Question

If $\dfrac{A}{\mu_0}$ has the dimensions $\left[MLT^{-4}\right]$,what is A?

Answer 1

Hint: Recall that $\mu_0$ is the magnetic permeability of free space. Think of how you would express magnetic permeability in terms of the magnetic flux density (B) and magnetic field strength (H). Reduce the units of the above two quantities to their base units and obtain their respective dimensional formulae following which you can put them together to arrive at the dimensions of magnetic permeability, to which end you’d obtain the dimensions of A.

Formula Used:
Magnetic permeability $\mu_0 = \dfrac{B}{H}$

Complete Step-by-Step Solution:
We are given that $\left[\dfrac{A}{\mu_0}\right] = \left[M^1L^1T^{-4}\right]$, where $\mu_0$ is the magnetic permeability of free space.
$\Rightarrow \dfrac{\left[A\right]}{ \left[\mu_0\right]} = \left[M^1L^1T^{-4}\right]$
$\Rightarrow \left[A\right] = \left[M^1L^1T^{-4}\right]. \left[\mu_0\right] $

Now, the magnetic permeability is defined as the ratio of the magnetic flux density and the magnetic field strength, i.e.,
$\mu_0 = \dfrac{B}{H} \Rightarrow \left[\mu_0\right] = \dfrac{\left[B\right]}{ \left[H\right]}$

We know that the magnetic flux density is given as:
$B = \dfrac{F}{I \times l}$, where F is the magnetic force (in newtons), I is the current (in amperes), and l is the length (in metres).
We know that $\left[F\right] = \left[M^1L^1T^{-2}\right]$, and $\left[I\right] = \left[I^1 \right]$ and $\left[l\right] = \left[L^1\right]$
$\Rightarrow \left[B\right] = \dfrac{\left[F\right]}{ \left[I\right] \left[l\right]} = \dfrac{\left[M^1L^1T^{-2}\right]}{ \left[I^1 \right] \left[L^1\right]} = \left[M^1L^0T^{-2}I^{-1}\right]$

Now, the magnetic field strength is usually measured in units of amperes per metre $Am^{-1}$:
$\left[H\right] = \left[L^{-1}I^1\right] = \left[M^0L^{-1}T^0I^1\right]$

Therefore, dimensions of the magnetic permeability
$\left[\mu_0\right] = \dfrac{\left[B\right]}{ \left[H\right]} = \dfrac{\left[M^1L^0T^{-2}I^{-1}\right]}{ \left[M^0L^{-1}T^0I^1\right]} = \left[M^1L^1T^{-2}I^{-2}\right]$

To this end, the dimensions of A will be:
$\left[A\right] = \left[M^1L^1T^{-4}\right]. \left[\mu_0\right] = \left[M^1L^1T^{-4}\right]. \left[M^1L^1T^{-2}I^{-2}\right] = \left[M^2L^2T^{-6}I^{-2}\right]$

Note:
The dimensional formula allows us to express the unit of a physical quantity in terms of the fundamental quantities mass (M), length (L), time (T) and temperature (K) in a standard form. We did not use temperature since they are most often concerned with thermodynamic equations.
However, while deducing the dimensional formula of the constants from the base units be careful in assigning the right power to the right fundamental quantity each time as any discrepancy in the dimension will result in an anomalous nature of the quantities involved.