Q.193. Motion of a charged particle in an electromagnetic field.

Question 193.
A velocity selector uses a 6o mT magnetic field perpendicular to 24 kN/C electric field. At what speed will a proton pass through the selector undeflected?
(a) After emerging from the velocity selector, the proton enters a mass spectrometer with a 1 T magnetic field perpendicular to the velocity. What is the diameter x of the semicircular trajectory of the proton just before it impinges on the detector?
(b) What would be deflection x be if the charged particle is an electron going through the same velocity selector and the mass spectrometer as above?

Answer 193.
v = E/B = 24000/0.06 = x * 10^5 m/s.

(a)
Force on proton of mass m moving on trajectory of radius r equals the force due to magnetic field
=> mv^2/r = qvB
=> diameter = 2r
= 2mv/qB ... [m = mass of proton and q = charge of proton]
= 2 (1.66 x 10^-27) (4 x 10^5) / (1.6 x 10^-19 x 1)
= 8.3 x 10^-3 m.

(b)
The force on the electron due to electric field and magnetic field will still be equal but opposite in direction to that on proton. Hence, electron will also move undeflected.
diameter of the trajectory of electron
= 8.3 x 10^-3 / 1836
... [because mass of electron is 1/1836 of the mass of proton and charge is the same as that of proton]
= 4.52 x 10^-6 m.

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