In SI units, the dimensions of $$\sqrt{\frac{\varepsilon_0}{\mu_0}}$$ is:

We have to find the dimensional formula of $$\sqrt{\dfrac{\varepsilon_0}{\mu_0}}$$ in the SI system. For this, we shall first obtain the dimensions of $$\varepsilon_0$$ (permittivity of free space) and $$\mu_0$$ (permeability of free space) separately, and then form their ratio.

1. Dimensions of $$\varepsilon_0$$

For a parallel-plate capacitor the well-known relation is stated first: $$C=\varepsilon_0\dfrac{A}{d}$$ where $$C$$ is capacitance, $$A$$ is plate area and $$d$$ is separation.

So $$\varepsilon_0=\dfrac{C\,d}{A}$$. We now write each quantity in dimensional symbols $$[M]$$ for mass, $$[L]$$ for length, $$[T]$$ for time and $$[A]$$ for electric current.

• Charge $$Q$$ has the dimension $$[Q]=A\,T$$. • Potential difference $$V$$ is work per unit charge. Work (or energy) has dimension $$M\,L^{2}\,T^{-2}$$, so $$[V]=\dfrac{M\,L^{2}\,T^{-2}}{A\,T}=M\,L^{2}\,T^{-3}\,A^{-1}.$$

Capacitance is defined by $$C=\dfrac{Q}{V}$$, hence $$[C]=\dfrac{A\,T}{M\,L^{2}\,T^{-3}\,A^{-1}}=A^{2}\,T^{4}\,M^{-1}\,L^{-2}.$$

Area has dimension $$L^{2}$$ and distance $$d$$ has dimension $$L$$. Therefore

$$[\varepsilon_0]=\dfrac{[C]\,[L]}{[L^{2}]} =A^{2}\,T^{4}\,M^{-1}\,L^{-2}\times L^{-1} =A^{2}\,T^{4}\,M^{-1}\,L^{-3}.$$

2. Dimensions of $$\mu_0$$

For a long solenoid, the inductance is given (stated formula) by $$L=\mu_0\,\dfrac{N^{2}A}{l},$$ where $$L$$ is inductance, $$N$$ is the (dimensionless) number of turns, $$A$$ is cross-sectional area and $$l$$ is its length. Thus

$$\mu_0=\dfrac{L\,l}{A}.$$

The energy stored in an inductor is $$U=\dfrac12\,L\,I^{2}.$$ Equating dimensions of both sides, the inductance $$L$$ must possess the dimensions of $$\dfrac{\text{energy}}{I^{2}}$$, i.e.

$$[L]=\dfrac{M\,L^{2}\,T^{-2}}{A^{2}}=M\,L^{2}\,T^{-2}\,A^{-2}.$$

Now substituting in $$\mu_0$$:

Area $$A\rightarrow L^{2}$$, length $$l\rightarrow L$$, so

$$[\mu_0]=\dfrac{M\,L^{2}\,T^{-2}\,A^{-2}\; \times L}{L^{2}} =M\,L\,T^{-2}\,A^{-2}.$$

3. Forming the ratio $$\dfrac{\varepsilon_0}{\mu_0}$$

We now divide the dimensions term-wise:

$$\left[\dfrac{\varepsilon_0}{\mu_0}\right] =\dfrac{A^{2}\,T^{4}\,M^{-1}\,L^{-3}}{M\,L\,T^{-2}\,A^{-2}} =A^{2-(-2)}\,T^{4-(-2)}\,M^{-1-1}\,L^{-3-1}$$

Performing the exponent arithmetic:

$$=A^{4}\,T^{6}\,M^{-2}\,L^{-4}.$$

4. Taking the square root

Because the required expression is the square root, we halve every exponent:

$$\left[\sqrt{\dfrac{\varepsilon_0}{\mu_0}}\right] =(A^{4}\,T^{6}\,M^{-2}\,L^{-4})^{1/2} =A^{2}\,T^{3}\,M^{-1}\,L^{-2}.$$

Re-ordering the symbols in the same sequence as in the options, we have $$A^{2}T^{3}M^{-1}L^{-2}.$$

Comparing with the given options, this exactly matches Option B.

Hence, the correct answer is Option B.