Question

A solenoidal 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 .

For this time, calculate;

a) the average magnetic flux through each turn of the innersolenoid;
b) the mutual inductance of the two solenoids;
c) the emf induced in the outer solenoid by the changing current inthe inner solenoid
Thanks for any help you can offer! Explanation would be wonderful!

Answer 1

`0.00027646\ T`

`2.33* 10^(-5)\ H`

-0.04194 V

Step-by-step explanation:

`N_2` = Number of turns in outer solenoid = 330

`N_1` = Number of turns in inner solenoid = 22

`I_1` = Current in inner solenoid = 0.14 A

`(dI_2)/(dt)` = Rate of change of current = 1800 A/s

`\mu_0` = Vacuum permeability =
`4\pi * 10^(-7)\ H/m`

r = Radius = 0.0115 m

Magnetic field is given by

`B=\mu_0(N_2)/(l)I\\\Rightarrow B=4\pi * 10^(-7)* (330)/(0.21)* 0.14\\\Rightarrow B=0.00027646\ T`

The average magnetic flux through each turn of the inner solenoid is
`0.00027646\ T`

Magnetic flux is given by

`\phi=BA\\\Rightarrow \phi=0.00027646* \pi 0.0115^2\\\Rightarrow \phi=1.14862* 10^(-7)\ wb`

Mutual inductance is given by

`M=(N_1\phi)/(I)\\\Rightarrow M=(22* 1.14862* 10^(-7))/(0.14)\\\Rightarrow M=2.33* 10^(-5)\ H`

The mutual inductance of the two solenoids is
`2.33* 10^(-5)\ H`

Induced emf is given by

`\epsilon=-M(dI_2)/(dt)\\\Rightarrow \epsilon=-2.33* 10^(-5)* 1800\\\Rightarrow \epsilon=-0.04194\ V`

The emf induced in the outer solenoid by the changing current inthe inner solenoid is -0.04194 V