# Induced EMF vs battery EMF

I am bit confused at whether induced emf and battery emf are same in nature differing just in their source. Does a potential gradient exist in case of induced emf as we have when a battery is connected to a circuit? If it so then between which two points in a current carrying coil (with changing flux) the induced emf exist (as we have well defined position for emf in case of a battery i.e. the terminals, do we have a similar one in case of induced emf)?

• The terminology can be awkward, but "potential," strictly speaking, implies that the electric field is of the form $E=-\nabla\phi$, which is not the case when induced fields are present. – user4552 Dec 4 '19 at 5:53

I am bit confused at whether induced emf and battery emf are same in nature differing just in their source

EMF is voltage regardless of how it is generated.

Does a potential gradient exist in case of induced emf as we have when a battery is connected to a circuit?

A voltage or potential gradient exists in a coil, but for a different reason than for a battery. For a coil Faraday's Law states for a coil of N turns enclosing flux $$ϕ$$:

$$v=-N\frac{dϕ}{dt}$$

So, for example, if a voltage $$v$$ of 100 volts is induced in a coil of 100 turns, the magnetic field induces one volt per turn.

According to Wikipedia, in the case of a battery the emf gradient is referred to as an electrochemical gradient. It consists of two parts, the chemical gradient, or difference in solute concentration across a membrane, and the electrical gradient, or difference in charge across a membrane.

If it so then between which two points in a current carrying coil (with changing flux) the induced emf exist

As I indicated above, the gradient in the coil is the turn to turn voltage produced by the changing magnetic flux which, when multiplied by the total number of turns, gives the total emf.

Hope this helps.

I am bit confused at whether induced emf and battery emf are same in nature differing just in their source.

Yes, though both the battery EMF and the induced EMF move the charges in the same manner by creating an electric field, the "EMFs" aren't of the same origin. In the first case, the EMF is generated due to the the electro-chemical reactions happening inside the cell while in the second case the electric field generated is due to a change in the magnetic field around the inductor.

Does a potential gradient exist in case of induced emf as we have when a battery is connected to a circuit?

But unlike a battery, there isn't any potential difference or gradient across the ends of an inductor because of excess charges at the ends of an inductor.

The potential gradient here will be due to the E- field here which is generated due to a different reason.

We just call it under the name of "induced EMF" because it behaves the same way as the battery EMF.

After all in both the cases, the charges are moving due to the E field generated and not by pushing on each other so it doesn't really matter. We can treat the EMFs in the same way.

• Are you saying there is no potential gradient in an inductor? – Bob D Dec 3 '19 at 18:02
• Per Faraday's law the induced emf is proportional to the number of turns. If the emf induced in a 100 turn coil is 100 volts then the magnetic flux induces one volt per turn. – Bob D Dec 3 '19 at 18:17
• @BobD I clarified what I meant. Hope it's clear now. – user8718165 Dec 3 '19 at 18:43

I am bit confused at whether induced emf and battery emf are same in nature differing just in their source

Yes with one important difference. Electric field produced by a battery is conservative in nature (does not form closed loops, path-independent, etc.) but electric field produced by a changing magnetic field will almost always form closed loops and thus be non-conservative.

Does a potential gradient exist in case of induced emf as we have when a battery is connected to a circuit?

A potential gradient (ie electric field) does exist in induced emf as mentioned above.

If it so then between which two points in a current carrying coil (with changing flux) the induced emf exist (as we have well defined position for emf in case of a battery i.e. the terminals, do we have a similar one in case of induced emf)?

Now here comes the tricky part. First of all we have Faraday's Law of Induction which states:-

$$\oint \vec E \cdot \vec dl = - \frac{d}{dt} \varphi$$

The left side of the equation $$\oint \vec E \cdot \vec dl$$ is the induced EMF in the coil. Notice that we are integrating the electric field over the entire coil loop and equating it to the time derivative of magnetic flux. What this means is that there are no two endpoints or terminals across which the EMF exists. The EMF exists across the entire loop.

However when we typically model and analyse circuits we look at a property of the current carrying-coil known as its inductance and a device known as the inductor. The EMF produced by a changing magnetic field across a current-carrying loop is proportional to the number of turns of the loop (The Faraday's law equation I have given should have an extra $$N$$ term on the right side to account for the number of turns of the coil). What this means is that for a normal circuit with just one turn the induced EMF will be insignificant.

We say that the inductance of the loop is small. For something like a solenoid however we do have to account for the inductance and induced EMF as due to the high number of turns it becomes significant.

An inductor is a device containing many turns of current-carrying coil which thus has a significant induced EMF and when we model circuits we assume that all the inductance of the circuit is due to an inductor connected in series with other components of the circuit. The majority of induced EMF will be across the inductor and thus, induced EMF is typically taken across the endpoints of the inductor.

So to answer your question, no induced EMF doesn't typically have fixed endpoints across it which exists such as for a battery but we model circuits so that it does, across a component known as an inductor.

• If high and low potential points are not defined for an induced electric field how can we state that there is a potential difference across the inductor? – Aditya Ahuja Dec 30 '19 at 19:08
• @AdityaAhuja We assume induced e-field to only be significant in the inductor components and zero elsewhere. For loops there are no high/low points but for inductors with two terminals, there is a sense of potential difference across the inductor. Does that make sense? – aditya_stack Dec 31 '19 at 10:41

$$\bf E$$ and $$\bf B$$ are most useful for describing static charges and currents. However when everything becomes time dependent these are no longer independent quantities. To think in terms of electric and magnetic fields then may become confusing. It is then necessary to use the four potential $$(\phi, {\bf A})$$, connected to the charge current $$(q, \bf j)$$ by the inhomogeneous wave equation. It then becomes clear that $$\bf E$$ has multiple origins: $${\bf \Delta} \phi$$ and $$\partial_t \bf A$$. The battery produces a Coulomb force and is an example of the former. Induction is an example of the latter, the time varying vector potential.

I am bit confused at whether induced emf and battery emf are same in nature differing just in their source.

No, they are very different. Battery EMF is due to chemical reactions inside, and acts only on particles inside the battery, so it drives the current only along the path inside the battery; outside the battery, the current is driven by other kinds of forces, such as electrostatic or induced electric field.

Induced EMF is due to induced electric field, which is present everywhere magnetic field changes in time, it drives current at all parts of the circuit, although how strongly depends on strength of the induced electric field at that place.

Does a potential gradient exist in case of induced emf as we have when a battery is connected to a circuit?

In practice it often does, but it is not directly due to induced field. See, induced field is a solenoidal field (goes in circles, has no source or sink) and that kind of field cannot be described by potential function. However, the induced field acts on mobile charges in conductors and is able to concentrate or deplete them at different regions of space, such as on the surface and terminals of an inductor. These surface charges produce Coulomb electric field which does have a potential. When we say there is difference of potential on a circuit element such as resistor, capacitor or inductor, we mean the potential of the Coulomb field of the surface charges in the system. It has nothing to do with induced field.

In a perfect inductor, the effect of the Coulomb field cancels the effect of the induced field. That's why when EMF for an inductor (net work of induced field per unit charge) is customarily written with minus sign as $$EMF = -L \frac{dI}{dt}~~~~\tag{*}$$ to conform to the Lenz law, potential drop in the same positive direction (often called voltage on element), which we use in writing down Kirchhoff's voltage law, is $$p.d. = L \frac{dI}{dt}$$ i.e. it has the opposite sign. For inductor with appreciable ohmic resistance $$R$$, the Coulomb field doesn't cancel the induced field completely and the potential drop is $$p.d. = L \frac{dI}{dt} +RI$$ while the induced EMF is still given by (*).

Induced field can be theoretically present without gradient of electric potential, if the setup is symmetric enough. For example, if the induced field is due to moving cylinder magnet that has negligible electric charge density, its Coulomb field is negligible and the potential is unchanged by the induced field. We can even add a conducting circular path; if we place it so the circular induced field acts equally on mobile charges in all parts of the circular conductor, then no charge separation will happen in the conductor, and potential will still be unaffected.