Answer:
(a) emf = 0.507 V
(b) emf = 0.0507 V
(c) emf = 0.00234 V
Explanation:
Given;
number of turns of the coil, N = 40 turns
diameter of the coil, d = 11 cm
radius of the coil, r = 5.5 cm = 0.055 m
magnitude of the magnetic field, B = 0.4 T
The magnitude of the induced emf is calculated as;
[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]
(a) when the time, t = 0.3 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]
(b) when the time, t = 3.0 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]
(c) when the time, t = 65 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]
