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Chapter 22: Problem 75

Energy carried by an EM wave coming through the air can be used to light a bulb that is not connected to a battery or plugged into an electric outlet. Suppose a receiving antenna is attached to a bulb and the bulb is found to dissipate a maximum power of 1.05 W when the antenna is aligned with the electric field coming from a distant source. The wavelength of the source is large compared to the antenna length. When the antenna is rotated so it makes an angle of \(20.0^{\circ}\) with the incoming electric field, what is the power dissipated by the bulb?

Short Answer

Expert verified
Answer: The power dissipated by the bulb when the antenna is rotated at an angle of \(20.0^{\circ}\) is approximately 0.979 W.

Step by step solution

01

Understanding the relationship with the angle and power dissipation

Since we know that the maximum power of 1.05 W is dissipated when the antenna is aligned with the electric field, the power will decrease when an angle is formed. The power dissipated will depend on the angle formed between the electric field and the antenna.
02

Using cosine function to relate the angle and power

The component of the electric field parallel to the antenna will be proportional to the cosine of the angle formed. The power dissipated by the bulb is proportional to the square of the electric field component along the antenna. Thus, we can write the power \(P\) at an angle of \(\theta\) as: \(P(\theta) = P_{max} \times \cos^2(\theta)\) Where \(P_{max}\) is the maximum power when the antenna is aligned with the electric field (1.05 W), and \(\theta\) is the angle formed.
03

Calculate the power at \(20.0^{\circ}\)

Now, we need to calculate the power \(P\) when the antenna is rotated by an angle of \(20.0^{\circ}\): \(P(20.0^{\circ}) = 1.05 \times \cos^2(20.0^{\circ})\) First, find the cosine of \(20.0^{\circ}\): \(\cos(20.0^{\circ}) \approx 0.9397\) Now, square the cosine and multiply it by the maximum power: \(P(20.0^{\circ}) = 1.05 \times (0.9397)^2 \approx 0.979\,\text{W}\) Therefore, the power dissipated by the bulb when the antenna is rotated at an angle of \(20.0^{\circ}\) is approximately 0.979 W.

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Most popular questions from this chapter

Polarized light of intensity \(I_{0}\) is incident on a pair of polarizing sheets. Let \(\theta_{1}\) and \(\theta_{2}\) be the angles between the direction of polarization of the incident light and the transmission axes of the first and second sheets, respectively. Show that the intensity of the transmitted light is $I=I_{0} \cos ^{2} \theta_{1} \cos ^{2}\left(\theta_{1}-\theta_{2}\right)$.
An unpolarized beam of light (intensity \(I_{0}\) ) is moving in the \(x\) -direction. The light passes through three ideal polarizers whose transmission axes are (in order) at angles \(0.0^{\circ}, 45.0^{\circ},\) and \(30.0^{\circ}\) counterclockwise from the \(y\) -axis in the \(y z\) -plane. (a) What is the intensity and polarization of the light that is transmitted by the last polarizer? (b) If the polarizer in the middle is removed, what is the intensity and polarization of the light transmitted by the last polarizer?
A microwave oven can heat 350 g of water from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\) in 2.00 min. (a) At what rate is energy absorbed by the water? (b) Microwaves pass through a waveguide of cross-sectional area \(88.0 \mathrm{cm}^{2} .\) What is the average intensity of the microwaves in the wave guide? (c) What are the rms electric and magnetic fields inside the wave guide?
Verify that the equation \(I=\langle u\rangle c\) is dimensionally consistent (i.e., check the units).
Prove that, in an EM wave traveling in vacuum, the electric and magnetic energy densities are equal; that is, prove that,$$\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2 \mu_{0}} B^{2}$$ at any point and at any instant of time.,
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