Question

A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5 m form the source will be

• A. 56.7 V/m
• B. 62.6 V/m
• C. 39.3 V/m
• D. 47.5 V/m

Answer 1

Answer: (B)

62.6 V/m

Answer 2

Intensity of electromagnetic wave given is by

$E _ { m } = \sqrt { \frac { \mu _ { 0 } c P _ { a v } } { 2 \pi r ^ { 2 } } }$ $= \sqrt { \frac { \left( 4 \pi \times 10 ^ { - 7 } \right) \times \left( 3 \times 10 ^ { 8 } \right) \times 800 } { 2 \pi \times 3.5 ^ { 2 } } } = 62.6 \mathrm {~V} / \mathrm { m }$