Question

A solenoid has a diameter d = 1.5 cm, length l = 30.0 cm, and 440 turns.

(a) Find the inductance of the solenoid.
Inductance = .143 mH

(b) If the current through the solenoid increases at a rate 730 A/s, what is the magnitude of the induced emf?
Induced emf = .105 V

(c) What is the magnetic flux through each turn of the solenoid when the current through it is 5.9 A?
Magnetic flux = _____T·m^2

I just need part three. Thanks!

3.26e-7 is not the correct answer.

Answer 1

magnetic flux= inductance * current
inductance=.143 mH for 440 turns
inductance per turn=.143 mH/440=3.25 * 10 ^-7 H

so flux per turn= 3.25 * 10 ^ -7 * 5.9= 1.9175 * 10 ^-6

Answer 2

B = u*n*I*A
by putting values

2.87e-4

Answer 3

sol:

You do the magnetic field inside a solenoid * area so B*A --> B = u*n*I*A --> 4*pi*10e-7(440/.30)[pi(.015/2)^2]. I get 3.26e-7 T·m^2