Physics - EMI/AC Question with Solution | TestHub

The magnetic field inside a long solenoid is given by $B={\mu }_{0}nI$, where $n$ is the number of turns per unit length and $I$ is the current.

The magnetic flux through a single turn of the solenoid is $\varphi =BA=({\mu }_{0}nI)(\pi r^2)$.

The total number of turns in the solenoid of length $l$ is $N=nl$.

The total magnetic flux linkage is $\Phi =N\varphi =(nl)({\mu }_{0}nI\pi r^2)={\mu }_0\pi n^2{r}^{2}lI$.

The self-inductance $L$ is defined by the relation $\Phi =LI$.

Substituting the value of $\Phi$, we get $L={\mu }_0\pi n^2{r}^{2}l$.

Answer: ${\mu }_0\pi n^2{r}^{2}l$