Question

The magnetic flux linking each loop of a 250-turn coil is given by φB (t) = A
+ Dt2
, where A = 3 Wb and D = 15 Wbs–2 are constants. Show that a) the
magnitude of the induced emf in the coil is given by ε = (2ND)t, and b)
evaluate the emf induced in the coil at t = 0s and t = 3.0s.

Answer 1

induced emf is the negative of rate of change of magnetic flux with respect to time.

i.e., ε = -N(dΦ/dt)

where N is number of turns of coil and (dΦ/dt) is the rate of change of magnetic flux.

here given, magnetic flux, Φ = A + Dt²

differentiating with respect to time,

dΦ/dt = 0 + 2Dt = 2Dt

so, ε = -N(dΦ/dt)

= -2NDt

so, magnitude of induced emf , ε = 2NDt

hence proved

(b) induced emf at t = 0s, ε = 2ND × 0 =0

induced emf at t = 3s , ε = 2 × 250 × 15 × 3 = 22500 Volts