Question

A solenoidal coil with 25 turns of wire is wound tightly aroundanother coil with 300 turns. The inner solenoid is 25.0 cm long andhas a diameter of 2.00 cm. At a certain time, the current in theinner solenoid is 0.120 A and is increasing at a rate of1.75x103A/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

Answer:

(a). The average magnetic flux through each turn of the inner solenoid is $5.68\times10^{-8}\ Wb$

(b). The mutual inductance of the two solenoids is $1.183\times10^{-5}\ H$

(c). The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.0207 V.

Explanation:

Given that,

Number of turns of coil = 25

Number of turns of another coil = 300

Length = 25.0 cm

Diameter = 2.00 cm

Current = 0.120 A

Rate $\dfrac{di_{2}}{dt}=1.75\times10^{3}\ A/s$

(a). We need to calculate the magnetic field due to inner solenoid

Using formula of magnetic field

$B=\mu_{0}(\dfrac{N_{2}}{l})I$

Put the value into the formula

$B=4\pi\times10^{-7}\times(\dfrac{300}{0.25})\times0.120$

$B=1.81\times10^{-4}\ T$

We need to calculate the average magnetic flux through each turn of the inner solenoid

Using formula of magnetic flux

$\phi=B\cdot A$

Put the value into the formula

$\phi=1.81\times10^{-4}\times\pi\times (1.00\times10^{-2})^2$

$\phi=5.68\times10^{-8}\ Wb$

The average magnetic flux through each turn of the inner solenoid is $5.68\times10^{-8}\ Wb$

(b). We need to calculate the mutual inductance of the two solenoids

Using formula of mutual inductance

$M=\dfrac{N_{1}\phi}{i_{1}}$

Put the value into the formula

$M=\dfrac{25\times5.68\times10^{-8}}{0.120}$

$M=0.00001183\ H$

$M=1.183\times10^{-5}\ H$

The mutual inductance of the two solenoids is $1.183\times10^{-5}\ H$

(c).  We need to calculate the emf induced in the outer solenoid by the changing current in the inner solenoid

Using formula of emf

$\epsilon=-M\dfrac{di_{2}}{dt}$

Put the value into the formula

$\epsilon=-1.183\times10^{-5}\times1.75\times10^{3}$

$\epsilon=-0.0207\ V$

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.0207 V.

Hence, (a). The average magnetic flux through each turn of the inner solenoid is $5.68\times10^{-8}\ Wb$

(b). The mutual inductance of the two solenoids is $1.183\times10^{-5}\ H$

(c). The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.0207 V.