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A beam of electron passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off, and the same magnetic field is maintained, the electrons move
A
in a circular orbit
B
along a parabolic path
C
along a straight line
D
in an elliptical orbit
Detailed Solution
The beam passes undeflected, so the electric force balances the magnetic force: $eE = evB$.
This is possible only when the velocity of the electrons is perpendicular to both the electric and the magnetic field (velocity selector), with $v = \dfrac{E}{B}$.
When the electric field is switched off, only the magnetic force $F = evB$ acts on the electrons.
This force is always perpendicular to the velocity, so it does no work and the speed stays constant; it only changes the direction of motion.
A force of constant magnitude that is always perpendicular to the velocity acts as a centripetal force: $evB = \dfrac{mv^2}{r}$, giving $r = \dfrac{mv}{eB}$.
Since the electrons travel in a magnetic field perpendicular to their velocity, they move in a circular orbit.
This is possible only when the velocity of the electrons is perpendicular to both the electric and the magnetic field (velocity selector), with $v = \dfrac{E}{B}$.
When the electric field is switched off, only the magnetic force $F = evB$ acts on the electrons.
This force is always perpendicular to the velocity, so it does no work and the speed stays constant; it only changes the direction of motion.
A force of constant magnitude that is always perpendicular to the velocity acts as a centripetal force: $evB = \dfrac{mv^2}{r}$, giving $r = \dfrac{mv}{eB}$.
Since the electrons travel in a magnetic field perpendicular to their velocity, they move in a circular orbit.