Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
bartleby

Videos

Question
Book Icon
Chapter 20, Problem 64P

(a)

To determine

The magnetic force on side 2 and 4 at the instant shown in Figure 20.7.

(a)

Expert Solution
Check Mark

Answer to Problem 64P

The magnetic force on side 2 at the instant shown in Figure 20.7 is F2=ωB2ALRsinωt,down_ and that on side 4 is F4=ωB2ALRsinωt,up_.

Explanation of Solution

In side 2, current flows into the page and in side 4, current flows out of the page.

Given that the current in the system is,

  I(t)=ωBARsinωt                                                                                                      (I)

Write the expression for the magnetic force due to the current carrying loop.

  F=IL×B                                                                                                                  (II)

Here, F is the magnetic force, I is the current, L is the side length of the loop, and B is the magnetic field.

L and B are always perpendicular for both sides. Thus, the magnitude of the magnetic force on each side is same and is obtained as,

  F=ILB (III)

Use equation (I) in (III).

  F=ωBARLBsinωt=ωB2ALRsinωt

The direction of the magnetic force is determined using the right-hand rule, and applying right hand rule yields that the magnetic force on side 2 is directed downward and that on side 4 is directed upward.

Conclusion:

Therefore, the magnetic force on side 2 at the instant shown in Figure 20.7 is F2=ωB2ALRsinωt,down_ and that on side 4 is F4=ωB2ALRsinωt,up_.

(b)

To determine

The reason for which the magnetic forces on side 1 and 3 do not cause a torque about the axis of rotation.

(b)

Expert Solution
Check Mark

Answer to Problem 64P

The magnetic forces on side 1 and 3 are always parallel to the axis of rotation, and hence they cannot produce a torque about the axis.

Explanation of Solution

If the forces acting on a body is parallel to the axis of rotation of the body, the forces cannot produce a torque on the body. In the given system, the magnetic forces on side 1 and 3 are parallel to the axis of rotation. Hence, they cannot cause a torque about the axis.

Conclusion:

Therefore, the magnetic forces on side 1 and 3 are always parallel to the axis of rotation, and hence they cannot produce a torque about the axis.

(c)

To determine

The torque on the loop about its axis of rotation.

(c)

Expert Solution
Check Mark

Answer to Problem 64P

The torque on the loop about its axis of rotation is τ=2ωrB2ALRsin2ωt,counterclockwise_.

Explanation of Solution

The diagram depicting the forces on the loop is shown in Figure 1.

Physics, Chapter 20, Problem 64P

Write the expression for the total torque on the loop using the diagram.

  τ=(F2sinθ)r+(F4sinθ)r                                                                              (IV)

Here, τ is the torque, θ is the angular displacement, and r is the half of side length of the loop.

The magnitude of magnetic force F is same on side 2 and 4 and thus replace F2 and F4 by F in equation (IV).

  τ=(Fsinθ)r+(Fsinθ)r=2Frsinθ                                                                                (V)

Write the expression for the angular displacement.

  θ=ωt                                                                                                                   (VI)

Here, ω is the angular speed, and t is the time.

From part (a) the magnetic force is obtained as,

  F=ωB2ALRsinωt                                                                                               (VII)

Use equation (VI) and (VII) in(V).

  τ=2(ωB2ALRsinωt)rsinωt=2ωrB2ALRsin2ωt

The direction of forces indicates that the torque on the loop is in the counter clockwise direction.

Conclusion:

Therefore, the torque on the loop about its axis of rotation is τ=2ωrB2ALRsin2ωt,counterclockwise_.

(d)

To determine

Whether the magnetic torque make the loop increase or decrease its angular velocity, in the absence of other torques.

(d)

Expert Solution
Check Mark

Answer to Problem 64P

In the absence of other torques the magnetic torque make the loop decrease its angular velocity.

Explanation of Solution

In the given system, the loop has the angular velocity in the clockwise direction whereas the torque exerted by the magnetic force is obtained as in the counterclockwise direction. Since the direction of magnetic torque is opposite to the direction of the angular velocity, the magnetic torque would tend to decrease the angular speed of the loop.

Conclusion:

Therefore, in the absence of other torques the magnetic torque make the loop decrease its angular velocity.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 20 Solutions

Physics

Ch. 20.6 - Practice Problem 20.7 An Ideal Transformer An...Ch. 20.7 - Conceptual Practice Problem 20.8 Choosing a Core...Ch. 20.9 - CHECKPOINT 20.9 Five solenoids are wound with...Ch. 20.9 - Practice Problem 20.9 Power in an Inductor The...Ch. 20.10 - Prob. 20.10CPCh. 20.10 - Prob. 20.10PPCh. 20 - Prob. 1CQCh. 20 - Prob. 2CQCh. 20 - Prob. 3CQCh. 20 - Prob. 4CQCh. 20 - Prob. 5CQCh. 20 - Prob. 6CQCh. 20 - Prob. 7CQCh. 20 - Prob. 8CQCh. 20 - Prob. 9CQCh. 20 - Prob. 10CQCh. 20 - Prob. 11CQCh. 20 - Prob. 12CQCh. 20 - Prob. 13CQCh. 20 - Prob. 14CQCh. 20 - Prob. 15CQCh. 20 - Prob. 16CQCh. 20 - Prob. 17CQCh. 20 - Prob. 18CQCh. 20 - Prob. 19CQCh. 20 - Prob. 1MCQCh. 20 - Prob. 2MCQCh. 20 - Prob. 3MCQCh. 20 - Prob. 4MCQCh. 20 - Prob. 5MCQCh. 20 - Prob. 6MCQCh. 20 - Prob. 7MCQCh. 20 - Prob. 8MCQCh. 20 - Prob. 9MCQCh. 20 - Prob. 10MCQCh. 20 - Prob. 1PCh. 20 - Prob. 2PCh. 20 - Prob. 3PCh. 20 - 4. In Fig. 20.2, what would the magnitude (in...Ch. 20 - Prob. 5PCh. 20 - 6. The armature of an ac generator is a circular...Ch. 20 - Prob. 7PCh. 20 - 8. A solid copper disk of radius R rotates at...Ch. 20 - 9. A horizontal desk surface measures 1.3 m × 1.0...Ch. 20 - 10. A square loop of wire, 0.75 m on each side,...Ch. 20 - 11. A long straight wire carrying a steady current...Ch. 20 - 12. A long straight wire carrying a current I is...Ch. 20 - Prob. 13PCh. 20 - 14. While I1 is increasing, what is the direction...Ch. 20 - 15. While I1 is constant, does current flow in...Ch. 20 - 16. A circular conducting coil with radius 3.40 cm...Ch. 20 - Prob. 17PCh. 20 - Prob. 18PCh. 20 - 19. In the figure, switch s is initially open. It...Ch. 20 - 20. Crocodiles are thought to be able to detect...Ch. 20 - 21. A bar magnet approaches a coil as shown, (a)...Ch. 20 - 22. Another example of motional emf is a rod...Ch. 20 - 23. Two loops of wire are next to each other in...Ch. 20 - 24. A dc motor has coils with a resistance of 16 Ω...Ch. 20 - Prob. 25PCh. 20 - Prob. 26PCh. 20 - Prob. 27PCh. 20 - Prob. 28PCh. 20 - 29. A doorbell uses a transformer to deliver an...Ch. 20 - Prob. 30PCh. 20 - 31. When the emf for the primary of a transformer...Ch. 20 - 32. A transformer with a primary coil of 1000...Ch. 20 - Prob. 33PCh. 20 - Prob. 34PCh. 20 - 35. A 2 m long copper pipe is held vertically....Ch. 20 - Prob. 36PCh. 20 - Prob. 37PCh. 20 - Prob. 38PCh. 20 - 39. A solenoid of length 2.8 cm and diameter 0.75...Ch. 20 - Prob. 40PCh. 20 - Prob. 41PCh. 20 - Prob. 42PCh. 20 - Prob. 43PCh. 20 - 44. The current in a 0.080 H solenoid increases...Ch. 20 - Prob. 45PCh. 20 - Prob. 46PCh. 20 - Prob. 47PCh. 20 - Prob. 48PCh. 20 - Prob. 49PCh. 20 - Prob. 50PCh. 20 - Prob. 51PCh. 20 - Prob. 52PCh. 20 - Prob. 53PCh. 20 - Prob. 54PCh. 20 - Prob. 55PCh. 20 - Prob. 56PCh. 20 - Prob. 57PCh. 20 - Prob. 58PCh. 20 - Prob. 59PCh. 20 - Prob. 60PCh. 20 - Prob. 61PCh. 20 - Prob. 62PCh. 20 - Prob. 63PCh. 20 - Prob. 64PCh. 20 - Prob. 65PCh. 20 - Prob. 66PCh. 20 - 67. Switch S2 has been closed for a long time, (a)...Ch. 20 - Prob. 68PCh. 20 - Prob. 69PCh. 20 - Prob. 70PCh. 20 - Prob. 71PCh. 20 - 72. A uniform magnetic field of magnitude 0.29 T...Ch. 20 - Prob. 73PCh. 20 - Prob. 74PCh. 20 - Prob. 75PCh. 20 - Prob. 76PCh. 20 - Prob. 77PCh. 20 - Prob. 78PCh. 20 - Prob. 79PCh. 20 - Prob. 80PCh. 20 - Prob. 81PCh. 20 - Prob. 82PCh. 20 - Prob. 83PCh. 20 - Prob. 84PCh. 20 - Prob. 85PCh. 20 - Prob. 86PCh. 20 - Prob. 87PCh. 20 - Prob. 88PCh. 20 - Prob. 89PCh. 20 - Prob. 90P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
What is Electromagnetic Induction? | Faraday's Laws and Lenz Law | iKen | iKen Edu | iKen App; Author: Iken Edu;https://www.youtube.com/watch?v=3HyORmBip-w;License: Standard YouTube License, CC-BY