A solenoidal coil with 21 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 21 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 23.0 cm long and has a diameter of 2.10 cm. At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1600 A/s. For this time, calculate the average magnetic flux through each turn of the inner solenoid. For this time, calculate the mutual inductance of the two solenoids. For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Explanation / Answer

N! = no of turns of outer solenoid = 21

N2 = no of turns of inner solenoid = 330

L = length  of inner solenoid = 23 cm

d = diameter  of inner solenoid = 2.10 cm

di/dt = rate of change of current  of inner solenoid = 1600 A/s

I = current  of inner solenoid = 0.100 A

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1) average magnetic flux  of inner solenoid = B.A

( B = magnetic field = u0 N2 *I / L, A ( area) = pi*r2 = pi ( d/2)2 )

average magnetic flux  of inner solenoid    = (  u0 N2 *I / L) * (  pi ( d/2)2 )

= (4*pi*10-7* 330 * 0.100 / 23*10-2) * ( pi* ( 2.10*10-2 / 2)2 )

= 6.23 * 10-8 Wb [ pi = 3.14]

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2) M = mutual inductance of two solenoids

= u0*N1*N2 *A / L

= 4*pi*10-7* 21*330 * pi* ( 2.10*10-2/ 2) / 23*10-2

= 1.31*10-5 H

---------------------------------------------------------------------

3)emf in outer solenoid

E = - M * dI / dt

= - 1.31*10-5 *1600 = - 0.02096 V

  

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