If $$E$$ and $$H$$ represents the intensity of electric field and magnetizing field respectively, then the unit of $$\frac{E}{H}$$ will be:

We start by recalling the standard SI units of the two given field quantities.

For the electric field intensity, we have the definition $$E \;=\; \frac{\text{Force}}{\text{Charge}}.$$ Since force has the unit $$\text{newton (N)}$$ and charge has the unit $$\text{coulomb (C)},$$ the unit of $$E$$ is $$\frac{\text{N}}{\text{C}}.$$ However, it is equally common to express this in terms of volts per metre, because $$1\;\text{V} = 1\;\frac{\text{J}}{\text{C}}$$ and $$1\;\text{N} = 1\;\frac{\text{J}}{\text{m}},$$ so $$\frac{\text{N}}{\text{C}} = \frac{\,\frac{\text{J}}{\text{m}}\,}{\text{C}} = \frac{\text{J}}{\text{C}\,\text{m}} = \frac{\text{V}\,\text{C}}{\text{C}\,\text{m}} = \frac{\text{V}}{\text{m}}.$$ Hence, the electric field intensity $$E$$ carries the unit $$\text{volt per metre (V m}^{-1}\text{)}.$$

For the magnetizing field (also called the magnetic field intensity) $$H$$, the SI definition is $$H \;=\; \frac{\text{current}}{\text{length}}.$$ Thus the unit of $$H$$ is $$\frac{\text{ampere (A)}}{\text{metre (m)}} = \text{A m}^{-1}.$$

Now we consider the ratio in question: $$\frac{E}{H}.$$ Writing in unit symbols, we substitute the individual units we have just derived: $$\frac{E}{H} \;=\; \frac{\text{V m}^{-1}}{\text{A m}^{-1}} \;=\; \frac{\text{V m}^{-1}}{\text{A m}^{-1}} \times \frac{\text{m}^{-1}}{\text{m}^{-1}} \;=\; \frac{\text{V}}{\text{A}}.$$

Next, we recognize that $$\frac{\text{volt}}{\text{ampere}} = \text{ohm}.$$ This follows directly from Ohm’s law, which states $$V = I R,$$ so $$R = \frac{V}{I}.$$ The symbol $$R$$ represents electrical resistance, and its SI unit is the ohm (Ω). Hence the unit $$\frac{\text{V}}{\text{A}}$$ is precisely the ohm.

Therefore, the unit of the given ratio $$\dfrac{E}{H}$$ is ohm.

Hence, the correct answer is Option B.