Question

The approximate formula expressing the formula of mutual inductance of two thin coaxial loops of the same radius $a$ when their centers are separated by a distance $l$ with $l>>a$ is

• A. $\frac{1}{2}\frac{{\mu }_0\pi a^4}{l^3}$
• B. $\frac{1}{2}\frac{{\mu }_0{a}^4}{l^2}$
• C. $\frac{{\mu }_0}{4\pi }\frac{\pi a^2}{l^2}$
• D. $\frac{{\mu }_0}{\pi }\frac{a^4}{l^3}$

Answer 1

The correct option is A $\frac{1}{2}\frac{{\mu }_0\pi a^4}{l^3}$

Let $I$= current in one loop. The magnetic field at the centre of the other co-axial loop at a distance l from the centre of the first loop is
$B=\frac{{\mu }_0}{4\pi }\frac{2I\pi a^2}{(a^2+l^2{)}^{3/2}}$
where, $p_m=I\pi a^2$ = magnetic moment of the loop
Flux through the other loop is
${\varphi }_{12}=B\pi a^2$ $=\frac{{\mu }_0}{4\pi }\frac{2\pi a^{2}I}{(a^2+l^2{)}^{3/2}}\pi a^2$
or $M=\frac{{\varphi }_{12}}{I}=\frac{{\mu }_0}{4\pi }\frac{2{\pi }^2{a}^4}{(a^2+l^2{)}^{3/2}}$
$=\frac{{\mu }_0\pi a^4}{2(a^2+l^2{)}^{3/2}}=\frac{{\mu }_0\pi a^4}{2l^3}(a<<l)$