Physics - Unit & Dimension Question with Solution | TestHub

Dimension of $\frac{\mathrm{Force}}{\mathrm{Length}\times \mathrm{Time}}$$=\frac{{\mathrm{MLT}}^{-2}}{\mathrm{LT}}={\mathrm{MT}}^{-3}$

Dimensions of $\frac{\mathrm{Power}}{\mathrm{Area}}=\frac{\mathrm{Force}\times \mathrm{Velocity}}{\mathrm{Area}}=\frac{{\mathrm{MLT}}^{-2}\times {\mathrm{LT}}^{-1}}{{\mathrm{L}}^2}={\mathrm{MT}}^{-3}$, 

Dimension of Electric field $\left(E\right), \left[E\right]={\mathrm{MA}}^{-1}{\mathrm{LT}}^{-3}$

Dimensions of Magnetic field $\left(B\right); \left[B\right]={\mathrm{MA}}^{-1}{\mathrm{T}}^{-2}$

Dimension of magnetic permeability $\left({\mu }_0\right); \left[{\mu }_0\right]={\mathrm{MA}}^{-2}{\mathrm{T}}^{-2}\mathrm{L}$

Therefore, dimensions of $S,$$\left[S\right]=\left[\frac{EB}{{\mu }_0}\right]=\frac{\left({\mathrm{MA}}^{-1}{\mathrm{LT}}^{-3}\right)\left({\mathrm{MA}}^{-1}{\mathrm{T}}^{-2}\right)}{{\mathrm{MA}}^{-2}{\mathrm{T}}^{-2}\mathrm{L}}={\mathrm{MT}}^{-3}$