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

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

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