A solenoidal coil with 25 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 345 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.108 A and is increasing at a rate of 1.90 times 10^3 A/s. For this time, calculate the average magnetic flux through each turn of the inner solenoid. _________________ Wb For this time, calculate the mutual inductance of the two solenoids. ________________H For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid. _________________V

Explanation / Answer

no turns of solenoid N1=25

no turns of inner coil N2=345

length of the inner solenoid l=25cm

diameter, d=2cm

current I=0.108 A

rate of increasing current, dI/dt=1.9*10^3 A/sec

a)

magnetic flux = B*A

=uo*(N2/l)*I*(pi*r^2)

=4pi*10^-7*(345/0.25)*(pi*0.01)^2

=1.71*10^-6 Wb

b)

mutual inductance, M=N1*flux/I

=25*1.71*10^-6/(0.108)

=3.96*10^-4 H

c)

induced emf( in outer solenoid) =M*di/dt

=3.96*10^-4*(1.9*10^3)

=0.75 v

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