# How to draw + and - and $\nabla \times E$ on a circular wire?

Faraday's Law: $$\vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$

Circulation of electric field:

1. For time-varying magnetic field and a closed wire, How can we add + and - pole signs that indicates that there is a voltage and current on that wire?

2. Is there a single + and - for every microscopic circulation on the wire?

There are no + or - poles. That is implicit in the statement that says:

So in other words going from the + to the - pole of a battery in a circuit should correspond to a fix energy drop for any path taken, which is not the case here. Consider for example a more realistic wire, such as one with two resistors in series, where would you put the poles in that case?.

• Yes. But + and - pole also indicates there is a charge difference somewhere in the circuit. And in that case it is in the battery. But for circular wire case, how can we indicate charge difference on wire? Even the current caused by time-varying magnetic field, there must be charge differences to make electrons flow. Right? Commented Dec 7, 2015 at 15:09
• Strictly speaking the poles of a battery stand for a difference in potential, not charge: An ideal battery has no parasitic capacitance. There is also no charge gradient in your circular wire, the electrons flow due to induced $E$-field.
– Rol
Commented Dec 7, 2015 at 18:34
• Yes but both of them are electric fields eventually. There must be attraction or repulsion. I think I found the answer. I am not sure. Can we say: According to relativity, magnet becomes + pole and wire become - pole. So there is a current on the wire. I thought + and - pole must be on the wire but in fact one of them is on the magnet and the other is on the wire. Right? Commented Dec 7, 2015 at 20:03
• I don't understand what you are talking about...
– Rol
Commented Dec 8, 2015 at 15:51