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Q) Consider a square loop of side 'l' made of a metallic wire moving with a velocity 'v' which is perpendicular to one of its sides. There exists a uniform magnetic field 'B' perpendicular to the loop's plane. Find the EMF induced in the loop.

I don't really get what is meant by 'EMF induced in the loop' as isn't potential difference taken to be between two points?

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The word "potential" is defined to be equal to the potential energy associated with the position of a charge divided by the charge itself, and has it's own unit -- volts. So volts = joules / coulombs. We usually use the words "potential difference" when talking about the flow of currents. It's really just the difference in voltage from one point in space to another. When potential/voltage differences exist between two points in space, conventional current (flow of + charges) always moves from the higher voltage to lower voltage. The electric field causing the charges to move always points from the higher to lower voltage. I tell students, "positive charges roll down the voltage hill".

Now let's address your question about the term "EMF".

In electromagnetic induction, it's natural to assume that induced currents must be caused by some kind of electric field associated with a potential difference, just like they do for Ohm's Law. Well, there IS indeed an electric field created inside the wire when the magnetic flux changes, and it does accelerate the charges. However, one should not think of this electric field as coming from a potential difference between 2 points. Rather, you should think of it as existing in the wire as a set of closed loops of electric field lines. You can see that there's no way to say that this closed looping electric field is due to a difference in voltage, because you could go around the loop once and end up right back at a point with the same voltage.

However, if a positive charge were allowed to be freely accelerated around the loop once without resistance, it would gain the same amount of energy as if it moved between two points in space having a voltage difference equal to what gets called an "EMF". Just as potential (voltage) is defined to be "potential energy divided by charge", an EMF is the same kind of thing, meaning "energy gained by going around the loop once without resistance divided by charge ". Once you add resistance to the situation, you can compute the current in the loop as EMF/Ohms if you know the EMF.

Remember, in electromagnetic induction, only changes in magnetic flux induce currents in loops of wire.

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If the whole loop is moving there will be no change in the flux linked with it, and no emf. If just one side of the loop is moving at speed $v$ (as a rod moving on conducting rails) the free electrons in the rod will have this velocity, and will therefore experience motor effect (magnetic Lorentz) force $Bev$ in a direction along the rod. The work done on each electron traversing the rod (length $L$) is therefore $Bev\ L$. What we mean by emf is work done by non-electrostatic forces per unit charge, so the emf in the moving rod is $$emf=\frac{BevL}{e}=BLv.$$ This is the emf in the loop, if the rails are connected together by a stationary conductor. This should give you a feel for what emf is all about.

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    $\begingroup$ So basically 'EMF in a loop' is the work done by non-electrostatic forces in moving a unit charge through a closed loop? $\endgroup$
    – User
    Commented Aug 7, 2017 at 0:44
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    $\begingroup$ That's what I'd say. $\endgroup$ Commented Aug 7, 2017 at 7:30
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Keep in mind also that the EMF does not depend on the presence of a physical metallic (or otherwise) loop of wire. The EMF exists as long as there is a changing magnetic flux, even in a vacuum. There is no movement of charge (whether in a loop or a "flying wire") unless charge is present.

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  • $\begingroup$ Agree with you in the case of an emf due to the electric field around a changing magnetic field (as per the Maxwell equation), but not so sure in the case of a conductor cutting magnetic flux. Without the conductor, how can one have any cutting? $\endgroup$ Commented Aug 6, 2017 at 21:18

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