Answer:
The induced emf in the coil is [tex]3.76\times 10^{-4}\ V[/tex].
Explanation:
Given that,
Number of turns in the coil, N = 40
Radius of the wire, r = 3 cm = 0.03 m
The field increase from 0 to 0.75 T at a constant rate in a time interval of 225 s. The field is perpendicular to the plane of the coil. We need to find the magnitude of induced emf in the coil. It is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=NA\dfrac{-dB}{dt}\\\\\epsilon=40\times \pi (0.03)^2\times \dfrac{0.75-0}{225}\\\\\epsilon=3.76\times 10^{-4}\ V[/tex]
So, the induced emf in the coil is [tex]3.76\times 10^{-4}\ V[/tex]
