ps10sol

# ps10sol - 8.02 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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Problem Set 10 Solutions p. 1 of 12 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 10 Solutions Problem 1: EMF Due to a Time-Varying Magnetic Field A long narrow coil is surrounded by a short wide coil as shown in the figure at the right. The short wide coil has a diameter d S , n S turns per unit length, and a length S . It has its ends connected through a resistor of resistance R . The long narrow inner coil has a diameter d L , n L turns per unit length, and a length L . It has its ends connected across a variable power source. For each of the partial sentences below, indicate whether they are correctly completed by the phrase greater than (>), less than (<), or the same as (=). If you cannot determine which is the case from the information given, indicate not sufficient information (NSI). The current through the inner coil is increased from 0 to1 Amps over a period of 10 seconds in a smooth fashion according to the rule I L ( t ) = (0.01 A/ 2 s ) 2 t . a. The magnitude of the current in the long narrow coil at time t = 5 s is greater than the current in that coil at time t = 1 s. The current in the long narrow (inner) coil is defined above as quadratically increasing with time. b. The magnitude of the current in the short wide coil at time t = 5 s is greater than the current in that coil at time t = 1 s. The current in the outer coil is mutually induced: ( ) ( ) S L I R M R dI dt ε = = . Since I L is changing quadratically, the induced current will increase at constant rate. c. The magnitude of the current in the long narrow coil at time t = 1 s is ??? (NSI) the current in the short wide coil at that same time. We don’t know what the mutual inductance (the ratio of the number of turns) is so we can’t tell which is bigger. d. If the long narrow coil was compressed to half its length before the current was turned on, the current in the short wide coil would be greater than it was without the compression. The compression will double the turn density in the inner coil and hence double the B field that it creates, doubling the flux through the outer coil and hence doubling the current induced in that coil.

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