JEE Main 2022 · 28 Jul · Shift 1 · Physics Q31

PhysicsSingle correct+4 / -1Question 31 of 90

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

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

[ML^2T^{-2}]

[MLT^{-2}]

[ML^{-1}T^{-2}]

[ML^2T^{-2}A^{-1}]

Answer and worked solution

Official answer: C

$\frac{B^2}{\mu_0}$ is twice the magnetic energy density, so it has the dimensions of energy per volume: $[ML^{-1}T^{-2}]$ .

(C)

Topics:Dimensional analysis Energy in an inductor

JEE Main 2022 · 28 Jul · Shift 1, Physics question 31