Find the dimensions of \(\frac{B}{\mu_0}\) (Shift I Memory Based)

Magnetic field B :
The dimensional formula for B is derived from the Lorentz force law:

\(F=q v B \quad \Rightarrow \quad B=\frac{F}{q v}\)

1. Force (F) has the dimensional formula:

\([F]=\left[\mathrm{MLT}^{-2}\right]\)

2. Charge (q) has the dimensional formula:

\([q]=[\mathrm{AT}]\)

3. Velocity (v) has the dimensional formula:

\([v]=\left[\mathrm{LT}^{-1}\right]\)

Thus, the dimensional formula for B is:

\([B]=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{AT}]\left[\mathrm{LT}^{-1}\right]}=\frac{[\mathrm{M}]\left[\mathrm{T}^{-2}\right]}{[\mathrm{A}]}=[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]\)

Magnetic permeability ( \(\mu_0\) ):
The dimensional formula of \(\mu_0\) is:

\(\left[\mu_0\right]=[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]\)

Dimensions of \(\frac{B}{\mu_0}\) :
Using the formula:

\(\frac{B}{\mu_0}=\frac{[\mathrm{M}]\left[\mathrm{A}^{-1}\right]\left[\mathrm{T}^{-2}\right]}{[\mathrm{M}][\mathrm{L}]\left[\mathrm{T}^{-2}\right]\left[\mathrm{A}^{-2}\right]}\)

Cancel the terms:

\(\frac{B}{\mu_0}=[\mathrm{A}]\left[\mathrm{L}^{-1}\right]\)