Question

Two concentric circular loops are placed as shown in the figure. The radius of the inner loop is $a$ and that of the outer loop is $b.$ The resistance of the wire of the outer loop is $R.$ If the current in the inner loop changes according to the equation $i=2t^2$ then, find the current in the capacitor as a function of time-

• A. $\frac{2{\mu }_0\pi b^{2}C}{a}[1-e^{\frac{-t}{RC}}]$
• B. $\frac{{\mu }_0\pi a^{2}C}{b}[1+e^{\frac{-t}{RC}}]$
• C. $\frac{2{\mu }_0\pi b^{2}C}{a}[1+e^{\frac{-t}{RC}}]$
• D. $\frac{{\mu }_0\pi b^{2}C}{a}[1-e^{\frac{-t}{RC}}]$

Answer 1

The correct option is A $\frac{2{\mu }_0\pi b^{2}C}{a}[1-e^{\frac{-t}{RC}}]$



As given current in the inner loop is time varying, $i=2t^2$, thus the magnetic field on its center will be,

$B=\frac{{\mu }_{0}i}{2a}$

Hence, flux linked with the outer loop is,

${\varphi }_2=B{A}_2=\frac{{\mu }_{0}i}{2a}\times \pi b^2$

We know that the mutual inductance between the two loop is,

$M=\frac{{\varphi }_2}{i}=\frac{{\mu }_0}{2a}\times \pi b^2$

$\therefore$ Emf induced in the larger coil,

$\left|\begin{array}{c}E\end{array}\right|=M\left(\frac{di}{dt}\right)[\because \left(\frac{di}{dt}\right)=\text{current in the smaller coil}]$

$=\frac{{\mu }_0\pi b^2}{2a}\times \frac{d}{dt}\left(2t^2\right)[\because i=2t^2]$

$=\frac{{\mu }_0\pi b^2}{2a}\times 4t=\frac{2{\mu }_0\pi b^{2}t}{a}$

Therefore, the equivalent circuit of the bigger loop can be drawn as,


Now, at any instant, $t$ charge on the capacitor is $q$ and current flowing in the bigger loop is $i_1.$

Applying $KVL$ to the loop,

$E-\frac{q}{C}-i_{1}R=0$

$\Rightarrow \frac{2{\mu }_0\pi b^{2}t}{a}-\frac{q}{C}-i_{1}R=0$

Differentiating the above equation with respect to time,

$\frac{2{\mu }_0\pi b^2}{a}-\frac{i_1}{C}-\frac{Rd{i}_1}{dt}=0$

$R\frac{d{i}_1}{dt}=\frac{2{\mu }_0\pi b^2}{a}-\frac{i_1}{C}$

$\Rightarrow \frac{d{i}_1}{\frac{2{\mu }_0\pi b^2}{a}-\frac{i_1}{C}}=\frac{dt}{R}$

$\frac{d{i}_1}{\frac{2{\mu }_0\pi b^{2}C}{a}-i_1}=\frac{dt}{RC}$

Integrating both sides, with proper limits we get,

${\int }_0^{i_1}\frac{d{i}_1}{\left(\frac{2{\mu }_0\pi b^{2}C}{a}\right)-i_1}={\int }_0^t\frac{dt}{RC}$

$\Rightarrow \ln \left(\frac{\frac{2{\mu }_0\pi b^{2}C}{a}-i_1}{\frac{2{\mu }_0\pi b^{2}C}{a}}\right)=\frac{-t}{RC}$

$\Rightarrow i_1=\frac{2{\mu }_0\pi b^{2}C}{a}[1-e^{\frac{-t}{RC}}]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.