Question

A long solenoid of radius $2cm$ has $100turns/cm$ and carries a current of $5A$ . A coil of radius $1cm$ having $100$ turns 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 through 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: In the given question, we have been given information about a solenoid that has a coaxially placed coil of a given resistance and given dimensions. The entire setup is connected to a galvanometer and then the current in the solenoid is reversed. We have been asked to find the charge flown through the galvanometer. Let’s see the detailed solution to the given question.

Complete step by step answer:
The number of turns of the solenoid has been given to us in terms of turns per centimetre. As a standard, we’ll convert it into the number of turns per meter.
$n=\dfrac{100 turns}{1cm}=\dfrac{100 turns}{{{10}^{-2}}m}={{10}^{4}}turns/m$
The current has a value of $5A$
The magnetic field density produced by the solenoid is given as $B={{\mu }_{0}}ni$ where \[{{\mu }_{0}}\] is the permeability constant.
Substituting the values, we get
\[\begin{align}
  & B=4\pi \times {{10}^{-7}}\times {{10}^{4}}\times 5T\left[ \because {{\mu }_{0}}=4\pi \times {{10}^{-7}} \right] \\
 & \Rightarrow B=2\pi \times {{10}^{-2}}T \\
\end{align}\]
Radius of the coil, \[r=1cm={{10}^{-2}}m\]
Area of the coil, \[A=\pi {{r}^{2}}=\pi \times {{\left( {{10}^{-2}} \right)}^{2}}=\pi \times {{10}^{-4}}{{m}^{2}}\]
The flux linked with the solenoid and the coil can be calculated as follows
\[\begin{align}
  & \phi =B\times {{n}_{2}}A=2\pi \times {{10}^{-2}}\times 100\times \pi \times {{10}^{-4}}T.{{m}^{2}} \\
 & \Rightarrow \phi =2\times {{\pi }^{2}}\times {{10}^{-4}}T.{{m}^{2}} \\
 & \Rightarrow \phi =2\times {{10}^{-3}}T.{{m}^{2}}\left[ \because {{\pi }^{2}}=10 \right] \\
\end{align}\]
Since the current in the solenoid is reversed, the total flux linkage would be twice the above calculated amount, that is, \[\Delta \phi =4\times {{10}^{-3}}T.{{m}^{2}}\]
The EMF developed by the arrangement can be given as the rate of change of the flux linkage, that is
\[E=\dfrac{\Delta \phi }{\Delta t}=\dfrac{4\times {{10}^{-3}}T.{{m}^{2}}}{\Delta t}\]
The current is given as the ratio of the EMF and the resistance of the coil and the charge is given as the product of the current and the change in time.
\[\begin{align}
  & I=\dfrac{E}{R} \\
 & \Rightarrow I=\dfrac{\Delta \phi }{\Delta t\times R}=\dfrac{4\times {{10}^{-3}}}{\Delta t\times 20\Omega }=\dfrac{2\times {{10}^{-4}}}{\Delta t}T{{m}^{2}}{{\Omega }^{-1}} \\
\end{align}\]
The charge can now be calculated as follows
\[\begin{align}
  & Q=I\times \Delta t \\
 & \Rightarrow Q=\dfrac{2\times {{10}^{-4}}}{\Delta t}\times \Delta t=2\times {{10}^{-4}}C \\
\end{align}\]

Hence we can analyse the options now and say that option (A) is the correct answer to the given question.

Note:
In the given solution, there are a lot of steps. We calculated some values and then substituted them into another expression. Questions like this are like a jigsaw puzzle. The pieces are simple but putting them together is the important thing. You might have noticed that we just let the change in time be $\Delta t$ and later in the calculation it got cancelled away. So when some quantities are not given to us or not known, we can just proceed with the calculations and if our pieces are right, the arrangement will give us the solution.