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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
