Question

A circular loop of radius $r$ is placed at the centre of current carrying conducting square loop of side $a$. If both loops are coplanar and $a>>r$, then the mutual inductance between the loops will be:

• A. $\frac{{\mu }_0{r}^2}{2\sqrt{2}\left(a\right)}$
• B. $\frac{{\mu }_0{r}^2}{4a}$
• C. $\frac{2\sqrt{2}{\mu }_0{r}^2}{\pi a}$
• D. $\frac{{\mu }_0{r}^2}{4\sqrt{2}a}$

Answer 1

Answer: (C)

$\frac{2\sqrt{2}{\mu }_0{r}^2}{\pi a}$

Both loops are coplanar.

Magnetic field at the center of outer square current carrying loop is

$B_1=\frac{2\sqrt{2}{\mu }_{0}I}{\pi a}$

where $a$= length of side of square loop and

$r$= radius of the circular loop.

Given $a>>r$,

The magnetic field through entire inner coil is $B_1$

Magnetic flux through inner coil, ${\varphi }_{21}=B_1{A}_2$

=$\frac{2\sqrt{2}{\mu }_{0}I}{\pi }\frac{r^2}{a}$------ (1)

Mutual induction, M= $\frac{\varphi }{I_1}$

From (1), M= $\frac{2\sqrt{2}{\mu }_0}{\pi }\frac{r^2}{a}$

Hence, $M\alpha \frac{r^2}{a}$