SATHEE: Optics Question 314

Optics Question 314

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

Show Answer

Answer:

Correct Answer: A

Solution:

$I = \frac{1}{2} ε_{0} C E_{0}^{2}$

Therefore $E_{0} = \sqrt{\frac{2 I}{ε_{0} C}} = \sqrt{\frac{2 \times 5 \times 10^{- 16}}{8.85}} = 0.61 \times 10^{- 6} \frac{V}{m}$ Also $E_{0} = \frac{V_{0}}{d}$

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

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Optics Question 314

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:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Optics Question 314

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 5×10−16W/m2 . The maximum instantaneous potential difference across the two ends of the antenna is

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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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