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

Last updated date: 07th Sep 2024
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Answer
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Hint:We are asked to find the amplitude of electric or magnetic field with the given data. From the given value of electrical power in volts, we find the value of power in watts by multiplying with the given current. We know the relation between power and intensity, so we find the value of intensity with respect to power and cross-sectional area. We find the magnitude of electric and magnetic fields by applying the formulas.

Formulas used:
The electric power of the given beam is given by, $P = VI$ and $P = \dfrac{{\rm E}}{t}$
Intensity is, $I = \dfrac{{\rm E}}{{At}}$
Intensity is also defined as the power per area, $I = \dfrac{P}{A}$
Intensity of the beam is the sum of electric and magnetic field, $I = {I_E} + {I_B}$
Intensity of electric field can be found from the formula, $I = {{\rm E}_0}{E^2}C$
Intensity of magnetic field can be found out from the formula, $C = \dfrac{E}{B}$
Where, ${I_E}$ is the intensity due to electric field, ${I_B}$ is the intensity due to the magnetic field and ${{\rm E}_0}$ is the permittivity of free space, ${{\rm E}_0} = 8.854 \times {10^{ - 12}}F/m$

Complete step by step answer:
Let us start by writing down the given information.
The cross-sectional area of the given beam is, $A = 100{m^2}$
The voltage of the bean is, $V = 500kV$
The current of the beam is, $I = {10^3}A$
We find the power associated with the given current and voltage using the formula,
$P = VI \\ \Rightarrow P = 500kV \times {10^3}A \\ \Rightarrow P = 500 \times {10^3} \times {10^3} \\ \Rightarrow P = 5 \times {10^8}W$
We convert the value of voltage from kilo volt to volt.

Now that we have the actual value of power, we find the relation between power and intensity.
Intensity is, $I = \dfrac{{\rm E}}{{At}}$ and power is $P = \dfrac{{\rm E}}{t}$
Hence, $I = \dfrac{P}{A}$
Substituting the values, we get
$I = \dfrac{P}{A} \\ \Rightarrow I = \dfrac{{5 \times {{10}^8}W}}{{100{m^2}}} \\ \Rightarrow I = 5 \times {10^6}W/{m^2}$
We also know that, $I = {I_E} + {I_B}$
In an electromagnetic wave, the electric and magnetic fields are equally oscillating, so we conclude ${I_E} = {I_B} = \dfrac{I}{2}$.

Now we apply the relation between intensity and electric field and find the intensity of electric field.
$E = \sqrt {\dfrac{I}{{{{\rm E}_0}C}}} \\ \Rightarrow E = \sqrt {\dfrac{{5 \times {{10}^6}}}{{8.85 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} \\ \Rightarrow E = 0.434 \times {10^5} \\ \Rightarrow E = 4.34 \times {10^4}N/C$
Since we have the value of electric field, we find the value of magnetic field using,
$B = \dfrac{E}{C} \\ \Rightarrow B = \dfrac{{4.34 \times {{10}^4}}}{{3 \times {{10}^8}}} \\ \therefore B = 1.45 \times {10^{ - 4}}T$

Hence, the value of electric field is $4.34 \times {10^4}N/C$ , and the value of amplitude of magnetic field is $1.45 \times {10^{ - 4}}T$.

Note:The value of power is directly not given. Thus, we have to calculate it using the formula relating the power with voltage and current. We can notice that the intensity of the field is inversely proportional to the area of the cross section and directly proportional to the power. The electric and magnetic fields are equally oscillating in a given electromagnetic field.