Question

A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.250 m2. The coil is now stretched to have no area in 0.100 s. What is the magnitude and direction of the average value of the induced emf if the uniform magnetic field points into the page and has a strength of 1.36 T

Answer 1

Answer:

The induced emf in the coil is 170 volt and its direction is clockwise direction.

Explanation:

Given that,

Number of turns of a coil, N = 50

Initial area, $A_i=0.25\ m^2$

Final area, $A_f=0$

Time, t = 0.1 s

The magnetic field points into the page and has a strength of 1.36 T.

We need to find the magnitude and direction of the average value of the induced emf. Due to change in area of the coil and emf will be induced in it. The induced emf is given by :

$\epsilon=-N\dfrac{d\phi}{dt}\\\\\text{since}\ \phi=BA\\\\\epsilon=-N\dfrac{d(BA)}{dt}\\\\\epsilon=-NB\dfrac{A_f-A_i}{dt}\\\\\epsilon=-50\times 1.36\dfrac{0-0.25}{0.1}\\\\\epsilon=170\ V$

So, the induced emf in the coil is 170 volt and its direction is clockwise direction. Hence, this is the required solution.