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

Answer 1

Answer: (B)

We know that
Electric field intensity $=\frac{\text { Electrostatic force }}{\text { Electric charge }}$
$\Rightarrow E=\frac{F}{q}$
$\Rightarrow [E]=\frac{[F]}{[q]}=\frac{\left[M^{1} L^{1} T^{-2}\right]}{\left[A^{1} T^{l}\right]}$
$=\left[M^{1} L^{1} T^{-3} A^{-1}\right] \ldots$ (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 r B}{I} \ldots . (i) $
Now, the magnetic force experienced by a current carrying conductor, when placed in a uniform external magnetic field is given by
$F=I L B \sin (\theta) $
$\Rightarrow B=\frac{F}{I L \sin (\theta)}$
Put this value of $B$ into Eq. (ii) to get
$\mu_{0}=\frac{2 \pi r\left[\frac{F}{I L \sin (\theta)}\right]}{I}=\frac{2 \pi r F}{I^{2} L \sin (\theta)}$
$\Rightarrow \left[\mu_{0}\right]=\frac{[2 \pi][r][F]}{\left[I^{2}\right][L][\sin (\theta)]}=\frac{\left[M^{0} L^{0} T^{0}\right]\left[L^{1}\right]\left[M^{1} L^{1} T^{-2}\right]}{\left[A^{2}\right]\left[L^{1}\right]\left[M^{0} L^{0} T^{0}\right]}$
$=\left[M^{1} L^{1} T^{-2} A^{-2}\right] \ldots$ (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
$\left[\frac{E^{2}}{\mu_{0}}\right]=\frac{\left[M^{1} L^{1} T^{-3} A^{-1}\right]^{2}}{\left[M^{1} L^{1} T^{-2} A^{-2}\right]}=\frac{\left[M^{2} L^{2} T^{-6} A^{-2}\right]}{\left[M^{1} L^{1} T^{-2} A^{-2}\right]}$
$=\left[M^{1} L^{1} T^{-4}\right]=\left[M L T^{-4}\right]$