In terms of basic units of mass (M), length ( | vedclass

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Question

(a)

Magnetic permeability of free space $=4 \pi 	imes 10^{-7} NA ^{-2}$

The dimension of force is $\left[ MLT ^{-2}ight]$

The dimension of

Vedclass · Accepted Answer

$\left[ MLQ ^{-2}\right]$

Vedclass · Answer

(a)

Magnetic permeability of free space $=4 \pi 	imes 10^{-7} NA ^{-2}$

The dimension of force is $\left[ MLT ^{-2}ight]$

The dimension of current is $\left[ T ^{-1} Q ight]$

Hence, the dimension of magnetic permeability of free space is $= N / A ^2=$

$\left[ MLT ^{-2}ight]\left[ T ^{-1} Q ight]^{-2}$

$=\left[ MLT ^{-2}ight]\left[ T ^{-1} Q ^{-2}ight.$

$=\left[ MLQ ^{-2}ight]$

In terms of basic units of mass $(M)$, length $(L)$, time $(T)$ and charge $(Q)$, the dimensions of magnetic permeability of vacuum $\left(\mu_0\right)$ would be

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