Question

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 speace)

• A. $\left[M^2{L}^3{T}^{-2}{A}^2\right]$
• B. $\left[ML{T}^{-4}\right]$
• C. $\left[M{L}^E{T}^{-2}\right]$
• D. $M{L}^4{T}^{-4}$

Answer 1

Answer: (B)

We know that
Electric field intensity $=\frac{\text{ Electrostatic force }}{\text{ Electric charge }}$
$\Rightarrow E=\frac{F}{q}$
$\Rightarrow [E]=\frac{[F]}{[q]}=\frac{\left[M^1{L}^1{T}^{-2}\right]}{\left[A^1{T}^l\right]}$
$=\left[M^1{L}^1{T}^{-3}{A}^{-1}\right]\dots$ (i)
Permeability of Free Space $\left({\mu }_0\right)$ Dimension Calculation Magnetic field intensity in at a point in space due to an infinitely long current carrying conducting wire
$B=\frac{{\mu }_{0}I}{2\pi r}$
$\Rightarrow {\mu }_0=\frac{2\pi rB}{I}\dots .(i)$
Now, the magnetic force experienced by a current carrying conductor, when placed in a uniform external magnetic field is given by
$F=ILB\sin (\theta )$
$\Rightarrow B=\frac{F}{IL\sin (\theta )}$
Put this value of $B$ into Eq. (ii) to get
${\mu }_0=\frac{2\pi r\left[\frac{F}{IL\sin (\theta )}\right]}{I}=\frac{2\pi rF}{I^{2}L\sin (\theta )}$
$\Rightarrow \left[{\mu }_0\right]=\frac{[2\pi ][r][F]}{\left[I^2\right][L][\sin (\theta )]}=\frac{\left[M^0{L}^0{T}^0\right]\left[L^1\right]\left[M^1{L}^1{T}^{-2}\right]}{\left[A^2\right]\left[L^1\right]\left[M^0{L}^0{T}^0\right]}$
$=\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]\dots$ (iii)
So, now $\left[\frac{E^2}{{\mu }_0}\right]=\frac{[E{]}^2}{\left[{\mu }_0\right]}$
On putting the values from Eqs. (i) and (iii), we will get
$\left[\frac{E^2}{{\mu }_0}\right]=\frac{{\left[M^1{L}^1{T}^{-3}{A}^{-1}\right]}^2}{\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]}=\frac{\left[M^2{L}^2{T}^{-6}{A}^{-2}\right]}{\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]}$
$=\left[M^1{L}^1{T}^{-4}\right]=\left[ML{T}^{-4}\right]$