Asolenoidal coil with 25 turns of wire is wound tightly around another coil with 300 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 0.120 a and is increasing at a rate of 1.75×103a/s. for this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

-0.020475 V

Explanation:

= Vacuum permeability =

= Number of turns of coil  = 25

= Number of turns of coil 2 = 300

= Rate of current increased =

d = Diameter = 2 cm

r = Radius =

A = Area =

Magnetic field in the solenoid is given by

Magnetic flux is given by

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

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 in the inner solenoid is -0.020475 V

0.863333

86%

hope this

i believe the answer would be:

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