Question

Using B = µ0 H, find the ratio E0/H0 for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.

Answer 1

Explanation:

To find the ratio $\frac{E0}{H0}$ for a plane electromagnetic wave propagating through vacuum:

Given in the question  

$B=\mu_{0} H$

We can rewrite this equation for vacuum as,

$\mathrm{B}_{0}=\mu_{0} \mathrm{H}_{0} \ldots \mathrm{eq}^{\mathrm{n}}(\mathrm{i})$

Relationship between magnetic field and electric vacuum field is given as,

$\mathrm{B}_{0}=\mu_{0} \in_{0} c \mathrm{E}_{0} \ldots(\mathrm{ii})$

Step 1:

From Equation (i) by (ii),

$\begin{aligned}\mu_{0} \mathrm{H}_{0} &=\mu_{0} \in_{0} c \mathrm{E}_{0} \\\frac{\mathrm{E}_{0}}{\mathrm{H}_{0}} &=\frac{1}{\epsilon_{0} c} \\\frac{\mathrm{E}_{0}}{\mathrm{H}_{0}}=\frac{1}{\left(8.85 \times 10^{-12} \times 3 \times 10^{8}\right)} & \\\frac{\mathrm{E}_{0}}{\mathrm{H}_{0}} \approx 377\end{aligned}$

Step 2:

Dimension of  $\frac{1}{\epsilon_{0} c}=\frac{1}{\left[L T^{-1}\right]\left[M^{-1} L^{-3} T^{4} A^{2}\right]}$

$\begin{aligned}&=\frac{1}{\left[M^{-1} L^{-2} T^{2} A^{2}\right]}\\&=\left[M^{4} L^{2} T^{-3} A^{-2}\right] |\\&=[R]\end{aligned}$