0
$\begingroup$

Hall effect says, when a magnetic field is applied perpendicular to current direction in a semiconductor bar, an electric field (and hence Hall voltage) will be created across the side which is mutually perpendicular to both current direction and applied magnetic field direction. However I have a doubt that if instead of applying magnetic field, if we apply electric field, will a magnetic field be created?

$\endgroup$

1 Answer 1

1
$\begingroup$

A perpendicular electric field applied to a current carrying conductor (or semi-conductor) Would cause the fee electron density (and current) to shift toward one side of the conductor. Any current associated with this shift would be at the edges of the electric field, and the changes in the magnetic field would probably be minimal.

$\endgroup$
2
  • $\begingroup$ Hi I did not understand clearly how the changes in magnetic field be minimal? according to hall effect, hall voltage induced will be proportional to the applied magnetic field. So if we manually apply that much electric field, why don't we get a magnetic field similar to the initial case? Please help me if you have enough time. Sorry, but I am really confused with this. $\endgroup$ May 7, 2021 at 17:09
  • 1
    $\begingroup$ A Hall voltage is very small (generally much less than 1 volt). Applied sideways across a current carrying conductor, it would cause very little displacement of the existing current. $\endgroup$
    – R.W. Bird
    May 8, 2021 at 13:17

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.