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A long, fine wire is wound into a coil with inductance 5 mH. . .

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A long, fine wire is wound into a coil with inductance 5 mH. The coil is connected across the terminals of a battery, and the current is measured a few seconds after the connection is made. The wire is unwound and wound again into a different coil with L = 10 mH. This second coil is connected across the same battery, and the current is measured in the same way. Compared with the current in the first coil, is the current in the second coil

A. unchanged

B. half as large

C. one-fourth as large

D. twice as large

E. four times as large

Relevant equation: L = N(phiB)/I

Solving for I, I = N(phiB)/L. So why is it that when L changes from 5 mH to 10mH that the current isn't half as large? Assuming N(phiB) doesn't change from one coil to the next, the ratio of I2(current in second coil) to I1(current in first coil) is 1/2.

Am I allowed to assume N(phiB) doesn't change from one coil to the next? Or would N also double since N and L are directly related in the original equation, which would make the current unchanged?

 

 

asked Oct 26, 2014 in Intro College Physics by SciGirl (180 points)

1 Answer

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For a conducting wire, the resistance is quite small, typically under 1 ohm. Therefore the time constant in this case is in the order of ms. After a few seconds, the current will be the steady state current so it only depends on the resistance of the wire. In both cases, the same length of wire is used and the resistance of the wire is the same. So the current would be the same.
answered Oct 26, 2014 by PhysicsDude (21,750 points)
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