In terms of resistance R and time T, the dime | vedclass

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Question

$\begin{array}{l}
{m{Dimensions of}}\,\mu  = \left[ {ML{T^{ - 2}}{A^{ - 2}}} ight]\
{m{Dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L

Vedclass · Accepted Answer

$\left[ {{R^2}} \right]$

Vedclass · Answer

$\begin{array}{l}
{m{Dimensions of}}\,\mu  = \left[ {ML{T^{ - 2}}{A^{ - 2}}} ight]\
{m{Dimensions of}}\, \in \, = \,\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} ight]\
{m{Dimensions of}}\,R\, = \left[ {M{L^2}{T^{ - 3}}{A^{ - 2}}} ight]\
	herefore \frac{{{m{Dimensions of}}\,\mu }}{{{m{Dimensions of}}\, \in }} = \left[ {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} ight]\
 = \left[ {{M^2}{L^4}{T^{-6}}{A^{ - 4}}} ight] = \left[ {{R^2}} ight]
\end{array}$

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

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