Question

A circular loop of radius $0.3\text{cm}$ lies parallel to a much bigger circular loop of radius $20\text{cm}$. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is $15\text{cm}$. If a current of $2.0\text{A}$ flows through the smaller loop, then the flux linked with bigger loop is :

• A. $9.1\times {10}^{-11}\text{weber}$
• B. $3.3\times {10}^{-11}\text{weber}$
• C. $6\times {10}^{-11}\text{weber}$
• D. $6.6\times {10}^{-9}\text{weber}$

Answer 1

The correct option is A $9.1\times {10}^{-11}\text{weber}$

As field duet ocurrent loop $1$ at and aixal point,

$B_1=\frac{{\mu }_0{I}_1{R}^2}{2(d^2+R^2{)}^{3/2}}$

Flux linked with smaller loop $2$ due to $B_1$,

${\varphi }_2=B_1{A}_2=\frac{{\mu }_0{I}_1{R}^2}{2(d^2+R^2{)}^{3/2}}\pi r^2$

The coefficient of mutual inductance between the loops is,

$M=\frac{{\varphi }_2}{I_1}=\frac{{\mu }_0{R}^2}{2(d^2+R^2{)}^{3/2}}\pi r^2$

Flux linked with bigger loop $1$ is,

${\varphi }_1=M{I}_2=\frac{{\mu }_0{R}^2{I}_2}{2(d^2+R^2{)}^{3/2}}\pi r^2$

Given: $I_2=2A,R=20cm=20\times {10}^{-2}\text{m},x=15cm=15\times {10}^{-2}m$

${\varphi }_1=\frac{{\mu }_0\left(2\right)(20\times {10}^{-2}{)}^2}{2\left[\right(0.2{)}^2+\left(0.15{)}^2\right]}\times \pi (0.3\times {10}^{-2}{)}^2$

On solving
$=9.216\times {10}^{-11}$

$\approx 9.2\times {10}^{-11}\text{weber}$

Hence, option $\left(A\right)$ is correct.