e = -L di/dt. During 0 to T/4, di/dt = const. ∴ e = -ve. During T/4 to T/2, di/dt = 0 ∴ e = 0. During T/2 to 3T/4, di/dt = const. ∴ e = +ve. Thus graph given in option (a) represents the variation of induced e.m.f. with time.

Question

Accepted Answer

Graph in option (a) represents the variation of induced e.m.f. with time.. The induced emf is proportional to the rate of change of current (di/dt). During 0 to T/4, di/dt is constant and negative, so e is negative. During T/4 to T/2, di/dt is zero, so e is zero. During T/2 to 3T/4, di/dt is constant and positive, so e is positive. This matches the graph in option (a).

e = -L di/dt. During 0 to T/4, di/dt = const. ∴ e = -ve. During T/4 to T/2, di/dt = 0 ∴ e = 0. During T/2 to 3T/4, di/dt = const. ∴ e = +ve. Thus graph given in option (a) represents the variation of induced e.m.f. with time.

Solution

Related NEET Physics questions