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Chapter 22: Problem 36

A high-voltage power line carries a current of \(110 \mathrm{A}\) at a location where the Earth's magnetic field has a magnitude of \(0.59 \mathrm{G}\) and points to the north, \(72^{\circ}\) below the horizontal. Find the direction and magnitude of the magnetic force exerted on a \(250-\mathrm{m}\) length of wire if the current in the wire flows (a) horizontally toward the east or (b) horizontally toward the south.

Short Answer

Expert verified
(a) 0.5 N, upward; (b) 1.6175 N, west.

Step by step solution

01

Understanding the Problem

We need to find the magnitude and direction of the magnetic force on a segment of wire given certain variables. The wire carries a current, and the Earth's magnetic field is at a specific orientation. We'll use the formula for the magnetic force on a current-carrying wire: \( F = I L B \sin \theta \), where \( I \) is the current, \( L \) is the length of wire, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the direction of the current and the magnetic field.
02

Convert Units

The Earth's magnetic field is given in gauss (G), but we need to convert this to teslas (T) for standard SI units. \( 1 \mathrm{G} = 10^{-4} \mathrm{T} \). So, \( B = 0.59 \mathrm{G} = 0.59 \times 10^{-4} \mathrm{T} = 5.9 \times 10^{-5} \mathrm{T} \).
03

Calculate the Force for Current Towards East

For a current flowing east, we need to determine \( \theta \), the angle between the wire (east direction) and the magnetic field. The field points north and 72 degrees below horizontal, making the angle with the east-west direction \( \theta = 90^\circ - 72^\circ = 18^\circ \). Hence, \( F = 110 \mathrm{A} \times 250 \mathrm{m} \times 5.9 \times 10^{-5} \mathrm{T} \times \sin(18^\circ) \). Compute this using \( \sin(18^\circ) \approx 0.309 \): \( F \approx 110 \times 250 \times 5.9 \times 10^{-5} \times 0.309 \approx 0.5 \mathrm{N} \). The direction of the force is perpendicular to both current direction and magnetic field, using the right-hand rule, it points upward.
04

Calculate the Force for Current Towards South

When the current runs south, the angle \( \theta \) between current (south direction) and the magnetic field north (72 degrees down from horizontal) is 90 degrees. This results in \( \sin(\theta) = \sin(90^\circ) = 1 \). Thus, \( F = 110 \mathrm{A} \times 250 \mathrm{m} \times 5.9 \times 10^{-5} \mathrm{T} \times 1 = 1.6175 \mathrm{N} \). The direction of the force, using the right-hand rule, is horizontally towards the west.

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Key Concepts

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

Right-Hand Rule
When dealing with magnetic forces, the right-hand rule is an essential tool to determine the direction of the force exerted on a current-carrying wire. It is simple to use and only requires your hand!
  • Stretch out your right hand.
  • Point your thumb in the direction of the current flow in the wire.
  • Align your fingers in the direction of the magnetic field lines.
The direction your palm faces will indicate the direction of the magnetic force. For instance, if a current flows towards the east and the magnetic field is oriented towards the north, your palm will face upwards, signaling that the force is perpendicular to both directions. Remember, this rule only works for conventional current, which means positive electric charge flow.
SI Units Conversion
Converting units to the standard International System of Units (SI) is crucial in physics to maintain consistency and clarity. In our context, the Earth's magnetic field is often measured in gauss, but in SI, we prefer teslas. Here鈥檚 how to convert:
  • Knowing that 1 gauss is equal to \( 10^{-4} \) teslas.
  • If the magnetic field was \( 0.59 \) gausses, convert it: \( 0.59 \times 10^{-4} = 5.9 \times 10^{-5} \) teslas.
This conversion allows us to use the formula for magnetic force correctly, as all components need to comply with SI units for accurate calculations. It is a small yet vital step in solving problems involving magnetic forces.
Angle Between Current and Magnetic Field
Understanding the angle between the current direction and the magnetic field is key to solving problems involving magnetic forces. This angle, denoted by \( \theta \), determines the component of the magnetic field that effectively interacts with the current. Here's how you find this angle:
  • Identify the direction of the magnetic field; for instance, 72 degrees below the horizontal north.
  • Determine the direction of the current, such as east or south.
  • The angle \( \theta \) in these scenarios is the difference in orientation between the current's direction and the magnetic field.
For most calculations, you will use \( \sin(\theta) \) in the force equation. For example, with a current flowing east and a magnetic field tilted from the north, \( \theta \) would be granted by complementary angles from 90 degrees orientation. This angle informs how much of the magnetic force is effectively utilized.

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Most popular questions from this chapter

Two long, straight wires are separated by a distance of \(9.25 \mathrm{cm} .\) One wire carries a current of \(2.75 \mathrm{A},\) the other carries a current of 4.33 A. (a) Find the force per meter exerted on the 2.75-A wire. (b) Is the force per meter exerted on the 4.33-A wire greater than, less than, or the same as the force per meter exerted on the \(2.75-\mathrm{A}\) wire? Explain.

Two power lines, each \(270 \mathrm{m}\) in length, run parallel to each other with a separation of \(25 \mathrm{cm}\). If the lines carry parallel currents of \(110 \mathrm{A},\) what are the magnitude and direction of the magnetic force each exerts on the other?

When at rest, a proton experiences a net electromagnetic force of magnitude \(8.0 \times 10^{-13} \mathrm{N}\) pointing in the positive \(x\) direction. When the proton moves with a speed of \(1.5 \times 10^{6} \mathrm{m} / \mathrm{s}\) in the positive \(y\) direction, the net electromagnetic force on it decreases in magnitude to \(7.5 \times 10^{-13} \mathrm{N}\), still pointing in the positive \(x\) direction. Find the magnitude and direction of (a) the electric field and (b) the magnetic field.

Two current loops, one square the other circular, have one turn made from wires of the same length. (a) If these loops carry the same current and are placed in magnetic fields of equal magnitude, is the maximum torque of the square loop greater than, less than, or the same as the maximum torque of the circular loop? Explain. (b) Calculate the ratio of the maximum torques, \(\tau_{\text {square }} / \tau_{\text {circle }}\).

A solenoid is made from a 25-m length of wire of resistivity \(2.3 \times 10^{-8} \Omega \cdot \mathrm{m} .\) The wire, whose radius is \(2.1 \mathrm{mm},\) is wrapped uniformly onto a plastic tube \(4.5 \mathrm{cm}\) in diameter and \(1.65 \mathrm{m}\) long. Find the emf to which the ends of the wire must be connected to produce a magnetic field of \(0.015 \mathrm{T}\) within the solenoid.
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