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In terms of resistance R and time T, the dimensions of ratio $\frac{\mu }{\epsilon }$ of the permeability $\mu$ and permittivity $\epsilon$ is :
Option: 1
$\left [ RT^{-2} \right ]$

Option: 2
$\left [ R^{2}T^{-1} \right ]$

Option: 3
$\left [ R^{2} \right ]$

Option: 4
$\left [ R^{2}T^{2} \right ]$

Answers (1)

best_answer

As we have learnt in

The permittivity of free space -

- wherein its unit is

Permeability of free space -

Dimension of permeability of free space $\mu_{o}$ - $M^{1}L^{1}T^{-2}A^{-2}$

- wherein

$\frac{newton}{ampere^{2}} \ or \ \frac{henry}{metre}$

Resistance (R) -

- wherein its unit is

ohm

Dimension of $\left [\mu \right ]= \left [ MLT^{-2}A^{-2} \right ]$

Dimension of $\in = \left [ M^{-1}L^{-3}T^{4}A^{2} \right ]$

Dimension of Resistance $R^{2} = \left [ M^{2}L^{4}T^{-6}A^{-4} \right ]$

$\left [ \frac{\mu }{\in } \right ] = \left [ M^{2}L^{4}T^{-6}A^{-4} \right ]$

$\therefore$ Dimension of $\left [ \frac{\mu }{\in } \right ]$ is same as $R^{2}$

Correct answer is $\left [ 3 \right ]$

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