Are they two different phenomena or same phenomena having different properties in different situations? In what scenario should we consider induced emf and in what scenario should we consider motional emf? And which of the following does faraday's law and lenz law mention about? Are both caused due to induced electric field?
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$\begingroup$ My voltmeter can't tell the difference and if my voltmeter can't, then so can't I. Whether an electric potential difference/electric field is being created by motion or by a change in a magnetic field, it will accelerate charges the same. $\endgroup$– CuriousOneCommented Feb 25, 2016 at 10:11
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$\begingroup$ thank you for replying sir,i asked about the underlying phenomena.voltmeter may have the same reading for induced emf from faraday's law and a simple battery but that doesn't mean both are same $\endgroup$– Ajay SabarishCommented Feb 25, 2016 at 10:19
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$\begingroup$ There is only one electromagnetic field. It is one phenomenon, albeit a complicated one. I understand that this may not be what you are trying to ask, but unfortunately this is how the question sounds in English to me. Faraday's law states that the emf induced in a closed path (or loop) is given by the change of magnetic flux trough a surface that is defined by that path (it's slightly more complicated, actually). As a result changes in the magnetic field, movement and changes in the enclosed loop area can result in an emf, but one can not decide what caused this from the emf itself. $\endgroup$– CuriousOneCommented Feb 25, 2016 at 10:33
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$\begingroup$ @CuriousOne An electric field isn't created when a mathematical curve changes in time. But that's fine since your voltmeter doesn't detect electric fields. It detects currents which respond to electromagnetic forces. $\endgroup$– TimaeusCommented Feb 26, 2016 at 17:42
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$\begingroup$ @Timeus: What the universe looks like depends on how you are moving in it. That's already true in classical mechanics and there is nothing magical about it. What I am talking about is not a mathematical curve moving, I am talking about a moving observer. What a moving observer sees of an electromagnetic field depends on how fast the observer is moving. $\endgroup$– CuriousOneCommented Feb 26, 2016 at 19:29
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2 Answers
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Motional EMF and induced electric fields are 100% totally different phenomena
An EMF is a force per unit charge line integrated along a path. If the force in question is purely electromagnetic then the EMF equals $$\int\left(\vec E+\vec v \times \vec B\right)\cdot \mathrm d \vec \ell.$$
When a wire is stationary then $\vec v,$ the motion of a charge staying in the wire, and $\mathrm d\vec \ell,$ the vector going in the direction of the wire, point in the same direction so the magnetic field doesn't contribute to the electromagnetic EMF so it's all electric EMF. In which case the current value of the magnetic field doesn't matter and the electromagnetic EMF is entirely electric. However it could still be nonzero around a closed loop and you get: $$\oint\vec E\cdot \mathrm d \vec \ell=-\iint \frac{\partial \vec B}{\partial t}\cdot \mathrm d\vec A.$$
But notice on the left side how its 100% an electric field? That means the EMF is 100% entirely due to actual electric fields in the wire.
When a loop is moving that magnetic term in $\int\left(\vec E+\vec v \times \vec B\right)\cdot \mathrm d \vec \ell$ can be nonzero and in addition to the electric EMF, $\int\vec E\cdot \mathrm d \vec \ell,$ there can be a magnetic EMF, $\int\left(\vec v \times \vec B\right)\cdot \mathrm d \vec \ell.$ The magnetic EMF is called a motional EMF because it was zero unless you are moving. When the wire is moving, if the wire is thin, the charges stay in the wire, and there are no magnetic monopoles then you can derive:
$$\oint\left(\vec E+\vec v \times \vec B\right)\cdot \mathrm d \vec \ell=-\frac{\mathrm d}{\mathrm d t}\iint \vec B\cdot \mathrm d\vec A.$$
This is because $\oint\vec E\cdot \mathrm d \vec \ell$ always equals $-\iint \frac{\partial \vec B}{\partial t}\cdot \mathrm d\vec A$ and because if the wire is thin, the charges stay in the wire, and there are no magnetic monopoles then you can derive: $$\oint\left(\vec v \times \vec B\right)\cdot \mathrm d \vec \ell=\iint \frac{\partial \vec B}{\partial t}\cdot \mathrm d\vec A-\frac{\mathrm d}{\mathrm d t}\iint \vec B\cdot \mathrm d\vec A.$$
The motional EMF is entirely due to the motion of the loop in the current magnetic field and the electric EMF is entirely due to the flux of the change in magnetic field in the current loop. The combination ewuals the change in the flux of the magnetic field. Something easier to compute. And the total is also what you actual measure.
Both kinds of EMF produce current flow, and you don't really need to worry about which is doing what. The total EMF depends on how the magnetic flux changes in time.
Are they two different phenomena or same phenomena having different properties in different situations?
They are physically different. But they combine to a total EMF that is easier to use than either by itself. And it's the total EMF that matters.
In what scenario should we consider induced emf and in what scenario should we consider motional emf?
If your loop doesn't move, the induced electric EMF is the only one there is. But using the total is fine as well since they are the same. If your loop is moving in an unchanging magnetic field there is a magnetic or motional EMF but it equals the total EMF, so again you can use the total since they are the same. If there is a moving wire in a magnetic field that is also changing in time then there are two EMFs. You could compute the induced electric EMF and the motional magnetic EMF and add them together or you could just find the change of magnetic flux and compute the total in one go since that's what you need anyway.
It is always fine to compute the total electromagnetic EMF when you have thin wires with charge staying in the wires and no magnetic monopoles.
And which of the following does faraday's law and lenz law mention about?
Unfortunately different people use the word Faraday's Law to mean different things. Some people use it to mean the thing that is always true, that the electric EMF equals the flux of changing magnetic fields: $$\oint\vec E\cdot \mathrm d \vec \ell=-\iint \frac{\partial \vec B}{\partial t}\cdot \mathrm d\vec A.$$ That one is an actual one of Maxwell's equations.
Unfortunately, other people use the same words (Faraday's Law) to refer to the flux rule, something that is not one of Maxwell's laws, something that depends on the wires being thin, the charges staying in the wire, and there being no magnetic monopoles. They use it to refer to the universal flux rule:
$$\oint\left(\vec E+\vec v \times \vec B\right)\cdot \mathrm d \vec \ell=-\frac{\mathrm d}{\mathrm d t}\iint \vec B\cdot \mathrm d\vec A.$$
And that rule is not universal in that it always works, it's just more general becasue it includes the full electromagnetic EMF instead of just the induced electric EMF or the motional magnetic EMF. Lenz's law applies to either (to both) since they both have that minus sign.
Are both caused due to induced electric field?
Note that both laws have an electric field in them, but one of them is a law of physics. They other one only holds sometimes, but it includes electric fields as well as magnetic fields, so if gives you the thing you probably want, the total electromagnetic EMF.
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$\begingroup$ Good answer, for loops. But what about for non-loops? There is still a motional emf in a single wire, but is there an induced emf? $\endgroup$– LutormCommented Apr 10, 2021 at 7:58
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The 2nd term above, $v\times B\cdot dl$, will be zero in a current-carrying wire since the drift velocity is parallel to the wire segment $dl$. The line integral being non-zero around a loop must be the non-conservative induced electric field, the electrostatic field due to any charges, will have a line integral equal to zero.
