1
$\begingroup$

In the Figure we can see a conducting loop of radius $r$ situated in a uniform magnetic field that is perpendicular to the plane of the loop. If we change the magnetic field, then according to the Faraday’s law we get induced electric field or emf.

My question is that, why the direction of the electric field is perpendicular to the radius of the ring? If we would use a square ring then how would we direct the direction of $E$ field?

enter image description here

$\endgroup$
0
$\begingroup$

I think I can answer in the case of the circular and square ring.

Firstly, by cylindrical symmetry, we know that the E field is the same all the way around the circle.

Why there is no radial component: Imagine rotating the system by 180 degrees about any diameter. At any location, the direction of the magnetic field would reverse, the direction of the current would reverse, all components of the E field would reverse, EXCEPT for the radial one. Now let's superimpose the two systems on each other. Without a magnetic field, we nevertheless have an induced radial E field! As this is clearly impossible, any radial component must be ruled out.

Why there is no component in or out of the page: Such a component would pull the ring into or out of the page. All the work put into changing the magnetic field goes into the work the electric field does in driving charges around the circuit. So it seems the ring's kinetic energy would increase without any source.

Similar arguments can be used to shown that in the case of the square ring, the induced electric field must once again be such that it is parallel to the ring at each point and drives around a current in it (direction determined by Lenz's law).

In general my intuition is that the induced electric field should be parallel to the wire at each point... but without these symmetry arguments in a general case I don't know how to go further.

$\endgroup$
0
$\begingroup$

The force of the changing magnetic field on the electrons in the conductor can be deduced by imagining field lines traversing the conductor (from outside the ring to inside). This results in a force that is not necessarily parallel to the wire - but if the wire is very thin, then a tiny motion of the electrons will cause a polarization of the wire, after which the remaining field (observable microscopically) will be along the wire.

In other words - there may be a small component of the electric field due to polarization of the wire, but since the wire is very thin you can hardly notice it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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