How is the loop law applicable when I connect a battery with an ideal copper wire?

We would have a potential rise but where is potential drop? Considering wire to be having negligible resistance.

• Possible duplicates: physics.stackexchange.com/q/8675/2451 , physics.stackexchange.com/q/80400/2451 and links therein. Commented Jun 15, 2017 at 14:38
• not answered there!! Commented Jun 15, 2017 at 15:15
• your answer will be appreciated Commented Jun 15, 2017 at 15:15
• As I answered at physics.stackexchange.com/a/271068/73490 the ideal voltage source is analogous to an "unstoppable force" and the ideal wire is analogous to an "immovable object" and hence what you're asking is directly, "what happens when an unstoppable force collides with an immovable object"? If you don't add something which helps to resolve the paradox physically, then you will get an unphysical answer: an infinite current flows over the wire so that Ohm's law says $V = I\cdot R = \infty\cdot 0,$ which is an indeterminate form and can be thought to have a finite value. Commented Jun 15, 2017 at 16:24
• This leads me to a next question- what is the work of voltage in a wire/ Commented Jun 15, 2017 at 17:47

There would be a potential drop. If the wire has literally $0$ resistance, then this is a nonphysical question and no physical answer can be provided.
Energy will be lost due to heat in the wire. The current will will be very high, as $I=\frac{V}{R}$ and $R$ is very low, and the power dissipated (heat) will be very high, as $P=IV$.