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Phy 122
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Set 2 Number 2
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Problem set 2 #2 states:
The potential difference is not given, so call it V. The electric field or potential gradient in the first wire is V / .4 m; in the second wire it is V / 2.06 m. The potential gradient in the second wire is therefore (V / 2.06) / (V / .4) = .4 / 2.06 = .1941 times that in the first.
It follows that the drift velocity of the electrons in the second wire is 3.6 times that in the first. The wires are identical: except for length they each have the same number of current carriers available per unit length. So the current, or the rate at which charges pass a point in the second wire, is 3.6 times that in the first. The current in the second wire is therefore .1941 * 3.6 amps = .6987 amps.
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Shouldn't it say...
It follows that the drift velocity of the electrons in the second wire is .1941 times that in the first.
If so, I get this problem. If the 3.6 is correct, I don't get it.
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@&
Yes, the drift velocity should be .19 times that in the first wire.
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