JEE PYQ: Units And Dimensions Question 12
Question 12 - 2020 (07 Jan Shift 2)
The dimensions of $\frac{B^2}{2\mu_0}$, where $B$ is magnetic field and $\mu_0$ is the magnetic permeability of vacuum, is:
(1) $MLT^{-2}$
(2) $ML^2T^{-1}$
(3) $ML^2T^{-2}$
(4) $ML^{-1}T^{-2}$
Show Answer
Answer: (4)
Solution
$\frac{B^2}{2\mu_0}$ is the energy density of magnetic field. $\frac{B^2}{2\mu_0} = \frac{\text{Energy}}{\text{Volume}} = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$.
BETA
