Concept explainers
Part (a)
Force experienced by electric charge at rest when placed in a magnetic field of a magnet.
Solution: Electric charge at rest will not experience any force.
Part (a)
Explanation of Solution
Static Electric charge will not experience any force because according to the Lorentz relation
Conclusion: An external magnetic field always exerts force of other magnets and charge particles.
Part (b)
To determine: - Force experienced by moving electric charge when placed in magnetic field of a magnet.
Part (b)
Answer to Problem 1OQ
Solution: - Moving electric charge may or may not experience a force.
Explanation of Solution
Explanation: - Dynamic Electric charge will not experience a force if its direction is parallel or anti-parallel to the direction of magnetic field. And with respect to any other direction it will experience a force.
According to Lorentz relation,
For parallel (
So,
Conclusion: - An external magnetic field always a force on other magnets and charge particles.
Part (c)
To Determine: - Force experienced by
Part (c)
Answer to Problem 1OQ
Solution: - Electric current in a wire will experience a force.
Explanation of Solution
Explanation: - Electric current in a wire will experience a force because current is only due to moving charges and magnetic field always exerts a force on moving charges so, conductor will also experience a force.
Conclusion: - Every conductor experiences a force given by a relation
Part (d)
To Determine: - Force experienced by another magnet when placed in magnetic field of a magnet.
Part (d)
Answer to Problem 1OQ
Solution: - Another magnet will experience a force.
Explanation of Solution
Explanation: - Another magnet always experiences a force because it is a natural property of a magnet, but this force can be repulsive or attractive depending on the nature of poles of a magnet.
Conclusion: - Magnetic force between the magnets is inversely proportional to the distance between them.
Want to see more full solutions like this?
Chapter 20 Solutions
Physics
- Make a plot of the acceleration of a ball that is thrown upward at 20 m/s subject to gravitation alone (no drag). Assume upward is the +y direction (and downward negative y).arrow_forwardLab Assignment #3 Vectors 2. Determine the magnitude and sense of the forces in cables A and B. 30° 30° 300KN 3. Determine the forces in members A and B of the following structure. 30° B 200kN Name: TA: 4. Determine the resultant of the three coplanar forces using vectors. F₁ =500N, F₂-800N, F, 900N, 0,-30°, 62-50° 30° 50° F₁ = 500N = 900N F₂ = 800Narrow_forwardLab Assignment #3 Vectors Name: TA: 1. With the equipment provided in the lab, determine the magnitude of vector A so the system is in static equilibrium. Perform the experiment as per the figure below and compare the calculated values with the numbers from the spring scale that corresponds to vector A. A Case 1: Vector B 40g Vector C 20g 0 = 30° Vector A = ? Case 2: Vector B 50g Vector C = 40g 0 = 53° Vector A ? Case 3: Vector B 50g Vector C 30g 0 = 37° Vector A = ?arrow_forward
- Three point-like charges are placed at the corners of an equilateral triangle as shown in the figure. Each side of the triangle has a length of 20.0 cm, and the point (A) is located half way between q1 and q2 along the side. Find the magnitude of the electric field at point (A). Let q1=-1.30 µC, q2=-4.20µC, and q3= +4.30 µC. __________________ N/Carrow_forwardNo chatgpt pls will upvotearrow_forwardNo chatgpt plsarrow_forward
- The position of a coffee cup on a table as referenced by the corner of the room in which it sits is r=0.5mi +1.5mj +2.0mk . How far is the cup from the corner? What is the unit vector pointing from the corner to the cup?arrow_forwardNo chatgpt plsarrow_forwardFind the total capacitance in micro farads of the combination of capacitors shown in the figure below. HF 5.0 µF 3.5 µF №8.0 μLE 1.5 µF Ι 0.75 μF 15 μFarrow_forward
- the answer is not 0.39 or 0.386arrow_forwardFind the total capacitance in micro farads of the combination of capacitors shown in the figure below. 2.01 0.30 µF 2.5 µF 10 μF × HFarrow_forwardI do not understand the process to answer the second part of question b. Please help me understand how to get there!arrow_forward
- College PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningUniversity Physics (14th Edition)PhysicsISBN:9780133969290Author:Hugh D. Young, Roger A. FreedmanPublisher:PEARSONIntroduction To Quantum MechanicsPhysicsISBN:9781107189638Author:Griffiths, David J., Schroeter, Darrell F.Publisher:Cambridge University Press
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningLecture- Tutorials for Introductory AstronomyPhysicsISBN:9780321820464Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina BrissendenPublisher:Addison-WesleyCollege Physics: A Strategic Approach (4th Editio...PhysicsISBN:9780134609034Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart FieldPublisher:PEARSON