Question

Consider a conducting wire of length $L$ bent in the form of a circle of radius $R$ and another conductor of length '$a$' ($a<<R$) bent in the form of a square. The two loops are then placed in the same plane such that the square loop is exactly at the centre of the circular loop. What will be the mutual inductance between the two loops?

• A. ${\mu }_0\frac{\pi a^2}{L}$
• B. ${\mu }_0\frac{\pi a^2}{16L}$
• C. ${\mu }_0\frac{\pi a^2}{4L}$
• D. ${\mu }_0\frac{a^2}{4\pi L}$

Answer 1

The correct option is B ${\mu }_0\frac{\pi a^2}{16L}$


Let i be the current through the circular loop.
Magnetic field due to the current,
$\overrightarrow{B}$ =$\frac{{\mu }_{0}i}{2R}$ where $2\pi R=L$
Hence flux through square loop,
$\Phi =B\times$ (Area of loop) $=B\times (x{)}^2$
where $x$ is the side of the square loop.
Given, $x=\frac{a}{4}$
Also, $\Phi =Mi$
$\Rightarrow Mi=\frac{{\mu }_{0}i}{2\left(\frac{L}{2\pi }\right)}{\left(\frac{a}{4}\right)}^2=\frac{{\mu }_{0}i{a}^2\pi }{16L}\Rightarrow M=\frac{{\mu }_0\pi a^2}{16L}$