Question

A long solenoid of the radius \[2cm\] has $100turns/cm$ and carries a current of $5A$ . A coil of radius $1cm$ having $100turns$ and a total resistance of $20\Omega $ is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown in the galvanometer.
(A) $2 \times {10^{ - 4}}C$
(B) $1 \times {10^{ - 4}}C$
(C) $4 \times {10^{ - 4}}C$
(D) $8 \times {10^{ - 4}}C$

Answer 1

Hint Find the magnetic field through the solenoid using the direct equation. Find the change in flux through the coil inside in forward and reverse current. Using this flux, compute the induced electromotive force using Faraday's law and then use Ohm’s law to compute the current. Use the relation between current and charge to finally find the value of charge flowing through the galvanometer.

Complete step by step answer
Let the radius of the solenoid be denoted by $r$ . Let the number of turns per centimeter for the outer coil be denoted by ${n_1}$ . Let the number of turns per centimeter of the inner coil be denoted by ${n_2}$ . Let the flux related to the outer coil be denoted by $\phi $ . Let the current flowing through the solenoid be denoted by $I$ . Then the magnetic field $B$ around the solenoid will be given by
$B = {\mu _o}nI$
Here, ${\mu _o}$ is the permeability of free space and is given by $4\pi \times {10^{ - 7}}$ $N/{A^2}$ .
The number of turns per centimeter is given as $100turns/cm$or $10000turns/m$ and the current flowing through the solenoid is given as $5A$ .
By substituting the values of permeability of free space, number of turns per centimeter in the solenoid, and the current flowing through the solenoid, we get
$B = 4\pi \times {10^{ - 7}} \times 10000 \times 5$
Simplifying the above equation, we get the value of the magnetic field as
$B = 2\pi \times {10^{ - 2}}T$
The flux on the inner coil that has radius value $1cm$ or $0.01m$ is given by
$\phi = - {n_2}B\pi {r^2}$
If the direction of the current is reversed, then the flux related to the inner coil will be given by
$\phi ' = - {n_2}B\pi {r^2}$
Therefore, the change of flux through the inner coil is given by
$\Delta \phi = \phi - \phi '$
Therefore we get
$\Delta \phi = 2 \times {n_2}B\pi {r^2}$
Substituting the values, we get
$\Delta \phi = 2 \times 100 \times 2\pi \times {10^{ - 2}} \times \pi \times {10^{ - 4}}$
$ \Rightarrow \Delta \phi = 4{\pi ^2} \times {10^{ - 4}}T{m^2}$
Now, the induced emf is given by
$e = \dfrac{{\Delta \phi }}{{\Delta t}}$
By substituting the values, we get
$e = \dfrac{{4{\pi ^2} \times {{10}^{ - 4}}}}{{\Delta t}}$
Therefore, induced current can be computed as
$I = \dfrac{e}{R}$
Here $R$ is the resistance of the coil, and substituting its value in the above equation, we get
$I = \dfrac{{4{\pi ^2} \times {{10}^{ - 4}}}}{{\Delta t}} \times \dfrac{1}{{20}}$
But as charge can be computed as
$Q = I\Delta t$
We can write the above equation for current as
$Q = \dfrac{{4{\pi ^2} \times {{10}^{ - 4}}}}{{20}}$
$ \Rightarrow Q = 1.9719 \times {10^{ - 4}}C$
This value can be approximated to $2 \times {10^{ - 4}}C$ .

Hence, option (A) is the correct option.

Note
Be careful while doing problems involving Faraday’s law or problems involving solenoid because it is very common to get confused with the substitution of values in the equation to find the magnetic field or induced emf. It is common to get confused regarding the substitution of the number of turns or the current flowing through the coil as they mostly involve two coils.