Answer:
The magnitude of the induced emf is 2.5 volts.
Explanation:
Given that,
The strength of the magnetic field, B = 10 T
Area of the wire, [tex]A=0.5\ m^2[/tex]
Number of turns in the conducting loop, N = 50
The magnetic field is linearly reduced to zero in 2 s. We need to find the magnitude of the induced emf voltage. Here, an emf will be induced due to the change in magnetic field. It is given by :
[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\\epsilon=-\dfrac{d(BA)}{dt}\\\\\epsilon=-A\dfrac{d(B)}{dt}\\\\\epsilon=-A\dfrac{B_f-B_i}{dt}\\\\\epsilon=-0.5\times \dfrac{0-10}{2}\\\\\epsilon=2.5\ V[/tex]
So, the magnitude of the induced emf is 2.5 volts. Hence, this is the required solution.
