Question

What amplitude of electric/magnetic field is required to be transmitted in a beam of cross-section area $100m^2$ so that it is comparable to electric power of $500kV$ and ${10}^{3}A$

Answer 1

Given data = Electrical power of $500kV=500\times {10}^{3}V$ voltage and ${10}^{3}A$ current, cross-section area of the beam, $A=100m^2$

To find the amplitude of electric/magnetic field is required to be transmitted in a beam

Solution:

We know that

electrical power is voltage times current by substituting the given values, we get

$P=500\times {10}^3\times {10}^3$ watt

$P=5\times {10}^{8}W.........\left(i\right)$

Now we know that Intensity is energy transferred per unit area per unit time, hence

$I=\frac{\epsilon }{At}..............\left(ii\right)$

Where $\epsilon =$energy, t= time and A= area

And power is energy transferred per unit time, i.e

$P=\frac{\epsilon }{t}..........\left(iii\right)$

So from equation (ii) and (iii) we can say that intensity is power transmitted per unit area, i.e.,

$I=\frac{P}{A}$

now by substituting the values of power (from eqn(i)) and area(given), we get

$I=\frac{5\times {10}^{8}W}{100m^2}⟹I=5\times {10}^{6}W/m^2...........\left(iv\right)$

Now in electromagnetic wave, the Intensity is produced by both oscillating electric field and magnetic field, hence

$I=I_{ϵ}+I_B......\left(v\right)$

where I is total intensity, $I_{ϵ}$ = Intensity due to electric field and $I_B=$Intensity due to magnetic field

but

$I_{ϵ}=I_B=\frac{I}{2}.............\left(vi\right)$ (from eqn(v))

i.e., intensity due to electric field is half of total intensity

And also we know

$I={ϵ}_0{E}^{2}C........\left(vii\right)$

where ${ϵ}_0=8.85\times {10}^{-12}$ is the permitivity in free space, $E$ is RMS values of electric field and $C=3\times {10}^{8}m/s$ is speed of electromagnetic wave.

Hence eqn(vii) in eqn(vi) we get

$\frac{I}{2}=\frac{1}{2}{ϵ}_0{E}^{2}C$

Now substituting the corresponding values, we get

$⟹\frac{5\times {10}^6}{2}=\frac{1}{2}8.85\times {10}^{-12}\times E^2\times 3\times {10}^8⟹E^2=\frac{5\times {10}^6}{8.85\times {10}^{-12}\times 3\times {10}^8}⟹E^2=0.188\times {10}^{6+12-8}⟹E^2=0.188\times {10}^{10}⟹E=0.434\times {10}^5⟹E=4.34\times {10}^{4}N/C$

Where N=Newton and C=coulomb

Now we know that

$C=\frac{E}{B}$

where C is speed of electromagnetic wave

$⟹B=\frac{E}{B}⟹B=\frac{4.34\times {10}^4}{3\times {10}^8}⟹B=1.45\times {10}^{-4}T$

Hence $E=4.34\times {10}^{4}N/C,B=1.45\times {10}^{-4}T$ are the amplitude of electric/magnetic field is required to be transmitted in a beam.