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A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 3.5 m from the source will b
- a)62.6 V/m
- b)32.3 V/m
- c)16.15 V/m
- d)8.08 V/m
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A point source of electromagnetic radiation has an average power outpu...
Intensity of EM wave is given by, I = P/4πR2
► vac⋅c = 1/2∈0E02 × c
► E0 = (P/2πR2∈0c)1/2
= (800 / (2 * 3.14 * (3.5)2 * 8.85 * 10-12 * 3 * 108))1/2
= 62.6 V/m
► vac⋅c = 1/2∈0E02 × c
► E0 = (P/2πR2∈0c)1/2
= (800 / (2 * 3.14 * (3.5)2 * 8.85 * 10-12 * 3 * 108))1/2
= 62.6 V/m
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Calculation of Electric Field from Power Output
Given:
- Average power output of electromagnetic radiation = 800 W
- Distance from the source = 3.5 m
- We need to find the maximum value of electric field at this distance
Formula:
The power per unit area carried by an electromagnetic wave is given by the formula:
P = (c * ε * E_0^2)/2
where,
- P = Power per unit area
- c = Speed of light in vacuum = 3 x 10^8 m/s
- ε = Permittivity of free space = 8.85 x 10^-12 F/m
- E_0 = Maximum value of electric field
We can rearrange the formula to find E_0:
E_0 = sqrt((2 * P)/(c * ε))
Calculation:
Substituting the given values in the formula:
E_0 = sqrt((2 * 800)/(3 x 10^8 * 8.85 x 10^-12))
E_0 = 62.6 V/m
Answer:
The maximum value of electric field at a distance 3.5 m from the source is 62.6 V/m. Therefore, the correct answer is option 'A'.
Given:
- Average power output of electromagnetic radiation = 800 W
- Distance from the source = 3.5 m
- We need to find the maximum value of electric field at this distance
Formula:
The power per unit area carried by an electromagnetic wave is given by the formula:
P = (c * ε * E_0^2)/2
where,
- P = Power per unit area
- c = Speed of light in vacuum = 3 x 10^8 m/s
- ε = Permittivity of free space = 8.85 x 10^-12 F/m
- E_0 = Maximum value of electric field
We can rearrange the formula to find E_0:
E_0 = sqrt((2 * P)/(c * ε))
Calculation:
Substituting the given values in the formula:
E_0 = sqrt((2 * 800)/(3 x 10^8 * 8.85 x 10^-12))
E_0 = 62.6 V/m
Answer:
The maximum value of electric field at a distance 3.5 m from the source is 62.6 V/m. Therefore, the correct answer is option 'A'.
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