Question

A solenoidal coil with 21 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.50 cm . At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of 1700 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;

Answer 1

A) We need to calculate the Magnetic flux of a Solenoid,

$\Phi = BA$

Where B is the magnetic field and A the Area.

$B=\frac{\mu_0 N_2 I}{l}$

Here \mu_0 is the permeability constant, I the current and N number of turns.

Replacing we have,

$\Phi = \frac{\mu_0 N_2 I}{l}(\pi r^2)$

$\Phi = \frac{(4\pi*10^{-7})(350)(0.12)}{25*10^{-2}} (\pi*\frac{2.5*10^{-2}}{2}^2)$

$\Phi = 1.036*10^{-7}Wb$

B) We calculate the mutual inductance, so

$M=\frac{N_1 \Phi}{i}$

$M=\frac{21*1.036*10^{-7}}{0.12}$

$M=1.813*10^{-5}H$

c) We calculate the emf

$\epsilon = -M \frac{dI}{dt}$

$\epsilon = -1.813*10^{-5}*1700$

$\epsilon= -0.030V$