Question

The long solenoid of diameter $0.1m$ has $2\times {{10}^{4}}$ turns per metre. At the centre of the solenoid, a coil of 100 turns and radius $0.01m$ 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 0.05s. 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

Use the electro-magnetic concept that when the magnetic flux through a coil changes with time, an emf is induced in the coil. Find an expression for the emf. Then use ohm’s law and find an expression for the current in the coil. Later, find the charge flown through the coil in the given time.
Formula used:
$B={{\mu }_{0}}ni$
$\phi =NBA$
$E=\dfrac{d\phi }{dt}$
$E=i'R$
$i'=\dfrac{\Delta q}{\Delta t}$

Complete step by step answer:
In this case, the current in the solenoid will produce a constant magnetic field at its centre which is equal to $B={{\mu }_{0}}ni$, where ${{\mu }_{0}}$ is permeability of free space, n is the number of turn per unit length of the solenoid and i is the current in the solenoid.
This magnetic field is passing through the coil which is located at the centre of the solenoid.
Therefore, there is some magnetic flux through this coil.
The flux through a coil due to a constant magnetic field is given as $\phi =NBA$, where N is number of turns of the coil, B is the magnetic field and A is the area enclosed by the coil.
In this case, $B={{\mu }_{0}}ni$ and the area enclosed by the solenoid is $A=\pi {{r}^{2}}$, where r is the radius of the small coil.
$\Rightarrow \phi =N{{\mu }_{0}}ni(\pi {{r}^{2}})=N{{\mu }_{0}}\pi n{{r}^{2}}i$ …. (i)
When the current in the solenoid changes, the flux through the coil changes. As a result, an emf is induced in the coil.
The induced emf in the coil is equal to $E=\dfrac{d\phi }{dt}$.
$\Rightarrow E=\dfrac{d\phi }{dt}=\dfrac{d}{dt}\left( N{{\mu }_{0}}\pi n{{r}^{2}}i \right)$
Here, all the variables are constant except the current (i) in the solenoid.
$\Rightarrow E=N{{\mu }_{0}}\pi n{{r}^{2}}\dfrac{di}{dt}$ …. (ii).
It is given that the current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05s.
$\Rightarrow \dfrac{di}{dt}=\dfrac{4}{0.05}=80A{{s}^{-1}}$.
And according to the given data $N=100$, $n=2\times {{10}^{4}}{{m}^{-1}}$, $r=0.01m$.
The value of ${{\mu }_{0}}=4\pi \times {{10}^{-7}}N{{A}^{-2}}$.
Substitute the values in (ii).
$\Rightarrow E=100(4\pi \times {{10}^{-7}})\pi (2\times {{10}^{4}}){{(0.01)}^{2}}(80)$.
And from ohm’s law we know that $E=i'R$ , where i’ is the current in the coil produced due to the emf and R is the resistance of the coil.
This means that $i'=\dfrac{E}{R}$. ….. (iii)
Since E is constant, i’ is constant.
Therefore, we can write $i'=\dfrac{\Delta q}{\Delta t}$, where $\Delta q$ is the flow of charge in time $\Delta t$.
Substitute this value in (iii).
$\Rightarrow \dfrac{\Delta q}{\Delta t}=\dfrac{E}{R}$
$\Rightarrow \Delta q=\dfrac{E}{R}\Delta t$
Here, $\Delta t=0.05s$ and $R=10{{\pi }^{2}}\Omega$.
Substitute these values and the value of E in the above equation.
$\Rightarrow \Delta q=\dfrac{\Rightarrow E=100(4\pi \times {{10}^{-7}})\pi (2\times {{10}^{4}}){{(0.01)}^{2}}(80)}{10{{\pi }^{2}}}(0.05)=32\times {{10}^{-6}}C=32\mu C$.
This means that the charge flown through the coil in the given time is $32\mu C$.
Hence, the correct option is C.

Note: You may have seen the emf induced in a coil due to the change in magnetic flux through the coil is given as $E=-\dfrac{d\phi }{dt}$.
Here, the negative sign simply indicates that the emf induced opposes the change in current or magnetic flux. This means that the emf is induced in such that a way that it opposes the change in magnetic flux that occurs. We do not need to use the negative sign while solving problems as we are only interested in the magnitude of the emf.