If B is magnetic field and $$\mu_o$$ is permeability of free space, then the dimensions of $$(B/\mu_o)$$ is

Find the dimensions of $$B/\mu_0$$, where $$B$$ is the magnetic field and $$\mu_0$$ is the permeability of free space.

From the Lorentz force equation $$F = qvB$$, we can write $$B = \frac{F}{qv}$$.

$$[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{(AT)(LT^{-1})} = \frac{MLT^{-2}}{ALT^{-1} \cdot T} = \frac{M}{AT^2} = MT^{-2}A^{-1}$$

From the relation $$B = \mu_0 nI$$ (magnetic field inside a solenoid), where $$n$$ is the number of turns per unit length and $$I$$ is the current:

$$[\mu_0] = \frac{[B]}{[n][I]} = \frac{MT^{-2}A^{-1}}{L^{-1} \cdot A} = MLT^{-2}A^{-2}$$

$$\left[\frac{B}{\mu_0}\right] = \frac{MT^{-2}A^{-1}}{MLT^{-2}A^{-2}} = \frac{M}{M} \cdot \frac{T^{-2}}{T^{-2}} \cdot \frac{A^{-1}}{A^{-2}} \cdot \frac{1}{L} = L^{-1}A$$

Physical significance: $$B/\mu_0$$ has the same dimensions as magnetic field intensity $$H$$, which has dimensions of $$L^{-1}A$$ (equivalent to A/m in SI units).

The correct answer is Option (3): $$L^{-1}A$$.