Question

A circular loop of radius $0.3cm$ lies parallel to a much bigger circular loop of radius $20cm$. The centre of the smaller loop is on the axis of the bigger drop. The distance between their centres is $15cm$. If a current of $10A$ flows through the bigger loop, then the flux linked with smaller loop is

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

Answer 1

Answer: (A)

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

$\begin{array}{c}B\left(smallloop\right)=\frac{{\mu }_{o}I{R}^2}{2{\left(R^2=d^2\right)}^{\frac{3}{2}}}\\ There=B.A=\frac{{\mu }_0\pi R^2{r}^2}{2{\left(R^2+d^2\right)}^{\frac{3}{2}}}\\ MutualInductance=\frac{{\mu }_0\pi R^2{r}^2}{2{\left(R^2+d^2\right)}^{\frac{3}{2}}}\\ Thusthroughcoil=\frac{{\mu }_0\pi R^2{r}^2}{2{\left(R^2+d^2\right)}^{\frac{3}{2}}}c\\ =9.1\times {10}^{-11}weber\end{array}$