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
- air is pushed steadily though a forced air pipe at a steady speed of 4.0 m/s. the pipe measures 56 cm by 22 cm. how fast will air move though a narrower portion of the pipe that is also rectangular and measures 32 cm by 22 cmarrow_forwardNo chatgpt pls will upvotearrow_forward13.87 ... Interplanetary Navigation. The most efficient way to send a spacecraft from the earth to another planet is by using a Hohmann transfer orbit (Fig. P13.87). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the depar- ture planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet. (a) For a flight from earth to Mars, in what direction must the rockets be fired at the earth and at Mars: in the direction of motion, or opposite the direction of motion? What about for a flight from Mars to the earth? (b) How long does a one- way trip from the the earth to Mars take, between the firings of the rockets? (c) To reach Mars from the…arrow_forward
- No chatgpt pls will upvotearrow_forwarda cubic foot of argon at 20 degrees celsius is isentropically compressed from 1 atm to 425 KPa. What is the new temperature and density?arrow_forwardCalculate the variance of the calculated accelerations. The free fall height was 1753 mm. The measured release and catch times were: 222.22 800.00 61.11 641.67 0.00 588.89 11.11 588.89 8.33 588.89 11.11 588.89 5.56 586.11 2.78 583.33 Give in the answer window the calculated repeated experiment variance in m/s2.arrow_forward
- How can i solve this if n1 (refractive index of gas) and n2 (refractive index of plastic) is not known. And the brewsters angle isn't knownarrow_forward2. Consider the situation described in problem 1 where light emerges horizontally from ground level. Take k = 0.0020 m' and no = 1.0001 and find at which horizontal distance, x, the ray reaches a height of y = 1.5 m.arrow_forward2-3. Consider the situation of the reflection of a pulse at the interface of two string described in the previous problem. In addition to the net disturbances being equal at the junction, the slope of the net disturbances must also be equal at the junction at all times. Given that p1 = 4.0 g/m, H2 = 9.0 g/m and Aj = 0.50 cm find 2. A, (Answer: -0.10 cm) and 3. Ay. (Answer: 0.40 cm)please I need to show all work step by step problems 2 and 3arrow_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