Question

A solenoid 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 .

Answer 1

The average magnetic flux through each turn of the inner solenoid is
`11.486*10^(-8)\ Wb`

Step-by-step 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

`\phi=BA`

`\phi=(\mu_(0)N_(2)I)/(l)*\pi r^2`

Put the value into the formula

`\phi=(4\pi*10^(-7)*330*0.140)/(21.0*10^(-2))*\pi*((2.30*10^(-2))/(2))^2`

`\phi=11.486*10^(-8)\ Wb`

Hence, The average magnetic flux through each turn of the inner solenoid is
`11.486*10^(-8)\ Wb`