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JEE Main 2022 (28 Jul, Shift 1) · Units & Dimensions

Units & Dimensions question from JEE Main 2022 (28 Jul, Shift 1)

Q48JEE Main 2022 · 28 JUL · SHIFT 1Easy

Units & Dimensions

The dimensions of $\left(\frac{\mathrm{B}^2}{\mu_0}\right)$ will be:

(if $\mu_0$ : permeability of free space and $B$ : magnetic field)

Show answer and solution

Answer: (c)

$B^2/(2\mu_0)$ is the magnetic energy density (energy per unit volume).
$[B^2/\mu_0] = \mathrm{ML^2T^{-2}}/\mathrm{L^3} = \mathrm{ML^{-1}T^{-2}}$
So the answer is (c) $[\mathrm{ML^{-1}T^{-2}}]$ .

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