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Chapter 21: Problem 28

At New York City, the earth's magnetic field has a vertical component of \(5.2 \times 10^{-5} \mathrm{~T}\) that points downward (perpendicular to the ground) and a horizontal component of \(1.8 \times 10^{-5} \mathrm{~T}\) that points toward geographic north (parallel to the ground). What is the magnitude and direction of the magnetic force on a 6.0 -m long, straight wire that carries a current of 28 A perpendicularly into the ground?

Short Answer

Expert verified
The force on the wire is approximately 9.285 mN at 18.9掳 from the vertical towards the north.

Step by step solution

01

Understanding the Problem

The exercise requires us to calculate the magnetic force on a wire that is 6.0 m long and carries a current of 28 A. The wire is placed perpendicularly into the ground, interacting with the Earth's magnetic field, which has both vertical and horizontal components.
02

Identifying Forces in the Magnetic Field

The force on the wire can be determined using the formula for magnetic force on a current-carrying wire: \[ F = I \cdot L \cdot B \cdot \sin(\theta) \]where:\( I = 28 \text{ A} \), the current in the wire,\( L = 6.0 \text{ m} \), the length of the wire,\( B \), the magnetic field component,\( \theta \) is the angle between the direction of the magnetic field and the direction of the current.
03

Calculating the Vertical Force Component

The current is perpendicular to the ground. Hence, for the vertical component of the Earth's magnetic field, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \).Let's calculate the force due to the vertical magnetic field component, \( B_v = 5.2 \times 10^{-5} \text{ T} \):\[ F_v = I \cdot L \cdot B_v \]\[ F_v = 28 \cdot 6 \cdot 5.2 \times 10^{-5} \]\[ F_v = 8.736 \times 10^{-3} \text{ N} \]
04

Calculating the Horizontal Force Component

The current flows perpendicularly into the ground. The horizontal component of Earth's magnetic field is parallel to the surface and perpendicular to the current's direction. So, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \).For the horizontal component, \( B_h = 1.8 \times 10^{-5} \text{ T} \):\[ F_h = I \cdot L \cdot B_h \]\[ F_h = 28 \cdot 6 \cdot 1.8 \times 10^{-5} \]\[ F_h = 3.024 \times 10^{-3} \text{ N} \]
05

Calculating Total Force Magnitude

The total magnetic force magnitude is calculated using the Pythagorean theorem, since the force components are perpendicular:\[ F = \sqrt{F_v^2 + F_h^2} \]\[ F = \sqrt{(8.736 \times 10^{-3})^2 + (3.024 \times 10^{-3})^2} \]\[ F \approx 9.285 \times 10^{-3} \text{ N} \]
06

Determining the Direction of Net Force

The direction of the force can be determined using trigonometric relations. The angle \( \phi \) is given by:\[ \tan(\phi) = \frac{F_h}{F_v} \]\[ \phi = \tan^{-1} \left(\frac{3.024 \times 10^{-3}}{8.736 \times 10^{-3}}\right) \]\[ \phi \approx 18.9^\circ \]The force is at approximately \( 18.9^\circ \) from the vertical toward the north.

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Key Concepts

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

Magnetic Field
The magnetic field is an invisible force that exists around magnetic materials and charged particles. It can influence other magnetic objects and moving electric charges within its vicinity. At a location such as New York City, the Earth's magnetic field is not uniform.
It consists of two main components:
  • A vertical component that points downward, perpendicular to the ground.
  • A horizontal component that points toward geographic north, parallel to the ground.
These components work together to create a specific magnetic force on objects, such as a current-carrying wire, that interact with them. The Earth's magnetic field strength is typically measured in teslas (T).
Understanding how these components influence objects helps us in determining the resulting force and its direction.
Current-Carrying Wire
A current-carrying wire is simply a wire through which electrical current flows. When the wire has a current moving through it, the wire generates its own magnetic field. This magnetic field can interact with external magnetic fields, like that of the Earth.
In the given exercise, the wire is 6.0 meters long and carries a significant current of 28 amperes (A). The current flows perpendicularly into the ground, meaning it is vertical with respect to the horizontal component of the magnetic field.
This orientation is crucial as it affects how the wire interacts with the Earth's magnetic field components and subsequently determines the magnetic force acting on the wire.
Components of Magnetic Force
The magnetic force on a current-carrying wire is influenced by the interaction between the wire's current and the external magnetic fields. The force can be calculated using the formula:
\[ F = I \cdot L \cdot B \cdot \sin(\theta) \]
  • I is the current flowing through the wire, measured in amperes (A).
  • L is the length of the wire, measured in meters (m).
  • B is the magnetic field strength, measured in teslas (T).
  • \( \theta \) is the angle between the current direction and the magnetic field direction.
In this exercise, there are vertical and horizontal components of force due to the Earth's magnetic field:
- The vertical component comes from the Earth's vertical magnetic field. With the current perpendicular to this, the angle is 90掳, giving a sine value of 1, maximizing the force.
- Similarly, the horizontal component is also maximized with an angle of 90掳.
By using these angles, we can calculate the force individually for each component before finding the total magnetic force.
Angle Between Magnetic Field and Current
The angle between the magnetic field and the current is crucial since it determines how much of the magnetic field contributes to the magnetic force. In magnetic force calculations, the angle affects the value of \( \sin(\theta) \), which can entirely change the force's magnitude.
For instance, in our problem, when the wire is perpendicular to the field lines:
  • The vertical component of the Earth's magnetic field creates a direct angle (90掳) with the vertical current line, resulting in the maximum possible interaction. This maximizes the force exerted.
  • Similarly, the horizontal component of the field also aligns perpendicularly to the current, ensuring full interaction.
Thus, correctly recognizing and applying these angles in calculations is essential to accurately determine the magnetic force's magnitude and direction.

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Most popular questions from this chapter

Consult Interactive Solution \(21.3421 .3\) at to explore a model for solving this problem. The drawing shows a thin, uniform rod, which has a length of \(0.45 \mathrm{~m}\) and a mass of \(0.094 \mathrm{~kg}\). This rod lies in the plane of the paper and is attached to the floor by a hinge at point \(P\). A uniform magnetic field of \(0.36 \mathrm{~T}\) is directed perpendicularly into the plane of the paper. There is a current \(I=4.1 \mathrm{~A}\) in the rod, which does not rotate clockwise or counterclockwise. Find the angle \(\theta\). (Hint: The magnetic force may be taken to act at the center of gravity.)

A particle of mass \(6.0 \times 10^{-8} \mathrm{~kg}\) and charge \(+7.2 \mu \mathrm{C}\) is traveling due east. It enters perpendicularly a magnetic field whose magnitude is \(3.0 \mathrm{~T}\). After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend traveling in the magnetic field?

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{~m} / \mathrm{s}\) through a 3.6-T magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{~N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{~m}^{2}\). What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

Two infinitely long, straight wires are parallel and separated by a distance of one meter. They carry currents in the same direction. Wire 1 carries four times the current as wire 2 does. On a line drawn perpendicular to both wires, locate the spot (relative to wire 1 ) where the net magnetic field is zero. Assume that wire 1 lies to the left of wire 2 and note that there are three regions to consider on this line: to the left of wire \(1,\) between wire 1 and wire \(2,\) and to the right of wire 2

Refer to Interactive Solution \(21.6221 .62\) at for help with problems like this one. A very long, hollow cylinder is formed by rolling up a thin sheet of copper. Electric charges flow along the copper sheet parallel to the axis of the cylinder. The arrangement is, in effect, a hollow tube of current \(I\). Use Amp猫re's law to show that the magnetic field (a) is \(\mu_{0} I /(2 \pi r)\) outside the cylinder at a distance \(r\) from the axis and (b) is zero at any point within the hollow interior of the cylinder. (Hint: For closed paths, use circles perpendicular to and centered on the axis of the cylinder.)
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