Question

In terms of basic units of mass $ (M) $ , length $ (L) $ , time $ (T) $ , and charge $ (Q) $ , the dimensions of magnetic permeability of vacuum $ ({\mu _0}) $ would be:
(A) $ [ML{Q^{ - 2}}] $
(B) $ [L{T^{ - 1}}{Q^{ - 1}}] $
(C) $ [M{L^2}{T^{ - 1}}{Q^{ - 2}}] $
(D) $ [LTQ{}^{ - 1}] $

Answer 1

Hint: Magnetic permeability is a property which supports the formation of magnetic fields by the material. It is defined mathematically as the ratio between magnetic induction to the magnetic intensity. It is the constant we use to calculate the magnetic field of materials, and is given by $ ({\mu _0}) $ .

Formulas used: We will be using the formula $ \mu = \dfrac{B}{H} $ where $ \mu $ is the magnetic permeability, $ B $ is the magnetic flux density and $ H $ is the magnetising field strength.

Complete Answer:
We know that the dimensional formula is a formula given by the basic units like mass, length, time, charge, and temperature. So, we need to simplify any given derived quantity in terms of these basic units in order to find the dimensional formula of the quantity.
We also know that magnetic susceptibility is given by, $ \mu = \dfrac{B}{H} $ . Similarly, magnetic susceptibility in free space or vacuum is given by, $ {\mu _0} = \dfrac{{{B_0}}}{H} $ where the terms mean the same but calculated in free space.
But we also know that the magnetic force, $ F = BIl $
 $ \Rightarrow B = \dfrac{F}{{Il}} $ .
We also know that, $ F = ma $ and $ a = \dfrac{v}{s} $ , $ v = \dfrac{d}{s} $ .
So, we can write force, $ F = [ML{T^{ - 2}}] $ ( since $ a = [L{T^{ - 2}}] $ and $ F = m \times a $ )
Using the dimensional formula of $ F $ in $ B = \dfrac{F}{{Il}} $ ,
 $ \Rightarrow B = \dfrac{{[ML{T^{ - 2}}]}}{{[QL{T^{ - 1}}]}} $
We also know that the magnetic field strength given by $ H $ has units $ A/m $ . So the dimensional formula of $ H $ is $ [Q{L^{ - 1}}{T^{ - 1}}] $ (since the current is given by $ I = \dfrac{q}{t} $ )
Substituting the dimensional formula of $ B $ and $ H $ in $ \mu $ we get,
 $ \mu = \dfrac{B}{H} = \dfrac{{[ML{T^{ - 2}}]}}{{[QL{T^{ - 1}}]}} \times \dfrac{1}{{[Q{L^{ - 1}}{T^{ - 1}}]}} $
Simplifying the above dimensional formula, we get,
 $ \mu = [M{L^1}{Q^{ - 2}}] $
Thus, the dimensional formula of magnetic susceptibility in vacuum is, $ \mu = [ML{Q^{ - 2}}] $
Hence the correct answer is option A.

Note:
You could also find the answer by an alternate solution, using the formula, $ B = \dfrac{{{\mu _0}I}}{{2\pi r}} $
 $ \Rightarrow {\mu _0} = \dfrac{{2\pi rB}}{I} $
We know that the dimensional formula of $ B = \dfrac{{[ML{T^{ - 2}}]}}{{[QL{T^{ - 1}}]}} $ and $ I = [Q{T^{ - 1}}] $
We also know that magnetic susceptibility is given by, $ \mu = \dfrac{B}{H} $ . Similarly, magnetic susceptibility in free space or vacuum is given by, $ {\mu _0} = \dfrac{{{B_0}}}{H} $ where the terms mean the same but calculated in free space.
But we also know that the magnetic force, $ F = BIl \Rightarrow B = \dfrac{F}{{Il}} $ .
We also know that, $ F = ma $ and $ a = \dfrac{v}{s} $ , $ v = \dfrac{d}{s} $ .
So, we can write force, $ F = [ML{T^{ - 2}}] $ ( since $ a = [L{T^{ - 2}}] $ and $ F = m \times a $ )
Using the dimensional formula of $ F $ in $ B = \dfrac{F}{{Il}} $ , $ \Rightarrow B = \dfrac{{[ML{T^{ - 2}}]}}{{[QL{T^{ - 1}}]}} $
We also know that the magnetic field strength given by $ H $ has units $ {\raise0.7ex\hbox{ $ A $ } \!\mathord{\left/
 {\vphantom {A m}}\right.}
\!\lower0.7ex\hbox{ $ m $ }} $ . So the dimensional formula of $ H $ is $ [Q{L^{ - 1}}{T^{ - 1}}] $ (since the current is given by $ I = \dfrac{q}{t} $ )
Substituting the dimensional formula of $ B $ and $ H $ in $ \mu $ we get,
 $ \mu = \dfrac{B}{H} = \dfrac{{[ML{T^{ - 2}}]}}{{[QL{T^{ - 1}}]}} \times \dfrac{1}{{[Q{L^{ - 1}}{T^{ - 1}}]}} $
Simplifying the above dimensional formula, we get,
 $ \mu = [M{L^1}{Q^{ - 2}}] $
Thus, the dimensional formula of magnetic susceptibility in vacuum is, $ \mu = [ML{Q^{ - 2}}] $ (option A).

Note:
You could also find the answer by an alternate solution, using the formula, $ B = \dfrac{{{\mu _0}I}}{{2\pi r}} $
 $ \Rightarrow {\mu _0} = \dfrac{{2\pi rB}}{I} $
We know that the dimensional formula of $ B = \dfrac{{[ML{T^{ - 2}}]}}{{[QL{T^{ - 1}}]}} $ and $ I = [Q{T^{ - 1}}] $
 $ \Rightarrow {\mu _0} = \dfrac{{Br}}{I} = [ML{Q^{ - 2}}] $ .