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

Answer:

The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

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

$\phi=BA$

$\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2$

Put the value into the formula

$\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2$

$\phi=11.486\times10^{-8}\ Wb$

Hence, The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$