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.
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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
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