Physics - Unit & Dimension Question with Solution | TestHub

Magnetic potential energy inside a solenoid is

$U=\frac{1}{2}L{I}^2$Where$L=$self-inductor$\Rightarrow U=\frac{1}{2}\left({\mu }_0{n}^{2}A l\right)I^2$

$\because n=$number of turns per unit length

$=\frac{N}{l}$

$U=\frac{1}{2}{\mu }_0{\left(\frac{N}{l}\right)}^{2}Al I^2$

$U=\frac{1}{2}{\mu }_0\frac{N^2}{l^2}Al I^2$

${\mu }_0=\frac{2Ul}{N^{2}A{I}^2}$

Dimensionally,$\left[U\right]=\left[\frac{hc}{\lambda }\right]$

$\left[I\right]=\left[\frac{e}{T}\right]$

$\left[A\right]={\left[l\right]}^2$

$\left[l\right]=\left[l^{\prime }\right]$

$\left[{\mu }_0\right]=\left[\frac{hc}{\lambda }\right] \frac{\left[l\right]}{\left[l^2\right]}\frac{\left[t^2\right]}{{\left[e\right]}^2}$

$=\left[\frac{h}{\lambda }\frac{t}{e^2}\right]=\left[\frac{h}{c{e}^2}\right]$

${\mu }_0=\frac{h}{c{e}^2}$