Question

a. What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric field and magnetic field of magnitude 8.8x10^3 V/m and 7.5x10^-3 T, respectively?

b. What is the radius of the electron orbit if the electric field is turned off? [hint: start with Newton’s 2nd law to derive an equation and then do the calculation]

Answer 1

a)

If the charge 'q' has a velocity 'v' in magnetic field 'B', then the magnetic force on the charge is given as -

F_m = qVB [This is because 'v' is perpendicular to the magnetic field which makes sine of the angle = 1]

And the electric force experienced by the charge is -

F_e = qE, Where 'E' is the electric field intensity.

As the charge moves undeflected, it means that the net force acting on it is zero. Which means that the electric force and magnetic force are balancing each other. So, we can say that -

F_b = F_e

=> qVB = qE

Or, V = E/B

So, V = (8.8 x 10³ V/m) / (7.5 x 10^-3 T)

= 1.1733 x 10⁶ m/s.

Thus, velocity of beam of electron = 1.1733 x 10⁶ m/s.

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b)

If the electric field is suddenly turned off, then the only force acting on charge would be magnetic force which will act tangentially at each point and hence particle will experience a circular motion.

If 'R' is the radius of circle, then the centripetal force must be balanced by the magnetic force.

So, the radius of the electron orbit is approximately 0.89mm.

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