Question

The 19th-century inventor Nikola Tesla proposed to transmit electric power via sinusoidal electromagnetic waves. Suppose power is to be transmitted in a beam of cross-sectional area 100 ${\mathbf{m}}^{\mathbf{2}}$What electric- and magnetic-field amplitudes are required to transmit an amount of power comparable to that handled by modern transmission lines (that carry voltages and currents of the order of 500 kV and 1000 A)?

Short answer

${\mathrm{E}}_{\max }=6.14\times {10}^4\mathrm{V}/\mathrm{m}\mathrm{and}{\mathrm{B}}_{\max }=2.05\times {10}^{-4}\mathrm{T}$

Step-by-step solution

Concept of the intensity of a sinusoidal electromagnetic wave in vacuum

The intensity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude E amplitude of magnetic field B and it is given in the form $\mathbf{l}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{cE}}_{\mathbf{max}}^{\mathbf{2}}$Also, the intensity I is proportional to the incident power P by area A.$\mathbf{l}\mathbf{=}\frac{\mathbf{P}}{\mathbf{A}}\mathbf{=}\frac{\mathbf{VI}}{\mathbf{4}{\mathbf{\tau \tau r}}^{\mathbf{2}}}$

Calculate the electric field

The intensity I is proportional to the incident power P by area A and given by$\mathrm{I}=\frac{\mathrm{P}}{\mathrm{A}}$ Where P= VI, the voltage times the current. Both equations (1) and (2) have the same left side, so we can get the

next expression of ${\mathrm{E}}_{\max }$as

$$\begin{gathered} \frac{\mathrm{VI}}{\mathrm{A}}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{cE}}_{\max }^2 \\ {\mathrm{E}}_{\max }=\sqrt{\frac{2\mathrm{VI}}{{\mathrm{A\epsilon }}_0\mathrm{c}}} \\ =\sqrt{\frac{2\left(5\times {10}^5\right)\left(1000\right)}{\left(100\right)\left(8.85\times {10}^{-12}\right)\left(3\times {10}^8\right)}} \\ =6.14\times {10}^4\mathrm{V}/\mathrm{m} \end{gathered}$$

Calculate the magnetic field

The maximum electric field is related to the maximum magnetic field and the relationship between both of them is given by ${\mathrm{B}}_{\max }=\frac{{\mathrm{E}}_{\max }}{\mathrm{c}}$ Substitute the values we have,

${\mathrm{B}}_{\max }=\frac{6.14\times {10}^4}{3\times {10}^8}=2.05\times {10}^{-4}\mathrm{T}$

Therefore,${\mathrm{B}}_{\max }=2.05\times {10}^{-4}\mathrm{T}$