Question

A long solenoid of diameter $0.1\,m$ has $2 \times {10^4}$ turns per meter. At the centre of the solenoid, a coil of $100$ turns and radius $0.01\,m$ is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to $0\,A$ from $4\,A$ in $0.05\,s$ . If the resistance of the coil is $10{\pi ^2}\,\Omega $ , the total charge flowing through the coil during this time is.
A) $32\pi \mu \,C$
B) $16\mu \,C$
C) $32\mu \,C$
D) $16\pi \mu \,C$

Answer 1

Hint:In this problem, we have a long solenoid and a coil. The current in the solenoid reduces at a constant rate. This induces emf in the inner coil. Calculate the value of mutual inductance and thus calculate the value of induced emf. Current can be calculated as emf divided by resistance. Charge is equal to current multiplied by time.

Complete step by step solution:The flowing current in the long solenoid will induce emf in the inner coil. As the inner coil has some resistance thus current will flow through the inner coil. The total charge is given as current flowing through the coil multiplied by the time for which the current is flowing.
Let us write down the given quantities.
Diameter of long solenoid, ${d_1} = 0.1\,m = \dfrac{1}{{10}}$ therefore the radius will be ${r_1} = \dfrac{1}{2} \times \left( {\dfrac{1}{{10}}} \right) = \dfrac{1}{{20}}m$
Radius of the inner coil, ${r_2} = 0.01\,m = \dfrac{1}{{100}}m$
Number of turns of the long solenoid, ${N_1} = 2 \times {10^4}$
Number of turns of the inner coil, ${N_2} = 100$
Resistance of the coil, $R = 10{\pi ^2}\,\Omega $
Time, $t = 0.05\,s$The rate of change of current is given as, $\dfrac{{di}}{{dt}} = \dfrac{{4 - 0}}{{0.05}} = 80\,A\,{s^{ - 1}}$
The mutual inductance $M$ on the second coil is given as:

$M = \dfrac{{{\mu _0}{N_1}{N_2}{A_2}}}{l}$

Here, ${\mu _0}$ is a constant having value ${\mu _0} = 4\pi \times {10^{ - 7}}$

${A_2}$ is the area of the inner coil; ${A_2} = \pi \times {\left( {{r_2}} \right)^2}$
$l$ is the length of the solenoid, $l = 1m$ as the number of turns is given per meter

The induced emf $e$ will be given as:

$e = M\dfrac{{di}}{{dt}}$

Substituting the given values in the equation of mutual inductance, we get

$e = M \times \dfrac{{di}}{{dt}} = \dfrac{{4\pi \times {{10}^{ - 7}} \times 2 \times {{10}^4} \times 100 \times \pi \times {{\left( {\dfrac{1}{{100}}} \right)}^2}}}{1} \times 80$
$ \Rightarrow e = 640 \times {\pi ^2} \times {10^{ - 5}}$

We need to find charge, the current is given as emf divided by resistance $i = \dfrac{e}{R}$ and we know that $i = \dfrac{q}{t}$ therefore current will be given as:

$q = \dfrac{e}{R} \times t$

Substituting the values of emf, $e = 640 \times {\pi ^2} \times {10^{ - 5}}$ , resistance $R = 10{\pi ^2}\,\Omega $ and time $t = 0.05\,s$ , we get

$q = \dfrac{{640 \times {\pi ^2} \times {{10}^{ - 5}}}}{{10{\pi ^2}}} \times 0.05$
$ \Rightarrow q = 64 \times {10^{ - 5}} \times 5 \times {10^{ - 2}}$
$ \Rightarrow q = 32 \times {10^{ - 6}}\,C$
$ \Rightarrow q = 32\mu \,C$

Therefore, total charge flowing through the coil during the given time is $q = 32\mu \,C$

Thus, option C is the correct option.

Note:The value ${10^{ - 6}}$ is known as micro and it is denoted using the Greek symbol as $\mu $ . The length of the solenoid is not given directly, instead it is to be considered as $1m$ as the number of turns per metre is given. The current in solenoid induced emf in the inner coil which leads to flow of charge in the inner coil.