In terms of basic units of mass (M) , length (L) , time (T) and charge (Q) , dimensions of magnetic

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1.Units, Dimensions and Measurement

medium

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

A

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

B

$\left[ LT ^{-1} Q ^{-1}\right]$

C

$\left[ ML ^2 T ^{-1} Q ^{-2}\right]$

D

$\left[ LTQ ^{-1}\right]$

(AIIMS-2015)

Solution

(a)

Magnetic permeability of free space $=4 \pi \times 10^{-7} NA ^{-2}$

The dimension of force is $\left[ MLT ^{-2}\right]$

The dimension of current is $\left[ T ^{-1} Q \right]$

Hence, the dimension of magnetic permeability of free space is $= N / A ^2=$

$\left[ MLT ^{-2}\right]\left[ T ^{-1} Q \right]^{-2}$

$=\left[ MLT ^{-2}\right]\left[ T ^{-1} Q ^{-2}\right.$

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

Standard 11

Physics

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