Asolenoid 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 .

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

Explanation:

Given that,

Number of turns = 22 turns

Number of turns another coil = 330 turns

Length of solenoid = 21.0 cm

Diameter = 2.30 cm

Current in inner solenoid = 0.140 A

Rate = 1800 A/s

Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid

We need to calculate the magnetic flux

Using formula of magnetic flux

Put the value into the formula

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

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