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Chapter 21: Problem 62

A small compass is held horizontally, the center of its needle a distance of \(0.280 \mathrm{m}\) directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is parallel to the earth's surface. When the current in the wire is \(25.0 \mathrm{A}\), the needle points \(23.0^{\circ}\) east of north. (a) Does the current in the wire flow toward or away from the earth's surface? (b) What is the magnitude of the horizontal component of the earth's magnetic field at the location of the compass?

Short Answer

Expert verified
(a) The current flows away from the Earth's surface. (b) The horizontal component of Earth's magnetic field is approximately \(4.21 \times 10^{-5} \, \text{T}\).

Step by step solution

01

Understanding the Problem

We have a compass placed horizontally, 0.280 m north of a vertical wire carrying a current of 25.0 A. The problem requires us to determine the direction and the effect of the current on the Earth's magnetic field at the location of the compass.
02

Determine Direction of Current

Using the right-hand rule: if the compass needle deflects 23.0掳 east when the current is applied, the magnetic field due to the wire must be pointing east. Hence, the wire's current must flow upward (away from the Earth's surface) to create a magnetic field going east at the location of the compass.
03

Calculate Magnetic Field due to Wire

The magnetic field due to a long straight wire is given by the formula: \[ B_{ ext{wire}} = \frac{\mu_0 I}{2\pi r} \]where \(\mu_0\) is the permeability of free space \(= 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\), \(I = 25.0 \, \mathrm{A}\), and \(r = 0.280 \, \mathrm{m}\). Substitute these values to find \(B_{\text{wire}}\): \[B_{\text{wire}} = \frac{4\pi \times 10^{-7} \times 25.0}{2\pi \times 0.280} = 1.79 \times 10^{-5} \, \text{T}. \]
04

Calculate Horizontal Component of Earth's Magnetic Field

When the current is turned on, the total horizontal magnetic field at the compass is at an angle of 23.0掳 east of north. Let Earth's horizontal field be \(B_e\) and the magnetic field from the wire be \(B_{\text{wire}}\). Use trigonometric relations:\[\tan(23.0^{\circ}) = \frac{B_{\text{wire}}}{B_e}\]Solving for \(B_e\):\[B_e = \frac{B_{\text{wire}}}{\tan(23.0^{\circ})} = \frac{1.79 \times 10^{-5}}{\tan(23.0^{\circ})} = 4.21 \times 10^{-5} \, \text{T}.\]

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Key Concepts

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

Compass
A compass is a simple navigational instrument that shows directions relative to Earth's magnetic poles. It has a magnetized needle that aligns itself with the Earth's magnetic field.
The compass needle has two ends, one pointing north (typically marked) and the other pointing south. When a compass is undisturbed by other magnetic influences, it aligns with the Earth's magnetic field and points towards the magnetic north, giving us a reliable direction to follow.
  • Compasses are commonly used in hiking and navigation.
  • They are affected by nearby magnetic fields, such as those produced by electrical currents in wires.
A key concept to understand is that magnetic fields influence how the compass needle aligns.
For instance, in our scenario, the current in the nearby wire generated its own magnetic field, causing the compass needle to deflect from its original direction.
Right-Hand Rule
The right-hand rule is an easy, effective method used to determine the direction of the magnetic field around a current-carrying wire. By using your right hand, you can quickly ascertain the field's direction as follows:
With your thumb pointing in the direction of the current flow, the curl of your fingers represents the circular pattern of the magnetic field around the wire.
In the exercise, the wire causes the compass needle to deflect 23掳 east of north, meaning the magnetic field produced by this current impacts the horizontal component of Earth's magnetic field.
  • If the wire鈥檚 current is directed upwards (away from Earth), the field circulates east at the compass's position.
  • This supports the initial deduction that the needle鈥檚 deflection resulted from the magnetic field produced by the upward current flow.
The right-hand rule aids in visualizing how currents create magnetic fields and how these fields impact nearby objects like compasses.
Current in Wire
An electrical current flowing through a wire produces a magnetic field surrounding the wire. The strength and direction of this magnetic field depend on several factors, including the magnitude of the current and the distance from the wire.
In this scenario, a current of 25.0 A flows through a vertical wire, creating a magnetic field that affects the nearby compass. This deflection occurs because the wire鈥檚 magnetic field combines with the horizontal component of the Earth's magnetic field, resulting in a noticeable change in direction.
  • As the problem states, the compass needle deflects due to the wire's magnetic influence, orienting 23掳 east of north.
  • The greater the current, the stronger the magnetic field surrounding the wire.
By understanding current in wires and their magnetic impacts, you can predict how they might affect devices like a compass.
Earth's Magnetic Field
The Earth's magnetic field is a natural field extending from the Earth's interior to where it meets the solar wind. This field is similar to a bar magnet field but varies in direction and strength at different locations worldwide.
The Earth's magnetic field has several components, with the horizontal component being of particular interest in navigation, such as using a compass.
This component lies parallel to the Earth's surface and is what typically aligns the compass needle to point north. In this exercise, this natural orientation was altered by the magnetic field created by the current in the wire.
  • Scientists are interested in the Earth's magnetic field as it affects aviation, communication, and navigation.
  • The strength and direction depend on both the Earth's field and additional magnetic fields, like those from nearby currents.
Understanding these interactions helps in determining both the magnitude and direction shifts, such as those experienced by a compass positioned near a current-carrying wire.

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Most popular questions from this chapter

In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5} \mathrm{T} .\) An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s} .\) What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

One component of a magnetic field has a magnitude of \(0.048 \mathrm{T}\) and points along the \(+x\) axis, while the other component has a magnitude of \(0.065 \mathrm{T}\) and points along the \(-y\) axis. A particle carrying a charge of \(+2.0 \times 10^{-5} \mathrm{C}\) is moving along the \(+z\) axis at a speed of \(4.2 \times 10^{3} \mathrm{m} / \mathrm{s} .\) (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the \(+x\) axis.

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2} .\) At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

The 1200 -turn coil in a de motor has an area per turn of \(1.1 \times 10^{-2} \mathrm{m}^{2}\) The design for the motor specifies that the magnitude of the maximum torque is \(5.8 \mathrm{N} \cdot \mathrm{m}\) when the coil is placed in a \(0.20-\mathrm{T}\) magnetic field. What is the current in the coil?
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