Q 11-01-236NEETNEET 2022Top questionEasy
The dimensions $[\mathrm{MLT^{-2}A^{-2}}]$ belong to the
Answer: (D) Magnetic permeability
Force between two parallel wires per unit length: $\dfrac{F}{l} = \dfrac{\mu_0 I^2}{2\pi d}$, so
$$[\mu_0] = \frac{[F]}{[I^2]} = \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{A^2}]} = [\mathrm{MLT^{-2}A^{-2}}]$$
For comparison: $[\varepsilon_0] = [\mathrm{M^{-1}L^{-3}T^4A^2}]$, $[\phi] = [\mathrm{ML^2T^{-2}A^{-1}}]$, $[L] = [\mathrm{ML^2T^{-2}A^{-2}}]$.
Solution by Sreeraj P, M.Sc Physics