# EMF induction in a coil placed in a variable magnetic field

Consider a conducting coil placed in a variable magnetic field. By the laws of nature, an EMF will be induced in the coil. Between which two points is the EMF induced in the coil?

If we consider it to be similar to a circuit in which a cell is connected, then in terms of 'energy height diagram' (a plot of amount of energy flowing in the circuit w.r.t the coordinates of the circuit,) it comes out to be the first image in the diagram, while the original coil has 'energy height diagram' as the second figure in the diagram, and the two are not compatible.

How is the EMF induced in the coil? (I agree that there is no such term as EMF in a non-conservative field, but just for the consistency with conservative field, I insist on defining such term. Also, we do usually find the EMF induced between any two random points in the given coil.)

Conservative fields, like gravity and electrostatics, allow you to define a uniquely-valued potential as a function of the coordinates. That's what you have more or less drawn for your battery example.

The electrical potential in a changing magnetic field is non-conservative. That is, the potential depends on the path you take, it's multi-valued with respect to coordinates. If you wind 10 times around a volume in which there is a changing field, then you will see 10 times the voltage than if you had wound only once.

This means you need to be precise in your description of where your meter is, and where you place your measuring leads, before you can talk about potential measurements, as they all form part of the circuit.

Between which two points is the EMF induced in the coil?

It's not. It's induced in closed circuits. Draw a closed circuit. Identify a surface enclosed by the circuit. Look at the rate that the flux crossing that surface changes, that's the emf round the circuit.

In certain simple asymmetric situations where either the flux is very high in some places and negligible in others, or the device under test has hundreds of turns and the DMM leads form only one, then you can approximate what happens to the 'emf between two points'. For instance, a transformer concentrates most of the flux in the core, and a search coil might have many turns, so in both cases it's reasonable to talk about the emf between the coil terminals.

How is the EMF induced in the coil?

First let's answer the question of what EMF would be induced by a magnetic field in the absence of any conductors

To do so, we will define an electric vector potential $$\\vec{W}\$$ and associated electric scalar potential $$\\phi\$$ as fields satisfying the two equations:

$$\vec{E} =-\nabla \phi + \nabla\times\vec{W}$$ $$\nabla\cdot\vec{W}=0$$

Now

$$\nabla\times\vec{E} = \nabla\times(-\nabla U)+\nabla\times(\nabla\times\vec{W})$$

but the curl of a gradient is $$\0\$$, so

$$\nabla\times\vec{E} = \nabla\times(\nabla\times\vec{W})$$

$$\nabla\times\vec{E} = \nabla(\nabla\cdot\vec{W})-\nabla^2\vec{W}$$

But since $$\nabla \cdot \vec{W}=0$$,

$$\nabla\times\vec{E} = -\nabla^2\vec{W}$$

Maxwell's version of Faraday's Law (in vector notation) says

$$\nabla\times\vec{E} = - \frac{\partial\vec{B}}{\partial t}$$

so

$$\nabla^2\vec{W} =\frac{\partial\vec{B}}{\partial t}$$

This is a Poisson equation, with solution:

$$\vec{W}(r)=\frac{1}{4\pi}\int_{R^3}\frac{\partial\vec{B}(r')}{\partial t}\frac{1}{|r-r'|}d^3r'$$

Define

$$\vec{E}_{\partial\vec{B}/\partial t} = \nabla\times\vec{W}$$

The emf induced along some curve $$\C\$$ from $$\a\$$ to $$\b\$$ is

$$\mathscr{E}_{\partial\vec{B}/\partial t} = \int_{C:a\rightarrow b} \vec{E}_{\partial\vec{B}/\partial t} \cdot d\vec{s}$$

Note that this emf is path-dependent.

Now, if we add a conductor to the system, which conductor lies along curve $$\C\$$, but possibly extends beyond $$\C\$$, we will still have the same emf induced as before. But the electric field inside a conductor is governed by the microscopic version of Ohm's Law:

$$\vec{E} = \sigma\vec{J}$$

The time-varying magnetic field wants to induce an electric field $$\\vec{E}_{\partial\vec{B}/\partial t}\$$, but the conductor insists on $$\\vec{E} = \sigma\vec{J}\$$. To reconcile these fields, a surface charge develops on the surface of the conductor. This surface charge ensures that, within the conductor, the microscopic version of Ohm's Law is obeyed.

If the conductor does not form a complete circuit, then there will be no steady-state current, and hence no steady state $$\\vec{E}\$$ field within the conductor. In that case, the surface charge completely cancels the $$\\vec{E}_{\partial\vec{B}/\partial t}\$$ field within the conductor, but creates a conservative $$\\vec{E}\$$ field external to the conductor. This reaction field creates a potential difference between one end of the conductor and the other, even though the ends may be far away from where the time-varying magnetic field is directly acting upon the conductor.

If the conductor does form a closed loop, or is part of a completed circuit, with perhaps other components such as resistors, then current will flow in the conductor. The surface charge doesn't exactly cancel the $$\\vec{E}_{\partial\vec{B}/\partial t}\$$ field within the conductor, but leaves enough $$\\vec{E}\$$ field to permit a current to flow within the circuit. However, like in the open-circuit case, the surface charge creates voltage drops that may be far away from the location where the time-varying magnetic field is directly acting upon conductor. For example, if circuit consists of only the conductor and a resistor, and the resistance of the conductor is very small, virtually all of the voltage drop corresponding to the induced emf will appear across the resistor, even though that resistor may be far from where the time-varying magnetic field directly acts upon the conductor.

Electrons have a net drift, i.e. a current, in an ordinary conductor because there is an electric field accelerating them in a particular direction. However, they only drift for a short distance before they collide with other things, and their motion becomes randomized. Thus, the electric field must be present all along the conductor for current to flow all along the conductor.

Your circuit is driven by a time-varying magnetic field, which generates an electric field (which has further consequences). It does not, however, contain any batteries or other possible sources of emf that might cause charges to migrate against an electric field. Using the analogy of altitude as a substitute for electric potential, your circuit has no ski-lifts.

As a result, the only way current can flow in such a circuit is for there to be a circulating electric field. The electric field in within the conductor is not conservative, and cannot be conservative, or there would be no current.

You wrote:

I agree that there is no such term as EMF in a non-conservative field

I'm not exactly sure how to interpret what you wrote, but it is most definitely the case that non-conservative nature of the electric field in your example (which lacks a battery, etc.) provides the only reason there is current.

Now the force that acts on charges as they flow around the circuit performs work on those charges. But the net work done on charges around the loop does not (for long) raise their energy level. The charges, while traversing the loop, collide with other things and convert the potential energy that they gained into heat. If an electron were to travel around the loop (in a short enough time period that we can ignore changes in the electric field), it would have the same "energy level" when it arrived at it's original starting point, as it did when it first left that starting point. The sum of the changes in potential, as one goes around the loop total zero. There is no spiral, as in your diagram.

Your confusion probably stems from assuming the change in "energy level" between two points can be found by finding the line integral of the electric field, i.e.

$$\int_{C:a\rightarrow b}\vec{E}\cdot d\vec{s}$$

But that is the work done on the charges as they move along the line from $$\a\$$ to $$\b\$$.

If we subtract the emf induced along a line from the work done on a unit charge as it moves along that line, we will find the change in potential energy.

$$\Delta \phi = \int_{C:a\rightarrow b}\vec{E}\cdot d\vec{s} - \int_{C:a\rightarrow b}\vec{E}_{\partial\vec{B}/\partial t}\cdot d\vec{s}$$

which turns out to be the same as

$$\Delta \phi=\int_a^b \left(\nabla\frac{1}{4\pi}\int_{R^3}\frac{-\rho(r')}{\epsilon_0}\frac{1}{|r-r'|}d^3r'\right)\cdot d\vec{s}$$

and that latter integral, when C is a closed loop, evaluates to 0 by the gradient theorem. That is, there is no "spiral" in the "energy height" diagram.