Jitender Pal
Last Activity: 11 Years ago
Sol. (a) During removal,
ϕ base 1 = B.A. = 1 x 50 x 0.5 x 0.5 – 25 x 0.5 = 12.5 Tesla-m^2
ϕ base 2 = 0, t = 0.25
e = - dϕ/dt = ϕ base 2 – ϕ base 1/dt = 12.5 /0.25 = 125x 10^-1 / 25x 10^-2 = 50V
(b) During its restoration
ϕ base 1 = 0 ; ϕ base 2 = 12.55 Tesla- m^22 ; t = 0.25 s
E = 12.5 -0 /0.25 = 50 V.
(c) During the motion
Φ base 1 = 0, ϕ base 2 = 0
E dϕ/dt = 0