Question

Suppose a 52-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.275 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.20 T

Answer 1

Explanation:

Given that,

Number of turns in the coil, N = 52

Original are of the coil, $A_1=0.275\ m^2$

It is stretched to have no area in 0.100 s, t = 0.1 s

The magnetic field strength, B = 1.2 T

To find,

The magnitude and the direction of average induced emf.

Solution,

The Faraday law gives the magnitude and the direction of induced EMF in the coil. Its formula is given by :

$E=-\dfrac{Nd\phi}{dt}$

$E=-\dfrac{Nd(BA)}{dt}$

$E=-NB\dfrac{d(A)}{dt}$

$E=-52\times 1.2\times \dfrac{(0-0.275)}{0.1}$

E = 171.6 volts

The direction of induced emf is given by Lenz law. It opposes the direction due to which it is produced. It is in counterclockwise direction.