Question

A long solenoid of diameter $0.1m$ has $2 \times {10^4}$turns per meter. At the center 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 $0A$ from $4A$ 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:
$\left( a \right)$ $26\pi \mu C$
$\left( b \right)$ $22\mu C$
$\left( c \right)$ $16\pi \mu C$
$\left( d \right)$ $32\pi \mu C$

Answer 1

Hint So in this question, the concept of faraday law will be used to solve it. By using the formula$\varepsilon = - N\dfrac{{d\phi }}{{dt}}$, and then formatting the formula to get the change in the charge, we will get the total charge flowing through the coil.
Formula used:
According to faraday law
$\varepsilon = - N\dfrac{{d\phi }}{{dt}}$
Here,
$N$, will be the number of turns

Complete Step By Step Solution In this question, we have the diameter and the number of turns is also given. We have to find the total charge flowing in the coil at that time.
As we know,
$\varepsilon = - N\dfrac{{d\phi }}{{dt}}$
Now dividing the above equation with$R$, we get
$ \Rightarrow \dfrac{\varepsilon }{R} = - \dfrac{N}{R}\dfrac{{d\phi }}{{dt}}$
So it can also be written as,
$ \Rightarrow \vartriangle I = - \dfrac{N}{R}\dfrac{{d\phi }}{{dt}}$
And here, the change in current can be written as
$ \Rightarrow \dfrac{{\vartriangle q}}{{\vartriangle t}} = - \dfrac{N}{R}\dfrac{{d\phi }}{{dt}}$
Now, taking the change in time to the right side, we get
$ \Rightarrow \vartriangle q = - \left[ {\dfrac{N}{R}\left( {\dfrac{{\vartriangle \phi }}{{\vartriangle t}}} \right)} \right]\vartriangle t$
Here, the negative sign shows us that the change in the flux is opposed by the induced emf.
So from here,
$ \Rightarrow \vartriangle q = \dfrac{{{\mu _0}ni\pi {r^2}}}{R}$
So, now on substituting the values given in the question, we get
$ \Rightarrow \vartriangle q = \dfrac{{4\pi \times {{10}^{ - 7}} \times 100 \times 4 \times \pi \times {{\left( {0.01} \right)}^2}}}{{10{\pi ^2}}}$
So on simplifying the equation, we get
$ \Rightarrow \vartriangle q = 32\mu C$
Therefore, $32\mu C$ a charge is required through the coil during this time.

Hence the option $D$ will be the correct choice.

Note When there is a change in attractive transition going through a curl, a prompted emf is delivered which contradicts the reason because of which it is created. That is if motion builds, at that point, there would be an electromotive power delivered in the framework which consistently contradicts the change in attractive flux. The same is genuine when the attractive motion is diminished. Continuously there is a restrict of change (Actually this is the law of nature)