Question

A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.140 m2. The coil is now stretched to have no area in 0.100 s. What are 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 1.25 T?

Answer 1

Answer:

Magnitude of induced emf will be 87.5 volt

Direction of induced emf will be clockwise  

Explanation:

We have given number of turns in the coil N = 50

Initial area $A=0.140m^2$

Time is given dt = 0.1 sec

Magnetic field B = 1.25 T

From Faraday's law of electromagnetic induction

n$e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}$

$e=-50\times 0.140\times \frac{1.25}{0.1}=-87.5volt$

So magnitude of induced emf is 87.5 volt

Direction of induced emf will be clockwise