Question

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of 1.75×103A/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:

$5.65487\times 10^{-8}\ Wb$

$1.17\times 10^{-5}\ H$

-0.020475 V

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

$N_1$ = Number of turns of coil  = 25

$N_2$ = Number of turns of coil 2 = 300

$\frac{di_2}{dt}$ = Rate of current increased = $1.75\times 10^3\ A/s$

d = Diameter = 2 cm

r = Radius = $\frac{d}{2}=\frac{2}{2}=1\ cm$

A = Area = $\pi r^2$

Magnetic field in the solenoid is given by

$B=\mu_0\frac{N_2}{l}I\\\Rightarrow B=4\pi\times 10^{-7}\frac{300}{0.25}\times 0.12\\\Rightarrow B=0.00018\ T$

Magnetic flux is given by

$\phi=BA\\\Rightarrow \phi=0.00018\times \pi\times 0.01^2\\\Rightarrow \phi=5.65487\times 10^{-8}\ Wb$

The average magnetic flux through each turn of the inner solenoid is $5.65487\times 10^{-8}\ Wb$

Mutual inductance is given by

$L=\frac{N_1\phi}{i_1}\\\Rightarrow L=\frac{25\times 5.65487\times 10^{-8}}{0.12}\\\Rightarrow L=1.17\times 10^{-5}\ H$

The mutual inductance of the two solenoids is $1.17\times 10^{-5}\ H$

Induced emf is given by

$V=-L\frac{di_2}{dt}\\\Rightarrow V=-1.17\times 10^{-5}\times 1.75\times 10^3\\\Rightarrow V=-0.020475\ V$

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.020475 V