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Chapter 21: Problem 31

A \(45-\mathrm{m}\) length of wire is stretched horizontally between two vertical posts. The wire carries a current of \(75 \mathrm{A}\) and experiences a magnetic force of \(0.15 \mathrm{N} .\) Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of \(60.0^{\circ}\) with respect to the wire.

Short Answer

Expert verified
The magnitude of the Earth's magnetic field is approximately \(5.92 \times 10^{-5} \ \mathrm{T}\).

Step by step solution

01

Understanding the Problem

We need to find the magnitude of the Earth's magnetic field at the location of a wire experiencing a magnetic force. The wire is carrying a current and makes a specific angle with the magnetic field.
02

Applying the Magnetic Force Formula

The magnetic force on a current-carrying wire can be calculated using the formula: \[F = I \cdot L \cdot B \cdot \sin(\theta)\]where \(F\) is the magnetic force, \(I\) is the current, \(L\) is the length of the wire, \(B\) is the magnetic field's magnitude, and \(\theta\) is the angle between the wire and the magnetic field.
03

Plugging in Given Values

Substitute the given values into the formula: \(F = 0.15 \ \mathrm{N}\), \(I = 75 \ \mathrm{A}\), \(L = 45 \ \mathrm{m}\), and \(\theta = 60.0^{\circ}\).\[0.15 = 75 \cdot 45 \cdot B \cdot \sin(60^{\circ})\]
04

Calculating the Sine Value

Calculate \(\sin(60^{\circ})\):\[\sin(60^{\circ}) = \frac{\sqrt{3}}{2} \approx 0.866\]
05

Solving for Magnetic Field - B

Rearrange the equation to solve for \(B\):\[B = \frac{0.15}{75 \cdot 45 \cdot 0.866}\]Simplify to find \(B\).
06

Final Calculation

Perform the final calculations:\[B = \frac{0.15}{2925 \cdot 0.866} \approx \frac{0.15}{2533.95}\]\(B \approx 5.92 \times 10^{-5} \ \mathrm{T}\).

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Key Concepts

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

Current-Carrying Wire
A current-carrying wire experiences a magnetic force when it is placed in a magnetic field. But how does this happen? A wire carrying an electric current creates its own magnetic field, and when it enters an external magnetic field, it experiences a force. This is known as the Lorentz force. The interaction between these two fields results in the force acting on the wire.

A current-carrying wire is relevant in many practical applications. For instance, it forms the basis of electric motors, where the interaction of magnetic fields and current-carrying wires results in rotational motion.
  • The direction of the current affects the direction of the force.
  • The strength of the force depends on the current, wire's length, and angle relative to the external magnetic field.
This is why it's important to understand how current-carrying wires operate within magnetic fields so effectively.
Earth's Magnetic Field
The Earth's magnetic field is a natural phenomenon similar to a giant bar magnet's field. It surrounds the planet and serves crucial functions such as navigation and protection from solar winds. This field is not uniform everywhere; it can vary in strength and direction depending on the location. In our context, the Earth's magnetic field plays a role in determining the force on the wire. It provides the external field in which the wire with current is placed.

When the problem mentions an angle of 60 degrees, it refers to the angle at which the magnetic field intersects the wire. Understanding this angle is essential for calculating how the Earth's magnetic field influences the magnetic force on the wire.
Magnetic Field Calculation
To calculate the magnetic field affecting the wire, we use the formula:\[F = I \cdot L \cdot B \cdot \sin(\theta)\]In this equation:
  • \(F\) represents the force on the wire (in Newtons, N).
  • \(I\) is the current flowing through the wire (in Amperes, A).
  • \(L\) stands for the length of the wire (in meters, m).
  • \(B\) denotes the magnetic field's magnitude (in Teslas, T).
  • \(\theta\) is the angle between the wire and the magnetic field (in degrees or radians).
In the given exercise, the formula allowed us to find the earth's magnetic field strength by manipulating the equation to solve for \(B\). Once all other values were substituted, and the sine of the angle was calculated, solving for \(B\) gave us the desired magnetic field intensity at the wire's location. This process demonstrates how one can apply known physical laws to calculate unseen forces like Earth's magnetic influence on wires.

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Most popular questions from this chapter

Two circular coils are concentric and lie in the same plane. The inner coil contains 140 turns of wire, has a radius of \(0.015 \mathrm{m},\) and carries a current of 7.2 A. The outer coil contains 180 turns and has a radius of \(0.023 \mathrm{m} .\) What must be the magnitude and direction (relative to the current in the inner coil) of the current in the outer coil, so that the net magnetic field at the common center of the two coils is zero?

A charged particle with a charge-to-mass ratio \(|q| / m=5.7 \times 10^{8} \mathrm{C} / \mathrm{kg}\) travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.72 T. How much time does it take for the particle to complete one revolution?

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude \(2470 \mathrm{N} / \mathrm{C},\) directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of \(6.50 \times 10^{3} \mathrm{m} / \mathrm{s},\) enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of \(+4.00 \times 10^{-12} \mathrm{C}\) enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is \(1.90 \times 10^{-9} \mathrm{N},\) pointing directly upward. What is the speed of this particle?

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{T},\) and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{A}\). What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu C\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is \(5.00 \mathrm{cm}\) from the charge and carries a current of \(67.0 \mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.
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