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Chapter 29: Problem 57

Antenna emf. A satellite, orbiting the earth at the equator at an altitude of 400 \(\mathrm{km}\) , has an antenna that can be modeled as a \(2.0-\mathrm{m}-\) long rod. The antenna is oriented perpendicular to the earth's surface. At the cquator, the earth's magnetic field is cssentially horizontal and has a value of \(8.0 \times 10^{-5} \mathrm{T}\) ; ignore any changes in \(B\) with altitude. Assuming the orbit is circular, determine the induced emf between the tips of the antenna.

Short Answer

Expert verified
The induced emf is approximately 1.23 volts.

Step by step solution

01

Understand the Given Problem

The problem involves finding the induced electromotive force (emf) on a satellite antenna as it orbits Earth. We are given information about the antenna's length, the value of the Earth's magnetic field, and the satellite's altitude and orientation.
02

Identify Relevant Formulas

To find the induced emf (\( \epsilon \)), we can use the formula: \( \epsilon = B \cdot v \cdot L \cdot \sin \theta \), where \(B\) is the magnetic field, \(v\) is the orbital velocity of the satellite, \(L\) is the length of the antenna, and \(\theta\) is the angle between the velocity and magnetic field. Since the magnetic field is horizontal and the antenna is vertical, \(\theta = 90^\circ\), making \(\sin \theta = 1\).
03

Calculate Orbital Velocity

The orbital velocity \(v\) of the satellite can be calculated with the formula \(v = \sqrt{\frac{GM}{R}}\), where \(G = 6.674 \times 10^{-11} \,\mathrm{m}^3/\mathrm{kg\cdot s}^2\), \(M = 5.972 \times 10^{24} \, \mathrm{kg}\) is the mass of Earth, and \(R = 6.371 \times 10^{6} \, \mathrm{m} + 400 \times 10^3 \, \mathrm{m}\) is the radius of the orbit. Calculate \(R\) first, then substitute values to find \(v\).
04

Calculate Radius of Orbit

The radius \(R\) of the orbit is the sum of Earth's radius and the altitude of the satellite: \(R = 6.371 \times 10^{6} + 400 \times 10^3 = 6.771 \times 10^6\, \mathrm{m}\). This is the distance from the center of the Earth to the satellite.
05

Compute the Orbital Velocity

Substitute the radius \(R\) into the velocity formula: \( v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.771 \times 10^6}} \), to find \(v\). Calculating gives \( v \approx 7.67 \times 10^3 \, \mathrm{m/s}\).
06

Calculate Induced Emf

Now use the formula to find the emf: \( \epsilon = B \cdot v \cdot L \cdot \sin 90^\circ \). Substitute \(B = 8.0 \times 10^{-5} \, \mathrm{T}\), \(v = 7.67 \times 10^3 \, \mathrm{m/s}\), and \(L = 2.0 \, \mathrm{m}\): \( \epsilon \approx 8.0 \times 10^{-5} \times 7.67 \times 10^3 \times 2.0 = 1.23 \, \mathrm{V} \).
07

Conclusion

The induced emf between the tips of the satellite's antenna as it orbits the Earth is approximately 1.23 volts.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orbital Velocity
Orbital velocity refers to the speed at which a satellite must travel to maintain orbit around a celestial body, like Earth. This velocity keeps the satellite moving in a stable, circular path due to the balance between gravitational pull and inertia.
  • To calculate orbital velocity, we use the formula: \[ v = \sqrt{ \frac{GM}{R} } \] where:
    • \( G \) is the gravitational constant \(6.674 \times 10^{-11} \, \mathrm{m^3/kg \cdot s^2}\)
    • \( M \) is the mass of Earth \(5.972 \times 10^{24} \, \mathrm{kg}\)
    • \( R \) is the orbital radius, the distance from Earth's center to the satellite.
  • The orbital radius accounts for Earth's radius and the altitude of the satellite, giving \[ R = 6.371 \times 10^6 + 400 \times 10^3 = 6.771 \times 10^6 \, \mathrm{m}. \]
Substituting values into the formula grants us the satellite's orbital velocity, \( v \), which in this case is approximately 7,670 meters per second. This speed carefully balances the gravitational force pulling the satellite inward and the inertia wanting it to fly straight.
Earth's Magnetic Field
Earth's magnetic field plays a significant role in the induced electromotive force experienced by antennas on orbiting satellites.
  • This geomagnetic field resembles that of a giant bar magnet centered on the Earth, influencing compasses and affecting satellites in orbit.
  • Near the equator, the magnetic field is essentially horizontal. In this context, the strength of Earth's magnetic field is given as \( 8.0 \times 10^{-5} \, \mathrm{T} \) or Tesla.
The magnetic field interacts with the satellite's motion to generate an electromotive force. Understanding this influence helps in determining key variables for satellite operations and communication systems.
Satellite Antenna
The satellite antenna in this scenario is modeled as a 2-meter long rod, oriented vertically perpendicular to the Earth's horizontal magnetic field.
  • As an important aspect of satellite design, antennas are pivotal in facilitating communication between the satellite and Earth.
  • The orientation is crucial because the induced electromotive force depends on the angle between the antenna and the magnetic field.
  • The formula used here for induced emf is: \[ \epsilon = B \cdot v \cdot L \cdot \sin \theta \] where \( \sin 90^\circ = 1 \), simplifying calculation as this angle results from the vertical antenna and horizontal field configuration.
This configuration ensures optimal interaction with Earth's magnetic field, maximizing the induced voltage.
Circular Orbit
A circular orbit is characterized by a path where a satellite maintains a constant distance from the Earth's center.
  • Satellites in circular orbits travel around Earth in a stable, predictable manner, making them ideal for communication, weather monitoring, and GPS systems.
  • The consistency of altitude around the Earth ensures uniform gravitational force and uniform speed, simplifying satellite operation.
  • In this exercise, the assumption of a circular orbit allows calculations to focus on constant speed and direction, maximizing the predictability of the satellite鈥檚 path.
Understanding circular orbits is essential for designing effective satellite systems and ensuring seamless data transmission and precision in applications. The stable nature of circular orbits aids in maintaining balance and efficiency in communication networks.

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Most popular questions from this chapter

A long, straight solenoid with a cross-sectional area of 8.00 \(\mathrm{cm}^{2}\) is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 \(\mathrm{A}\) . A second winding of 12 tums encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average indnced emf in the second winding?

A \(1.41-\mathrm{m}\) bar moves through a uniform, 1.20 . \(T\) magnetic field with a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) (Fig, 29.40\()\) . In cach case, find the emf induced between the ends of this bar and identify which, if any, end \((a \text { or } b)\) is at the higher potential. The bar moves in the direction of (a) the \(+x\) -axis; (b) the \(-y\) -axis; (c) the \(+z\) -axis. (d) How should this bar move so that the emf across its ends has the greatest possible value with \(b\) at a higher potential than \(a\) , and what is this maximum emf?

In a physics laboratory experiment, a coil with 200 tums enclosing an area of 12 \(\mathrm{cm}^{2}\) is rotated in 0.040 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is \(6.0 \times 0^{-5} \mathrm{T}\) . (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? b) What is the average emf induced in the coil?

A circular wire loop of radius \(a\) and resistance \(R\) initially has a magnetic flux through it due to an external magnetic field. The extermal field then decreases to zero. A current is induced in the loop while the external field is changing; however, this current does not stop at the instant that the external field stops changing. The reason is that the current itself generates a magnetic field, which gives rise to a flux through the loop. If the current changes, the flux through the loop changes as well, and an induced emf appears in the loop to oppose the change. (a) The magnetic field at the center of the loop of radius a produced by a current i in the loop is given by \(B=\mu_{0} i / 2 a\) . If we use the crode approximation that the field has this same value at all points within the loop, what is the flux of this field through the loop? (b) By using Faraday's law, Eq. \((29.3),\) and the relationship \(\mathcal{E}=i R,\) show that after the external field has stopped changing, the current in the loop obeys the differential equation $$ \frac{d i}{d t}=-\left(\frac{2 R}{\pi \mu_{0} a}\right) i $$ (c) If the current has the value \(i_{0}\) at \(t=0\) , the instant that the external field stops changing. snive the equation in part \((b)\) to find \(i\) as a function of time for \(t>0 .\) (Hint: In Section 26.4 we encountered a similar differential equation, Eq. \((26.15),\) for the quantity \(q .\) This equation for \(i\) may be solved in the same way. (d) If the loop has radius \(a=50 \mathrm{cm}\) and resistance \(R=0.10 \Omega,\) how long after the external field stops changing will the current be equal to 0.010 \(\mathrm{o}\) (that is, \(\frac{1}{100}\) of its initial value)? (e) In solving the examples in this chapter, we ignored the effects described in this problem. Explain why this is a good approximation.

A slender rod, 0.240 \(\mathrm{m}\) long, reates with an angular speed of 8.80 \(\mathrm{rad} / \mathrm{s}\) about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of 0.650 \(\mathrm{T}\) (a) What is the induced emf in the rod? (b) What is the potential difference between its ends? (c) Suppose instead the rod rotates at 8.80 \(\mathrm{rad} / \mathrm{s}\) about an axis through its center and perpendicular to the rod, In this case, what is the potential difference between the ends of the rod? Between the center of the rod and one end?
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