Question

A solenoidal coil with 23 turns of wire is wound tightly around another coil with 310 turns. The inner solenoid is 20.0 cm long and has a diameter of 2.20 cm. At a certain time, the current in the inner solenoid is 0.130 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 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.

Answer 1

Step-by-step explanation:

From the given information:

(a)

the average magnetic flux through each turn of the inner solenoid can be calculated by the formula:

`\phi _ 1 = B_1 A`

`\phi _ 1 = ( \mu_o (N_i)/(l) i_1)(\pi ( (d)/(2))^2)`

`\phi _ 1 = ( 4 \pi *10^(-7) \ T. m/A ) ( (310)/(20*10^(-2) \ m )) (0.130 \ A) ( \pi ( (2.20*10^(-2) \ m )/(2))^ 2`

`\phi_1 = 9.625 * 10^(-8) \ Wb`

(b)

The mutual inductance of the two solenoids is calculated by the formula:

`M = 23 *(9.625*10^(-8) \ Wb)/(0.130 \ A)`

M =
`1.703 *10^(-5)` H

(c)

the emf induced in the outer solenoid by the changing current in the inner solenoid can be calculate by using the formula:

`\varepsilon = -N_o (d \phi_1)/(dt)`

`\varepsilon = -M (d i_1)/(dt)`

`\varepsilon = -(1.703*10^(-5) \ H) * (1800 \ A/s)`

`\varepsilon = -0.030654 \ V`

`\varepsilon = -30.65 \ V`