Question

A 450-turn solenoid, 26 cm long, has a diameter of 2.1 cm . A 12-turn coil is wound tightly around the center of the solenoid.

If the current in the solenoid increases uniformly from 0 to 4.7 A in 0.57 s , what will be the induced emf in the short coil during this time?

Answer 1

Given that,

No of turns of solenoid = N1 = 450 ; Length = 26 cm = 0.26 m

diameter = 2.1 cm so radius = r = 2.1/2 = 1.05 cm = 0.0105 m

So area A = pi r² = 3.46 x 10^-4 m²

No of turns of coil = N2 = 12 ; Io = 0 and I1 = 4.7 A in time t = 0.57 s.

We need to find the induced emf in the short coil. Let it be .

We know that, the B field at the center of a long solenoid is given by:

B = o N I / L (N- no of turns, I is current and L is length)

B1 =  o N1 I / L (Bo is 0 as current is 0)

And flux is given by

= B A (B - magnetic field and A is Area)

So, o = 0 (as current is o intially as per the question)

1 = B1 A

From Faraday's Law of induction we know that, the induced emf is equal to the rate of change of magnetic flux. and if there are N numbers of turns then it will be N times the rate of change of magnetic flux. So

= - N d/dt =

= -N2 [o - 1]/[t - 0] = N2 x  o N1 I A / L x t =  o N1 N2 I A / L x t

= 4 pi x 10^-7 x 450 x 12 x 4.7 x 3.46 x 10^-4 / 0.26 x 0.57

=1102953.888 x 10^-11 / 0.1482 = 7442334 x 10^-11 =7.4 x 10^-5 Volts

Hence, induced emf in the sort coil = = 7.4 x 10^-5 Volts