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Question

Physics Question on Dimensional Analysis

In terms of basic units of mass (M)(M), length (L)(L), time (T)(T) and charge (Q)(Q), the dimensions of magnetic permeability of vacuum (μ0)(\mu_0) would be

A

[MLQ2]\left[ ML Q ^{-2}\right]

B

[LT1Q1]\left[ L T^{-1} Q ^{-1}\right]

C

[ML2T1Q2]\left[ ML ^{2} T ^{-1} Q ^{-2}\right]

D

[LTQ1]\left[ L TQ ^{-1}\right]

Answer

[MLQ2]\left[ ML Q ^{-2}\right]

Explanation

Solution

The force per unit length experienced due to two wires in which current is flowing in the same direction is given by dFdl=μ04π2I1I2d\frac{d F}{d l} =\frac{\mu_{0}}{4 \pi} \frac{2 I_{1} I_{2}}{d} [MLT2][L]=[μ0][A2][L]\Rightarrow \frac{\left[ ML T ^{-2}\right]}{[ L ]} =\left[\mu_{0}\right] \frac{\left[ A ^{2}\right]}{[ L ]} [MLT2][L]=μ0[O2T2L]\Rightarrow \frac{\left[ MLT ^{-2}\right]}{[ L ]} =\mu_{0}\left[\frac{ O ^{2}}{ T ^{2} L }\right] μ0=[MLQ2]\Rightarrow \mu_{0} =\left[ ML Q ^{-2}\right]