Question:
The dimensions of $\frac{B^{2}}{2 \mu_{0}}$, where $B$ is magnetic field and $\mu_{0}$
is the magnetic permeability of vacuum, is:
Correct Option: , 4
Solution:
(4) The quantity $\frac{\mathrm{B}^{2}}{2 \mu_{0}}$ is the energy density of magnetic field.
$\Rightarrow\left[\frac{B^{2}}{2 \mu_{0}}\right]=\frac{\text { Energy }}{\text { Volume }}=\frac{\text { Force } \times \text { displacement }}{(\text { displacement })^{3}}$
$=\left[\frac{M L^{2} T^{-2}}{L^{3}}\right]=M L^{-1} T^{-2}$
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