Question

Two coaxial circular loops of radii $0.5\text{m}$ and $0.05\text{m}$ are separated by a distance of $0.5\text{m}.$ The current in the coil of radius $0.5\text{m}$ varies as $i=\left(2t\right)\text{A}.$ If the emf induced in the small coil is $e$ and the mutual inductance between the coils is $M,$ then which of the following is correct-

$\left(i\right).M=2.5\times {10}^{-9}\text{H}$

$\left(ii\right).e=2.5\times {10}^{-9}\text{V}$

$\left(iii\right).M=5\times {10}^{-9}\text{H}$

$\left(iv\right).e=5\times {10}^{-9}\text{V}$

• A. $\left(i\right)$ and $\left(ii\right)$
• B. $\left(ii\right)$ and $\left(iii\right)$
• C. $\left(i\right)$ and $\left(iv\right)$
• D. $\left(ii\right)$ and $\left(iv\right)$

Answer 1

The correct option is C $\left(i\right)$ and $\left(iv\right)$

Given, $r_1=0.5\text{m};r_2=0.05\text{m};x=0.5\text{m}{i}_1=2t\text{A};e_2=e;$

Where, $r_1=\text{Radius of loop-}1r_2=\text{Radius of loop-}2i_1=\text{Current flowing through loop-}1e_2=\text{The emf induced in the loop -}2x=\text{Seperation between the two loops}$

The magnetic field at the location of the smaller loop due to current in the larger loop is,

$B_1=\frac{{\mu }_{0}i{r}_1^2}{2(x^2+r_1^2{)}^{\frac{3}{2}}}$

Flux through the smaller loop is,

${\varphi }_2=B_1\times \pi r_2^2$

${\varphi }_2=\frac{{\mu }_{0}i{r}_1^2}{2(x^2+r_1^2{)}^{\frac{3}{2}}}\times \pi r_2^2$

$\therefore$ The mutual inductance between the pair of coils is,

$M=\frac{{\varphi }_2}{i}$

$M=\frac{{\mu }_{0}i{r}_1^2}{2(x^2+r_1^2{)}^{\frac{3}{2}}}\times \pi r_2^2$

$\Rightarrow M=\frac{{\mu }_0{r}_1^2{r}_2^2\pi }{2(x^2+r_1^2{)}^{\frac{3}{2}}}.......\left(1\right)$

Putting the values in (1) we get,

$\Rightarrow M=\frac{4\pi \times {10}^{-7}\times 0.25\times 25\times {10}^{-4}\times \pi }{2(0.25+0.25{)}^{\frac{3}{2}}}$

$M=246.7\times {10}^{-11}\text{H}\text{(or)}$

$M\approx 2.5\times {10}^{-9}\text{H}$

Using the principle of mutual inductance,

${\varphi }_2=Mi$

$\therefore$ Emf induced in smaller loop is,

$\left|\begin{array}{c}e\end{array}\right|=\frac{d{\varphi }_2}{dt}=M\left(\frac{d{i}_1}{dt}\right)$

$e=M\times \frac{d}{dt}\left(2t\right)=2M$

$\Rightarrow e=5\times {10}^{-9}\text{V}$

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