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Physics · 2020
JEE Main · 7 January 2020, Shift 2 · Q1
The dimension of B^2/(2 μ_0), where B is magnetic field and μ_0 is the magnetic permeability of vacuum, is:
The dimension of $\displaystyle \frac{B^2}{2 \mu_0}$, where $\displaystyle B$ is magnetic field and $\displaystyle \mu_0$ is the magnetic permeability of vacuum, is :
Official answer
(4)
$\displaystyle \mathrm{ML}^{-1} \mathrm{~T}^{-2}$
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JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.