Question

If E is the electric field intensity and ${\mu }_0$ is the permeability of free space, then the quantity $\frac{E^2}{{\mu }_0}$ has the dimensions of

• A. $\left[M^0{L}^1{T}^{-1}\right]$
• B. $\left[M^1{L}^1{T}^{-4}\right]$
• C. $\left[M^1{L}^0{T}^{-4}\right]$
• D. $\left[M^2{L}^2{T}^0\right]$

Answer 1

The correct option is B $\left[M^1{L}^1{T}^{-4}\right]$

Write $\frac{E^2}{{\mu }_0}as\frac{\left(E^2{\in }_0\right)}{\left({\mu }_0{\in }_0\right)}$ and note that the numerator has the dimensions of energy per unit volume whereas the denominator has the dimensions of square of reciprocal of speed.
$\left[\frac{E^2}{{\mu }_0}\right]=\frac{\left[E^2\right]}{{\mu }_0}$
We have,
$\left[E\right]=ML{T}^{-3}{I}^{-1}(\text{using}E=\frac{V}{d})$
$\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2}(\text{using}dB=\frac{{\mu }_0}{4\pi }.\frac{\text{Id sin}\theta }{r^2})$
$\therefore \left[\frac{E^2}{{\mu }_0}\right]=\frac{M^2{L}^2{T}^{-6}{I}^{-2}}{ML{T}^{-2}{I}^{-2}}=ML{T}^{-4}$