $\phi$ is given by product of the area and the component of the magnetic field perpendicular to it. Thus

$\phi_{B} =B_{\perp}A=BA$                    ...(2)

Combining both the equations, the emf in a coil can be written as

$\mathcal{E} = \dfrac{NBA}{t}$

Substituting the given values,

$\mathcal{E} = \dfrac{1000\times0.0125\times5\times10^{-3} }{0.05}$

$= \dfrac{0.0125\times5 }{0.05}$

$=1.25V$

Therefore, the emf produced in the coil is 1.25V.