If a conducting loop is placed in an increasing magnetic field, current starts flowing through it due to induced EMF. But similar to a conducting wire, if the loop is considered to have 0 resistance (and should also have same potential at any point) why should current flow through it?
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
The electric potential can only be defined for a conservative electric field, and by Faraday's law, the electric field is not conservative in a region with a time-varying magnetic field. So, it does not make sense to talk about the potential in this case.
For a loop of finite volume and conductivity, current flows according to Ohm's Law, $\mathbf{J} = \sigma \mathbf{E}$, where $\mathbf{E}$ is the induced electric field satisfying $\oint \mathbf{E} \cdot d\mathbf{l} = -\dfrac{d}{dt}\oint \mathbf{B} \cdot d\mathbf{s}$. As the conductivity is increased, the current concentrates near the boundaries of the conductor, in what's known as the "skin effect". A small electric field, mostly tangential to the boundary, "pushes" the current in the thin skin region. In the limit of infinite conductivity, any time-varying EM field vanishes inside the conductor and all the current flows in the zero-thickness surface that bounds the conductor. The current that flows in this surface needs no electric field to push it along, as the resistance is zero. What determines the surface current is the condition that the EM fields it produces exactly cancel the EM fields that would exist in the region of the conductor if the conductor were absent. From the boundary condition for the magnetic field $\mathbf{H}$, this surface current is $\mathbf{J}_s = \mathbf{\hat{n}} \times \mathbf{H}|_\text{on boundary}$, where $\mathbf{\hat{n}}$ is the outward-pointing unit normal to the conductor boundary. Note that since the electric field is zero inside the conductor, there can be no time-varying magnetic flux through the loop. The infinite conductivity of the loop (an idealization of course), allows immediate cancellation of the magnetic flux through the loop, which makes sense in view of Lenz's law.
$\begingroup$
$\endgroup$
If magnetic field is changing in time, it is associated with so-called induced electric field. This kind of electric field is not conservative field, it cannot be described by the concept of potential.
This electric field, if alone, would push the mobile charge carriers and start current. Without ohmic resistance, the current would keep on increasing. The rate of current increase would depend on self inductance of the circuit L. The higher the L, the slower the current increase rate.
answered Sep 8, 2019 at 12:43
$\begingroup$
$\endgroup$
An increasing magnetic field is a changing field, and will produce some current in a conducting loop. A stationary conducting loop in an unchanging magnetic field will produce no current.
answered Sep 8, 2019 at 8:52
$\begingroup$
$\endgroup$
Its more or less like an ideal wire connected to an ideal battery. The charges rush to make both the terminals at the same potential but fail because the battery is constantly trying to maintain the potential difference by means of re-dox reactions.
Same in this case, the charges move in response to the induced E-field (EMF) to make two points at the same potential but the EMF is again generated due to the relative movement of the two (magnet and wire loop).
Remember you can't use Ohm's law here because your wires have no resistance. However in reality, the current will always be limited due to resistance of the wires.
answered Sep 9, 2019 at 8:23
$\begingroup$
$\endgroup$
The correct answer is that in general there is a scalar - energy - potential that depends on charge. In this case there is no charge. This potential is conservative. Then there is a vector or momentum potential that depends on current. It is not conservative. The time derivative of this potential contributes to the force in charges, hence by the definition implied in the Lorentz field must be called electrical field even though it is caused by currents.
