Question

Mutual inductance of between coaxial coils of same radius a and number of turns $N_1\&N_2$ if there centers are seperated by a distance $l$ with $l>>a$ is

• A. $\frac{{\mu }_0{N}_1{N}_2\pi a^4}{2l^3}$
• B. $\frac{{\mu }_0{N}_1{N}_2\pi a^3}{l^2}$
• C. $\frac{\mu \pi a^4}{2l^3{N}_1{N}_2}$
• D. $\frac{{\mu }_0\pi N_1{a}^3}{l^2{N}_2}$

Answer 1

The correct option is A $\frac{{\mu }_0{N}_1{N}_2\pi a^4}{2l^3}$


$B=\int N_i\frac{{\mu }_0}{4\pi }i\frac{\overrightarrow{dl}\times \overrightarrow{r}}{r^3}\cos \theta$
$=N_i\frac{{\mu }_{0}i\cos \theta }{4\pi }\int \frac{\overrightarrow{dl}\times \overrightarrow{r}}{r^3}=N_i\frac{{\mu }_{0}i\cos \theta }{4\pi }\frac{2\pi al}{l^3}$
Since, $\cos \theta \simeq \frac{a}{l}$
$=\frac{{\mu }_0}{2}\frac{N_i{a}^2}{l^3}$

$\varphi =BA{N}_2=\frac{{\mu }_0}{2}\frac{N_i{a}^2}{l^3}\times \pi a^2{N}_2$
$\varphi =Mi\Rightarrow M=\frac{{\mu }_0}{2}\frac{N_1{N}_2\pi a^4}{l^3}$