- #1
Nick O
- 158
- 8
Homework Statement
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) twice as large
b) one-fourth as large
c) unchanged
d) half as large
e) four times as large
Homework Equations
(1) εL = -L(dI/dt)
(2) ε - IR - L(dI/dt) = 0
The Attempt at a Solution
So, coming to a symbolic solution here is no problem. The answers include no exponential term, so the wire is obviously assumed to have zero resistance. Therefore, |ε| = εL, and I = ε/L * t. Therefore, the current in the second case is half as large as the current in the first case.
What bothers me here is this: R = 0, so by equation (2) the back-emf is equal to the battery's emf at all times. This would imply zero current at all times, wouldn't it? But if current is zero at all times, then the equation would false because dI/dt would be zero.
Could someone clear this up for me? How would an ideal inductor interact with an ideal source of emf with no resistors in between?