Answer:
The magnitude of the induced emf in the coil is 6.13 V.
Explanation:
Given that,
Number of turns in a coil, N = 140
Side of a square frame, s = 35 cm = 0.35 m
A uniform magnetic field is turned on perpendicular to the plane of the coil. The the field changes linearly from 0 to [tex]0.426\ Wb/m^2[/tex] in a time of 1.19 s.
We need to find the magnitude of the induced emf in the coil while the field is changing. Due to change in field, an emf is induced in it. It is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\\epsilon=-\dfrac{d(NBA)}{dt}\\\\\epsilon=-NA\dfrac{dB}{dt}[/tex]
A is area of square frame, [tex]A=s^2=(0.35)^2=0.1225\ m^2[/tex]
[tex]\epsilon=-NA\dfrac{dB}{dt}\\\\\epsilon=-140\times 0.1225\times \dfrac{(0.426-0)}{1.19}\\\\\epsilon=-6.13\ V[/tex]
So, the magnitude of the induced emf in the coil is 6.13 V.
