Question

Doctor Crusher arrives at the new Enterprise Sick Bay and sees a brand new mass spectrometer has been installed. Captain Picard says the technicians did not get time to calibrate it, so Doctor Crusher will have to calibrate it herself. Commander LaForge provides Doctor Crusher with some antiprotons to test the mass spectrometer. Antiprotons have the same mass as a proton, but are negatively charged.

Doctor Crusher sets the electric and magnetic fields such that the antiprotons leave the velocity selector at a speed of 3.0 x 106 m/s. The magnetic field, as viewed from overhead of the mass spectrometer is 0.40 Tesla pointing toward you, and the antiprotons are initially moving to the right within the velocity selector.

a) If the magnetic field continues at the same magnitude and direction within the detection area, how long will it take to hit the sensor (it needs to make a semi-circle)?

b) As viewed from above, which direction will the antiproton initially deflect?

c) What is the voltage required in the velocity selector to select for the velocity stated?

Answer 1

Here we can assume the mass spectrometer works similar to a cyclotron. A cyclotron is a device that is used to accelerate charged particles by moving them in circular paths on the presence of oscillating electric field and constant magnetic field or vice versa.

In this case, the particle is an antiproton

mass = $m_{p} = 1.6605\times 10^{-27} kg$

charge = $q = -e = -1.6\times 10^{-19} C$

velocity $v = 3\times 10^{6}m/sec$

magnetic field

direction of B = $+ \hat{z}$

direction of v = $+ \hat{x}$

from the concept of cyclotron we know that

radius of the path = $r = \frac{mv}{qB}$

Time taken to cross a semicircular path

$T = \frac{\pi r}{v} = \frac{\frac{\pi mv}{qB}}{v}= \frac{\pi m}{qB}$

$T = \frac{\pi m}{qB} = \frac{3.14\times 1.6605\times 10^{-27}}{1.6\times 10^{-19}\times 0.4}= 8.147\times 10^{-8 }Sec = 81.47 ns$

The direction of Lorentz force due to magnetic field

$F_{B} = q\left ( \vec{v}\times \vec{B} \right ) = -e \left ( \hat{x}\times \hat{z} \right ) = -e\times -\hat{y} = e\hat{y}$

The particle will be deflected towards positive y-axis or towards the above direction.

The relationship between the accelerator voltage and the kinetic energy of the particle is

$eU = E_{K} = \frac{1}{2}mv^{2}$

$U = \frac{1}{2e}mv^{2} = \frac{1.6605\times 10^{-27}\times \left ( 3\times 10^{6} \right )^{2}}{2\times 1.6\times 10^{-19}} = 4.67\times 10^{4} Volts$