Asolenoidal coil with 22 turns of wire is wound tightly around another coil with 330 turns. the inner solenoid is 21.0 cm long and has a diameter of 2.30 cm . at a certain time, the current in the inner solenoid is 0.140 a and is increasing at a rate of 1800 a/s .
for this time, calculate;

a) the average magnetic flux through each turn of the innersolenoid;
b) the mutual inductance of the two solenoids;
c) the emf induced in the outer solenoid by the changing current inthe inner solenoid
for any you can offer! explanation would be wonderful!

-0.04194 V

Explanation:

= Number of turns in outer solenoid = 330

= Number of turns in inner solenoid = 22

= Current in inner solenoid = 0.14 A

= Rate of change of current = 1800 A/s

= Vacuum permeability =

r = Radius = 0.0115 m

Magnetic field is given by

The  average magnetic flux through each turn of the inner solenoid is

Magnetic flux is given by

Mutual inductance is given by

The mutual inductance of the two solenoids is

Induced emf is given by

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V

volts times amps 6 x 0.5 3w