Question

In terms of resistance $R$ and time $T$, the dimensions of ratio $\frac{\mu }{ϵ}$ of the permeability $\mu$ and permittivity $ϵ$ is:

• A. $\left[R^2{T}^2\right]$
• B. $\left[R{T}^{-2}\right]$
• C. $\left[R^2{T}^{-1}\right]$
• D. $\left[R^2\right]$

Answer 1

Answer: (D)

The electrostatic force can be given by:

$F_e=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

So, permittivity is:

So, ${\epsilon }_0=\frac{q^2}{4\pi r^2{F}_e}$

So, dimensions : ${\epsilon }_0=\frac{\left[Q^2\right]}{\left[L^2\right]\left[ML{T}^{-2}\right]}$

The dimensions of the permittivity $\epsilon$ is $=\left[M^{-1}{L}^{-3}{T}^2{Q}^2\right]$

The magnetic force can be given by:

$F_m=\frac{{\mu }_0}{4\pi }\frac{q_{m_1}{q}_{m_2}}{r^2}$

So, permeability is:

So, ${\mu }_0=\frac{q_{m_1}{q}_{m_2}}{4\pi r^2{F}_m}$

The dimensions of the permeability $\mu$ is $=\left[M^1{L}^1{Q}^{-2}\right]$

So, the ratio of the two dimensions is:

$\frac{\mu }{\epsilon }=\frac{\left[M^1{L}^1{Q}^{-2}\right]}{\left[M^{-1}{L}^{-3}{T}^2{Q}^2\right]}$

$\Rightarrow M^2{L}^4{T}^{-2}{Q}^{-4}$

The dimension of the resistance is $R=\left[M{L}^2{T}^{-1}{Q}^{-2}\right]$

comparing the dimension with the dimesion of resistance:

$\therefore \frac{\mu }{\epsilon }=\left[R^2\right]$