0
$\begingroup$

Consider a circular homogeneous metallic coil sliding on a smooth horizontal surface in a region of uniform magnetic field $B$ which is perpendicular to the face of the coil. By Faraday's law, the net flux is constant and hence there is no electro-motive force (EMF) induced in the coil.

But now look at individual electrons present in the coil. Due to the Lorentz force, all electrons will experience a net force in the same direction. Since they are constrained to move on the ring, they will move on one of the half side of the ring. Thus, an electric field will be setup and an EMF is induced. Hence, there will be induced motional EMF.

Please explain whether the EMF will be induced in the coil or not with the appropriate reasoning. Thanks!

$\endgroup$
2
  • $\begingroup$ Hall effect in establishing the equilibrium. But let's wait for other answers $\endgroup$
    – basics
    Oct 26, 2022 at 18:16
  • $\begingroup$ @basics Unfortunately, I haven't learnt hall effect. $\endgroup$ Oct 27, 2022 at 4:29

2 Answers 2

3
$\begingroup$

Faraday's law in this situation predicts $$ \epsilon = \oint \left(\vec{E} + \vec{v} × \vec{B}\right) \cdot \vec{dl} = 0. $$ This does not imply that $$ \int_{a}^{b} \left(\vec{E} + \vec{v} × \vec{B}\right) \cdot \vec{dl} = 0. $$ The closed line integral of the Lorentz force is equal to zero in your case; however, an open line integral can be non-zero.

$\endgroup$
2
$\begingroup$

One thing you need to understand is that the closed loop which you talking about has the electrons distributed whole across the loop. Now imagine two electrons, first one which is present at the one side of the loop and another one at diametrically opposite side of the loop. Now as you have mentioned above that the magnetic field is directed perpendicularly to the loop (let us be more specific that it is perpendicularly directed inwards to the plane as I have shown in the figure).

Now let the loop move with some velocity in the right direction. Now both the electrons have fulfilled the conditions to get influenced by lorentz force i.e they both have charge, they both have velocities and they are in the plane where the magnetic field is present. Now both the electrons will experience the same magnitude of force in the same direction which is upwards as they are moving with the same velocities, have same charge and are in a same constant magnetic field. Now both the electrons will move upwards inside the loop as shown below in the imaɡe, so these moving charges will form a current in opposite directions and hence the net current will be zero and there will be no accumulation of the charges which won't let you to induce an EMF inside the wire of that loop. That's why you won't get any EMF induced inside the wire which also obey the Faraday's law perfectly.

enter image description here

$\endgroup$
6
  • $\begingroup$ But won't the electrons experiencing the force will accumulate upward? $\endgroup$ Oct 27, 2022 at 4:26
  • $\begingroup$ Did you meant that electrons are trying to accumulate at same place and due to repulsion they won't? $\endgroup$ Oct 27, 2022 at 4:32
  • $\begingroup$ Yes they will accumulate upwards, this still does not mean faradays law is false, the close line integral about the loop is zero. $\endgroup$ Oct 27, 2022 at 16:30
  • $\begingroup$ @An_Elephant yes, they will but if you notice one thing very closely. The EMF generated by the accumulation of electrons is not inside the loop, it's about the two sides of the loop that means the whole loop can be used as a battery for another external circuit. Whereas Faraday's law tells us about the EMF which is generated inside the loop not across its side. $\endgroup$ Oct 27, 2022 at 17:48
  • $\begingroup$ ...And even if you connect the external circuit about which we talking earlier in this comment then again the whole system will now can be used as a battery for an another ExternalCircuit but the current won't flow inside those 2 loops Hence Faraday's Law is still valid. (Don't forget to read my first comment which was made just before this comment due to lack of space) $\endgroup$ Oct 27, 2022 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.