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

Answer:

Explanation:

From the given information:

(a)

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

$\phi _ 1 = B_1 A$

$\phi _ 1 = ( \mu_o \dfrac{N_i}{l} i_1)(\pi ( \dfrac{d}{2})^2)$

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

$\phi_1 = 9.625 * 10^{-8} \ Wb$

(b)

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

$M = 23 *\dfrac{9.625*10^{-8} \ Wb}{0.130 \ A}$

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:

$\varepsilon = -N_o \dfrac{d \phi_1}{dt}$

$\varepsilon = -M \dfrac{d i_1}{dt}$

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

$\varepsilon = -0.030654 \ V$

$\varepsilon = -30.65 \ V$