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Chapter 27: Problem 38

A straight, vertical wire carries a current of 2.60 A downward in a region between the poles of a large superconducting electromagnet, where the magnetic field has magnitude \(B =\) 0.588 T and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00-cm section of the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) 30.0\(^\circ\) south of west?

Short Answer

Expert verified
(a) 0.0153 N north, (b) 0.0153 N west, (c) 0.0132 N northeast.

Step by step solution

01

Understand the Problem

The problem involves calculating the magnetic force on a vertical wire in a magnetic field. The wire carries a current of 2.60 A, and we need to find the force on a 1.00 cm (0.01 m) segment of the wire. The magnetic force is affected by the direction of the magnetic field.
02

Use the Force Formula

The force on a current-carrying wire in a magnetic field is given by the formula: \[ F = I L B \sin(\theta) \]where:- \( F \) is the force in Newtons,- \( I = 2.60 \) A is the current,- \( L = 0.01 \) m is the length of the wire segment,- \( B = 0.588 \) T is the magnetic field,- \( \theta \) is the angle between the wire and the magnetic field.
03

Solve for Magnetic Field East

If the magnetic field is directed east, and given that the current is downward, the angle \( \theta \) between the wire and the magnetic field is 90 degrees (since they are perpendicular). Thus, \( \sin(90^\circ) = 1 \). Substitute into the formula:\[ F = 2.60 \times 0.01 \times 0.588 \times 1 = 0.015288 \] N.Using the right-hand rule, the direction of the force will be north.
04

Solve for Magnetic Field South

For a magnetic field directed south, the direction makes the angle \( \theta = 90^\circ \) with the vertical wire. Thus, \( \sin(90^\circ) = 1 \). Substitute into the formula:\[ F = 2.60 \times 0.01 \times 0.588 \times 1 = 0.015288 \] N.Using the right-hand rule, the force direction will be west.
05

Solve for Magnetic Field 30 Degrees South of West

For a magnetic field directed 30 degrees south of west, the direction makes an angle \( \theta = 90^\circ - 30^\circ = 60^\circ \) with the vertical wire. Thus, \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \). Substitute into the formula:\[ F = 2.60 \times 0.01 \times 0.588 \times \frac{\sqrt{3}}{2} = 0.013242 \] N.Using the right-hand rule, the force is directed toward the north-east direction, given the 30掳 offset.

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Key Concepts

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

Superconducting Electromagnet
Superconducting electromagnets are magnets made from coils of superconducting wire. These magnets are unique because they can carry electricity without losing any energy to resistance. Unlike regular electromagnets, they can generate significantly stronger magnetic fields.
In our exercise, the wire is placed between the poles of a superconducting electromagnet. This means that the magnetic field is both strong and steady. As a result, the interaction between the wire's current and the magnet's magnetic field creates a noticeable force.
Superconducting electromagnets are often used in scientific research and various industrial applications. These magnets are particularly beneficial when a high and consistent magnetic field strength is necessary for precision work or large-scale applications.
Right-Hand Rule
The right-hand rule is a helpful guideline in physics for determining the direction of a magnetic force on a current-carrying wire. It's easy to use, just follow these steps:
  • Point your thumb in the direction of the current's flow. In our case, that's downward.
  • Extend your fingers in the direction of the magnetic field. Depending on the scenario, this could be east, south, or 30 degrees south of west.
  • Your palm will indicate the direction of the magnetic force.

So, in our solutions:
  • For a magnetic field pointing east, the force points north.
  • If the field is south, the force is directed west.
  • At 30 degrees south of west, the force is toward the north-east.
The right-hand rule is a reliable tool for remembering how forces act in electromagnetic contexts.
Angle Between Wire and Magnetic Field
In calculating magnetic force, the angle between the wire and the magnetic field is crucial because it affects the intensity of the force. This is represented mathematically by \( \sin(\theta) \), where \( \theta \) is the angle.
Consider these scenarios:
  • If the wire and the field are perpendicular, like in parts (a) and (b) of the solution, \( \theta = 90^\circ \) leading to \( \sin(90^\circ) = 1 \). This means the force is at its maximum possible strength.
  • When the field moves to 30 degrees south of west, \( \theta = 60^\circ \), as the wire is 90 degrees minus the 30-degree deviation. This alignment gives \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \).
To find the magnetic force, use \[ F = I L B \sin(\theta) \]. This shows how the orientation of the magnetic field directly influences the result by modulating the effective strength of interaction.

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Most popular questions from this chapter

\(\textbf{Determining Diet.}\) One method for determining the amount of corn in early Native American diets is the \(stable\) \(isotope\) \(ratio\) \(analysis\) (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \(^{12}\)C and \(^{13}\)C isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed 8.50 km /s, and you want to bend them within a uniform magnetic field in a semicircle of diameter 25.0 cm for the \(^{12}\)C. The measured masses of these isotopes are 1.99 \(\times\) 10\(^{-26}\) kg (\(^{12}\)C) and 2.16 \(\times\) 10\(^{-26}\) kg (\(^{13}\)C). (a) What strength of magnetic field is required? (b) What is the diameter of the \(^{13}\)C semicircle? (c) What is the separation of the \(^{12}\)C and \(^{13}\)C ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

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The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes \(^{15}\)N to \(^{14}\)N in bone from human remains. Carnivores concentrate \(^{15}\)N, so this ratio tells archaeologists how much meat was consumed. For a mass spectrometer that has a path radius of 12.5 cm for \(^{12}\)C ions (mass 1.99 \(\times\) 10\(^{-26}\) kg), find the separation of the \(^{14}\)N 1mass 2.32 \(\times\) 10\(^{-26}\) kg2 and 15N (mass 2.49 \(\times\) 10\(^{-26}\) kg) isotopes at the detector.
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