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Chapter 20: Problem 88

An airplane is flying due north at \(180 \mathrm{m} / \mathrm{s}\). Earth's magnetic field has a northward component of \(0.30 \mathrm{mT}\) and an upward component of \(0.38 \mathrm{mT}\). (a) If the wingspan (distance between the wingtips) is \(46 \mathrm{m},\) what is the motional emf between the wingtips? (b) Which wingtip is positively charged?

Short Answer

Expert verified
(b) Which wingtip is positively charged? Answer: (a) The magnitude of the motional emf between the wingtips of the airplane is approximately 2.484 V. (b) The wingtip that is farthest from the ground (the one above it) is the positively charged one.

Step by step solution

01

Understanding the given information

We are given the airplane's speed \(v = 180 \mathrm{m/s}\), the northward component of Earth's magnetic field \(B_N = 0.30\ \mathrm{mT}\), the upward component of Earth's magnetic field \(B_U = 0.38\ \mathrm{mT}\), and the wingspan of the airplane \(l = 46 \mathrm{m}\). We have to calculate the motional emf generated between the wingtips of the plane.
02

Calculating the magnetic field vector

To find the magnetic field vector, we need to consider both the northward and upward components given. The magnetic field vector \(\vec{B}\) can be expressed as \(\vec{B} = B_N\hat{n} + B_U\hat{u}\). We can convert the given magnetic field components into Tesla: \(B_N = 0.30\times 10^{-3} \mathrm{T}\) and \(B_U = 0.38 \times 10^{-3} \mathrm{T}\).
03

Finding the motional emf

The formula for motional emf in a straight conductor moving perpendicular to a magnetic field is \(emf = Bvl\), where \(B\) is the magnetic field, \(v\) is the velocity, and \(l\) is the length of the conductor. Since the airplane is flying north, it's moving perpendicular to its wingspan. We will use the northward component of the magnetic field, \(B_N\), to calculate the motional emf: \(emf = B_Nvl = (0.30\times 10^{-3} \mathrm{T})(180 \mathrm{m/s})(46 \mathrm{m})\).
04

Calculating the motional emf

Now, we can calculate the motional emf: \(emf = (0.30\times 10^{-3} \mathrm{T})(180 \mathrm{m/s})(46 \mathrm{m}) = 2.484 \mathrm{V}\). So, the motional emf between the wingtips is approximately \(2.484\ \mathrm{V}\).
05

Determining the positively charged wingtip

To determine which wingtip is positively charged, we need to analyze the direction of the magnetic forces acting on the charges moving with the airplane. The moving charges feel a force due to the magnetic field, given by \(\vec{F} = q\vec{v}\times \vec{B}\). Since the charges are moving northward and the northward component of the magnetic field acts in the same direction, the cross product will produce a force on the positive charges in the upward direction (using the right-hand rule). Thus, the negatively charged wingtip will be the one closest to the ground, and the positively charged wingtip will be the one above it.
06

Summary

The motional emf between the wingtips of the airplane is approximately \(2.484\ \mathrm{V}\). The wingtip that is farthest from the ground is the positively charged one.

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Most popular questions from this chapter

The armature of an ac generator is a rectangular coil \(2.0 \mathrm{cm}\) by \(6.0 \mathrm{cm}\) with 80 turns. It is immersed in a uniform magnetic field of magnitude 0.45 T. If the amplitude of the emf in the coil is $17.0 \mathrm{V},$ at what angular speed is the armature rotating?
In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
The alternator in an automobile generates an emf of amplitude $12.6 \mathrm{V}\( when the engine idles at \)1200 \mathrm{rpm}$. What is the amplitude of the emf when the car is being driven on the highway with the engine at 2800 rpm?
A 100 -turn coil with a radius of \(10.0 \mathrm{cm}\) is mounted so the coil's axis can be oriented in any horizontal direction. Initially the axis is oriented so the magnetic flux from Earth's field is maximized. If the coil's axis is rotated through \(90.0^{\circ}\) in \(0.080 \mathrm{s},\) an emf of $0.687 \mathrm{mV}$ is induced in the coil. (a) What is the magnitude of the horizontal component of Earth's magnetic field at this location? (b) If the coil is moved to another place on Earth and the measurement is repeated, will the result be the same?
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