A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity 5 × 10 − 16 W / m 2 . The maximum instantaneous potential difference across the two ends of the antenna is Options: A . &nbsp1.23 mV B . &nbsp1.23 mV C . &nbsp1.23 V D . &nbsp12.3 mV

Answer: Option A : A I = 1 2 ϵ 0 C E 2 0 ⇒ E 0 = √ 2 I ϵ 0 C = √ 2 × 5 × 10 − 16 8.85 = 0.61 × 10 − 6 V m Alsop E 0 = V 0 d ⇒ V 0 = E 0 d = 0.61 × 10 − 6 × 2 = 1.23 μ V

A radio receiver antenna that is 2 m long is oriented along the direction of the

Question

A radio receiver antenna that is 2 m long is oriented along the direction of the electromagnetic wave and receives a signal of intensity

5 \times 10^{- 16} W / m^{2}

. The maximum instantaneous potential difference across the two ends of the antenna is

Options:

A . 1.23 mV

B . 1.23 mV

C . 1.23 V

D . 12.3 mV

Answer: Option A
:
A

I = \frac{1}{2} ϵ_{0} C E_{0}^{2}

\Rightarrow E_{0} = \sqrt{\frac{2 I}{ϵ_{0} C}} = \sqrt{\frac{2 \times 5 \times 10^{- 16}}{8.85}} = 0.61 \times 10^{- 6} \frac{V}{m}

Alsop

E_{0} = \frac{V_{0}}{d} \Rightarrow V_{0} = E_{0} d = 0.61 \times 10^{- 6} \times 2 = 1.23 μ V

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