Question

A technician wraps wire around a tube of length 34 cm having a diameter of 7.8 cm. When the windings are evenly spread over the full length of the tube, the result is a solenoid containing 575 turns of wire.

(a) Find the self-inductance of this solenoid.
________ mH

(b) If the current in this solenoid increases at the rate of 2 A/s, what is the self-induced emf in the solenoid?

________ mv

I got 6.75 for (a) but it was wrong and I can't tell what I'm doing wrong.

Answer 1

length of the tube l=34 cm

diameter d=7.8cm

radius r=3.9 cm

no of turns N=575

a)

self inductance is,

L=uo*n^2*A*l

L=uo*(N/l)^2*A*l

L=4pi*10^-7*(575/0.34)^2*pi*0.039^2*0.34

L=5.84*10^-3 H

L=5.84 mH

b)

given thet, di/dt=2 A/sec

induced emf E=L*di/dt

E=5.84*10^-3*2

E=11.68*10^-3 v

E=11.68 mv