HW9solutions-10 - Physics 2213 Q29.6 Homework 9 Solutions...

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Physics 2213 Homework 9 Solutions Spring 2010 Q29.6 [Magnet & Copper Pipe] Imagine the copper pipe to be made of thin horizontal circular rings piled on top of each other. In the drawing to the right, we show the effect of the falling magnet on a segment above the magnet as well as the effect on one below, as discussed in the lecture. The drawing considers the case that the magnet's north pole is on top and its south pole on the bottom. The ring above the magnet has an upward pointing magnetic field going through it. Since the magnet is falling away from this ring, this field is decreasing, so by Lenz’s law, there is an induced emf in the upper ring which creates an induced electric current CCW around the pipe as seen from above and an upward pointing induced magnetic field through the middle of the pipe. This is essentially the same as having a magnetic dipole at this ring with north pole on top. This dipole will attract the falling magnet, slowing its fall. We can do a similar analysis to a ring below the magnet. In this case, there is an increasing upward magnetic field at the ring due to the approaching falling magnet. This induces a CW electric current as seen from above and a magnetic field with south pole facing up. This field will repel the falling magnet, again slowing its fall. The force on the magnet will be proportional to the induced magnetic field, which is proportional to the induced emf, which is proportional to the rate of change of magnetic flux, which is proportional to the magnet’s fall speed. Thus, we have an upward force F mag =cv , where v is the magnet’s speed, and c is a constant. The gravitational force on the magnet is F grav =gm , a constant. For small v , F grav > F mag . However, as the velocity continues to increase, eventually F grav = F mag (when v=gm/c ), and the acceleration will be zero. This is the terminal velocity. 29.17. [Coils & Currents] The next three problems can be solved by applying Lenz's law, which states that the flux of the induced current tends to oppose the change in flux. (a) With the switch closed the magnetic field of coil A is to the right at the location of coil B. When the switch is opened the magnetic field of coil A goes away. Hence by Lenz's law the field of the current induced in coil B is to the right, to oppose the decrease in the flux in this direction. To produce magnetic field that is to the right the current in the circuit with coil B must flow through the resistor in the direction a to b . (b) With the switch closed the magnetic field of coil A is to the right at the location of coil B.
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This note was uploaded on 06/08/2010 for the course PHYS 213 taught by Professor Perelstein during the Spring '08 term at Cornell University (Engineering School).

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HW9solutions-10 - Physics 2213 Q29.6 Homework 9 Solutions...

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