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Chapter 29: Problem 68

Your friend who works for the college radio station must make electric-field measurements for a report to be filed with the station's application for license renewal. The measurement is made \(4.6 \mathrm{km}\) from the antenna, where your friend measures the electric field at \(380 \mathrm{V} / \mathrm{m}\). The station is allowed to broadcast at no more than \(55-\mathrm{kW}\) power. Assuming power spreads equally in all directions, is the station in compliance with its license?

Short Answer

Expert verified
Yes, the radio station is in compliance with its license as it's broadcasting at approximately 47 kW, which is within the allowed 55 kW limit.

Step by step solution

01

Electric Field Formula

Given that the electric field \(E = 380 V/m\) is measured at a distance \(r = 4.6km\) from the source, the power \(P\) in watts (W) from a point source that spreads its power equally in all directions is given by: \(P = E² \cdot 4πr² \cdot \epsilon_0\), where \(ε_0 = 8.85 x 10^{-12} C²/N·m²\) is the vacuum permittivity.
02

Calculation of Power

Let's plug the given values into the equation. \[P = (380^2) \cdot 4 \cdot π \cdot (4.6 \times 10^3)^2 \cdot (8.85 \times 10^{-12})\]After calculating the above expression: \[P \approx 47 \times 10^3 W = 47 kW\].
03

Comparison with License Requirements

The power calculated in step 2 is 47 kW. According to the problem statement, the station is allowed to broadcast at no more than 55 kW. Therefore, the station is in compliance with its license as the measured power is less than the allowed maximum power.

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Most popular questions from this chapter

A parallel-plate capacitor has square plates \(10 \mathrm{cm}\) on a side and \(0.50 \mathrm{cm}\) apart. The voltage across the plates is increasing at 220 V/ms. What's the displacement current in the capacitor?

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Show that it's impossible for an electromagnetic wave in vacuum to have a time-varying component of its electric field in the direction of its magnetic field. (Hint: Assume \(\vec{E}\) does have such a component, and show that you can't satisfy both Gauss's and Faraday's laws.)

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