Skip to main content
Ch. 27 - Magnetism
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 26, Problem 40

How much work is required to rotate the current loop (Fig. 27–23) in a uniform magnetic field B\(\overrightarrow{B}\) from (a) θ = 0° (μ\(\overrightarrow{\mu}\) ∣∣ B\(\overrightarrow{B}\)) to θ = 180°, (b) θ = 90° to θ = -90°.

Verified step by step guidance
1
Understand the problem: The work required to rotate a current loop in a uniform magnetic field is related to the change in potential energy of the magnetic dipole moment (μ) in the field. The potential energy is given by U = -μ·B·cos(θ), where θ is the angle between the magnetic moment (μ) and the magnetic field (B).
Set up the work equation: The work done to rotate the loop is equal to the change in potential energy, W = ΔU = U_final - U_initial. For each part of the problem, calculate the initial and final potential energies using U = -μ·B·cos(θ).
For part (a): When θ changes from 0° to 180°, calculate U_initial = -μ·B·cos(0°) and U_final = -μ·B·cos(180°). Substitute these values into W = U_final - U_initial to find the work required.
For part (b): When θ changes from 90° to -90°, calculate U_initial = -μ·B·cos(90°) and U_final = -μ·B·cos(-90°). Again, substitute these values into W = U_final - U_initial to determine the work required.
Simplify the expressions: Use the trigonometric values for cos(0°), cos(180°), cos(90°), and cos(-90°) to simplify the calculations. Note that cos(0°) = 1, cos(180°) = -1, cos(90°) = 0, and cos(-90°) = 0. This will help you finalize the expressions for the work in both parts (a) and (b).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
11m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Work in a Magnetic Field

In physics, work is defined as the energy transferred when a force is applied over a distance. When dealing with magnetic fields, the work done on a magnetic dipole (like a current loop) in a magnetic field is related to the angle between the dipole moment and the magnetic field. The work can be calculated using the formula W = -μ·B·cos(θ), where μ is the magnetic moment, B is the magnetic field strength, and θ is the angle between them.
Recommended video:
Guided course
05:30
Magnetic Fields and Magnetic Dipoles

Magnetic Moment

The magnetic moment is a vector quantity that represents the strength and direction of a magnetic source. For a current loop, the magnetic moment (μ) is given by the product of the current (I) flowing through the loop and the area (A) of the loop, expressed as μ = I·A. The orientation of the magnetic moment relative to an external magnetic field influences the torque and work done when the loop is rotated.
Recommended video:
Guided course
03:08
Intro To Dipole Moment

Torque in a Magnetic Field

Torque (τ) is the rotational equivalent of force and is crucial when analyzing the motion of a current loop in a magnetic field. The torque experienced by a magnetic dipole in a magnetic field is given by τ = μ × B, where the cross product indicates that the torque depends on the angle between the magnetic moment and the magnetic field. This torque causes the loop to rotate, and understanding it is essential for calculating the work done during the rotation.
Recommended video:
Guided course
05:30
Magnetic Fields and Magnetic Dipoles
Related Practice
Textbook Question

\(\What\) is the value of q/m for a particle that moves in a circle of radius 8.0 mm in a 0.46-T magnetic field if a crossed 320-V/m electric field will make the path straight?

1311
views
Textbook Question

For a particle of mass m and charge q moving in a circular path in a magnetic field B, (a) show that its kinetic energy is proportional to r², the square of the radius of curvature of its path. Show that its angular momentum is L=qBr² , around the center of the circle.

1414
views
Textbook Question

A long copper strip is 3.0 cm wide and thick. When it carries a steady 42-A current in a 0.80-T magnetic field it produces a 6.5-μV Hall emf. Determine:

(a) the Hall field in the conductor;

(b) the drift speed of the conduction electrons;

(c) the density of free electrons in the metal.

1264
views
Textbook Question

A particle of charge q moves in a circular path of radius r in a uniform magnetic field B\(\overrightarrow{B}\). If the magnitude of the magnetic field is doubled, and the kinetic energy of the particle remains constant, what happens to the angular momentum of the particle?

1838
views
Textbook Question

A 720-KeV (kinetic energy) proton enters a 0.20-T field, in a plane perpendicular to the field. What is the radius of its path? See Section 23–8.

1505
views
Textbook Question

A circular coil 18.0 cm in diameter and containing twelve loops lies flat on the ground. The Earth’s magnetic field at this location has magnitude 5.50 x 10⁻⁵ T and points into the Earth at an angle of 54.0° below a line pointing due north. If a 7.10-A clockwise current passes through the coil, (a) determine the torque on the coil; (b) which edge of the coil rises up : north, east, south, or west?

1211
views