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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 ..

Vedclass · Accepted Answer

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

Vedclass · Answer

(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).

Similar Questions

The fundamental physical quantities that have the same dimensions in the dimensional formulae of torque and angular momentum are

If $\varepsilon_0$ is the permittivity of free space and $E$ is the electric field,then $\varepsilon_0 E^2$ has the dimensions

The dimension of $\frac{B^{2}}{2 \mu_{0}}$,where $B$ is magnetic field and $\mu_{0}$ is the magnetic permeability of vacuum,is

If $\varepsilon_0$ and $\mu_0$ represent the permittivity and permeability of free space respectively, then the dimension of the product $\varepsilon_0 \mu_0$ is,

The dimensions of permeability of free space $(\mu_0)$ can be given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module