Question

A flat coil is wrapped with 200 turns of very thin wire on a square frame with sides 18 cm long. A uniform magnetic field is applied perpendicular to the plane of the coil. If the field changes uniformly from 0.50 T to 0.00 T in 8.0 s, find the emf induced in the coil.

Answer 1

Answer:

0.405 V

Explanation:

Using Faraday-Lenz law, the emf induced in the coil is given by

$\epsilon=-\frac{\Delta \Phi}{\Delta t}$

where

$\Delta \Phi$ is the change in flux linkage through the coil

$\Delta t = 8.0 s$

is the time interval

The change in flux linkage can be written as

$\Delta \phi = NA(B_f - B_i)$

where

N = 200 is the number of turns

$A=(18 cm)^2 = (0.18 m)^2=0.0324 m^2$ is the area of the squared loop

Bi = 0.50 T is the initial magnetic flux density

Bf = 0.00 T is the final magnetic flux density

Substituting everything into the first equation, we find

$\epsilon=-\frac{(200)(0.0324)(0.00-0.50}{8.0}=0.405 V$