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Ch. 28 - Sources of Magnetic Field
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 27, Problem 77

The power cable for an electric trolley (Fig. 27–60) carries a horizontal current of 330 A toward the east. The Earth’s magnetic field has a strength 5.0 x 10-5 T and makes an angle of dip of 22° at this location. Calculate the magnitude and direction of the magnetic force on a 15-m length of this cable.


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1
Identify the relevant formula for the magnetic force on a current-carrying conductor in a magnetic field: \( F = I L B \sin \theta \), where \( F \) is the magnetic force, \( I \) is the current, \( L \) is the length of the conductor, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the current direction and the magnetic field.
Determine the angle \( \theta \) between the current direction (east) and the magnetic field. The Earth's magnetic field makes an angle of dip of 22° with the horizontal, so the angle between the current and the magnetic field is \( 90° - 22° = 68° \).
Substitute the given values into the formula: \( I = 330 \; \text{A} \), \( L = 15 \; \text{m} \), \( B = 5.0 \times 10^{-5} \; \text{T} \), and \( \sin \theta = \sin 68° \). The formula becomes \( F = 330 \cdot 15 \cdot (5.0 \times 10^{-5}) \cdot \sin 68° \).
Calculate the magnitude of the magnetic force by evaluating the product of the terms in the formula. Ensure that the sine of the angle is computed correctly.
Determine the direction of the magnetic force using the right-hand rule. Point your fingers in the direction of the current (east), curl them toward the direction of the magnetic field (downward at an angle of 22° to the horizontal), and your thumb will point in the direction of the force. The force will be perpendicular to both the current and the magnetic field.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Magnetic Force on a Current-Carrying Conductor

When a current flows through a conductor in a magnetic field, it experiences a magnetic force. This force can be calculated using the formula F = I * L * B * sin(θ), where F is the force, I is the current, L is the length of the conductor, B is the magnetic field strength, and θ is the angle between the current direction and the magnetic field.

Earth's Magnetic Field

The Earth's magnetic field is a vector field that extends from the Earth's interior out into space, characterized by its strength and direction. At any location, it can be described by its magnitude (measured in teslas) and its angle of dip, which indicates how steeply the field lines enter the Earth. Understanding the local magnetic field is crucial for calculating the forces on current-carrying conductors.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic force on a current-carrying conductor. By pointing the thumb of the right hand in the direction of the current and the fingers in the direction of the magnetic field, the palm will face the direction of the force. This rule helps visualize the interaction between current, magnetic field, and force.
Related Practice
Textbook Question

Helmholtz coils are two identical circular coils having the same radius 𝑅 and the same number of turns N, separated by a distance equal to the radius 𝑅 and carrying the same dc current I in the same direction. (See Fig. 28–61.) They are used in scientific instruments to generate nearly uniform magnetic fields. (They can be seen in the photo, Fig. 27–19.) (a) Determine the magnetic field B at points 𝓍 along the line joining their centers. Let 𝓍 = 0 at the center of one coil, and 𝓍 = 𝑅 at the center of the other. (b) Show that the field midway between the coils is particularly uniform by showing that dB/d𝓍 = 0 and d²B/d𝓍² = 0 at the midpoint between the coils. (c) If 𝑅 = 10.0 cm, N = 85 turns and I = 3.0 A, what is the field at the midpoint between the coils, 𝓍 = 𝑅/2?

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Textbook Question

Part of a long, thin insulated straight wire is formed into a single circular loop of radius 𝑅 (Fig. 28–68) and carries a current I. (a) What is the magnitude and direction of the magnetic field at the center of the loop? (b) If the plane of the loop is twisted 90 degrees so that the plane is perpendicular to the straight part of the wire (i.e., in the yz plane) what is the magnitude and direction of the field now at the center of the loop?

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Textbook Question

You want to get an idea of the magnitude of magnetic fields produced by overhead power lines. You estimate that a transmission wire is about 12 m above the ground. The local power company tells you that the lines operate at 145 kV and provide a maximum of 45 MW to the local area. Estimate the maximum magnetic field you might experience walking under one such power line, and compare to the Earth’s field. [For an ac current, values are rms, and the magnetic field will be changing.]

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Textbook Question

A toroid is fabricated with a circular shape and loops with a square cross section as shown in Fig. 28–69. The cross-section of a loop is a square of side 6.0 cm. The inner radius of the whole circular toroid is 3.0 m. There are 320 loops of wire which carry a 45-A dc current using a nearby power supply at 20.0 V. The arrows show the current flow in and out of the toroid. The current flows up at the inner diameter and down at the outer diameter. (a) Calculate the strength of the magnetic field at the center of the square’s cross section at point P. (b) Is the magnetic field pointing clockwise or counterclockwise? (c) The square cross-sectional area of the wire is uniformly 0.10 cm2. What is the resistivity of the wire?

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Textbook Question

A set of Helmholtz coils (see Problem 62, Fig. 28–61) have a radius 𝑅 = 10.0 cm and are separated by a distance 𝑅 = 10.0 cm . Each coil has 85 loops carrying a current I = 2.0 A. Graph B as a function of 𝓍.

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Textbook Question

Two long straight aluminum wires, each of diameter 0.42 mm, carry the same current but in opposite directions. They are suspended by 0.50-m-long strings as shown in Fig. 28–66. If the suspension strings make an angle of 3.0° with the vertical and are hanging freely, what is the current in the wires?

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