Question

The dimension of$B^2/2{\mu }_0$where $B$ is magnetic field and ${\mu }_0$ is the magnetic permeability of vacuum, is

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

Answer 1

The correct option is A

$\left[M{L}^{-1}{T}^{-2}\right]$

Step 1: Given

The energy density in a magnetic field$=B^2/2{\mu }_0$

Step 2: Calculate the dimension of the $B^2/2{\mu }_0$

The energy density in a magnetic field = $\frac{Energy}{Volume}=\frac{Force\times displacement}{{\left(displacement\right)}^3}$

Therefore,

$$\begin{gathered} =\left[ML{T}^{-2}\right]\left[L\right]/\left[L^3\right] \\ =\left[M{L}^{-1}{T}^{-2}\right] \end{gathered}$$

Therefore the dimension of $B^2/2{\mu }_0$is $\left[M{L}^{-1}{T}^{-2}\right]$.

Hence, the correct option is (A)