Question

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

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

Answer 1

Answer: (D)

$9.1\times {10}^{-11}Wb$

As field due to current loop 1 at an axial 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 $B_1$ is

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

The coefficient of mutual inductance between the loops is

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

Flux linked with bigger loop 1 is

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

Substituting the given values, we get

${\varphi }_1=\frac{4\pi \times {10}^{-7}\times (20\times {10}^{-2}{)}^2\times \pi \times (0.3\times {10}^{-2}{)}^2\times 2}{2\left[\right(15\times {10}^{-2}{)}^2+(20\times {10}^{-2}{)}^2{]}^{3/2}}$

${\varphi }_1=9.1\times {10}^{-11}weber$
Hence, the correct answer is $\left(D\right)$