The dimension of frac{E^2}{mu_0} in mass (M), | vedclass

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(B) The dimension of the electric field $E$ is given by:$[E] = \frac{[F]}{[q]} = \frac{[M^1 L^1 T^{-2}]}{[A^1 T^1]} = [M^1 L^1 T^{-3} A^{-1}] \quad ..

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$\left[MLT^{-4}\right]$

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(B) The dimension of the electric field $E$ is given by:$[E] = \frac{[F]}{[q]} = \frac{[M^1 L^1 T^{-2}]}{[A^1 T^1]} = [M^1 L^1 T^{-3} A^{-1}] \quad ... (i)$The dimension of the permeability of free space $\mu_0$ is derived from the relation $B = \frac{\mu_0 I}{2 \pi r}$ for a long straight wire, or more fundamentally from the energy density of a magnetic field $u_B = \frac{B^2}{2 \mu_0}$. Using the force law $F = qvB$, we find:$[\mu_0] = [M^1 L^1 T^{-2} A^{-2}] \quad ... (ii)$Now, we calculate the dimension of $\frac{E^2}{\mu_0}$:$\left[\frac{E^2}{\mu_0}\right] = \frac{[E]^2}{[\mu_0]} = \frac{[M^1 L^1 T^{-3} A^{-1}]^2}{[M^1 L^1 T^{-2} A^{-2}]}$$= \frac{[M^2 L^2 T^{-6} A^{-2}]}{[M^1 L^1 T^{-2} A^{-2}]}$$= [M^{2-1} L^{2-1} T^{-6-(-2)} A^{-2-(-2)}]$$= [M^1 L^1 T^{-4}] = [MLT^{-4}]$

The dimension of $\frac{E^2}{\mu_0}$ in mass $(M)$, length $(L)$ and time $(T)$ is ($E=$ electric field, $\mu_0=$ permeability of free space).

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