A sheet of copper is placed between the poles of an
The reason why a force is required to pull or insert a copper sheet placed in between the poles of an electromagnet.
Explanation of Solution
A force is required to pull or insert a copper sheet between the electromagnet poles as per Lenz law due to an induced current formed due to the displacement of the sheet.
As per Lenz law, an induced emf gives rise to a current whose magnetic field opposes the cause inducing a change. This can be applied to the given situation.
As per the given data, when the copper sheet is pulled from the poles of electromagnet there will be a change in the magnetic flux. This magnetic flux change induces currents in the copper sheet.
This induced current opposes the change which induces it to flow. That is as per Lenz law, induced current opposes the moving copper sheet which is the primary cause for these currents. Hence there is a difficulty in pulling the copper sheet and thereby requires a force to pull the copper sheet.
A similar sequence arises when we insert a copper sheet between the electromagnet poles. This means there is an induced current due to magnetic flux which in turn is due to this moving copper sheet. Hence, this current opposes the moving of this copper sheet, both insertion and exertion, between two electromagnet poles.
Conclusion:
Thus, a force is required to pull or insert a copper sheet between the electromagnet poles as per Lenz law.
Want to see more full solutions like this?
Chapter 29 Solutions
University Physics with Modern Physics (14th Edition)
Additional Science Textbook Solutions
Applied Physics (11th Edition)
Lecture- Tutorials for Introductory Astronomy
University Physics (14th Edition)
Physics for Scientists and Engineers with Modern Physics
Conceptual Physics (12th Edition)
Physics: Principles with Applications
- (a) What is the angle between a wire carrying an 8.00-A current and the 1.20-T field it is in if 50.0 cm of the wire experiences a magnetic force of 2.40 N? (b) What is the force on the wire if it is rotated to make an angle of 90° with the field?arrow_forward(a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat part (a), assuming the moving particle is an electron. Figure P19.2 Problems 2 and 22.arrow_forwardA packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R = 0.500 cm. If each wire carries 2.00 A, what are (a) the magnitude and (b) the direction of the magnetic force per unit length acting on a wire located 0.200 cm from the center of the bundle? (c) What If? Would a wire on the outer edge of the bundle experience a force greater or smaller than the value calculated in parts (a) and (b)? Give a qualitative argument for your answer.arrow_forward
- Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46arrow_forwardDetermine the magnetic field (in terms of I, a, and d) at the origin due to the current loop in Figure P29.9. The loop extends to infinity above the figure. Figure P29.9arrow_forwardA long, straight wire carries a current I (Fig. OQ30.8). Which of the following statements is tine regarding the magnetic field due to the wire? More than one statement may be correct, (a) The magnitude is proportional to I/r, and the direction is out of the page at P. (b) The magnitude is proportional to I/r2, and the direction is out of the page at P. (c) The magnitude is proportional to I/r, and the direction is into the page at P. (d) The magnitude is proportional to I/r2, and the direction is into the page at P. (e) The magnitude is proportional to I, but does not depend on r.arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning