An electron with kinetic energy 4.2 keV moves horizontally into a region of spac
ID: 1438817 • Letter: A
Question
An electron with kinetic energy 4.2 keV moves horizontally into a region of space in which there is a downward-directed electric field of magnitude 11 kV/m.
(a) What are the magnitude and direction of the (smallest) magnetic field that will cause the electron to continue to move horizontally? Ignore the gravitational force, which is rather small.
(magnitude)
Choose the direction
The magnetic field must be parallel to the electric field.The magnetic field must be perpendicular to both the electric field and the velocity vector of the electron. The magnetic field must be perpendicular to the velocity vector of the electron.The magnetic field must be parallel to the velocity vector of the electron.
(b) Is it possible for a proton to pass through this combination of fields undeflected? If so, under what circumstances?
No, a proton cannot pass this field combination undeflected.Yes, if its velocity is perpendicular to that of the electron. Yes, if its velocity is the same as that of the electron.Yes, if its velocity is opposite that of the electron.
Explanation / Answer
a) Here, Force due to Electric Field = Force due to magnetic field
=> qE = qvB
=> v = E/B
= 11000/B
v = sqrt(2 * 6.729 * 10-16/9.11 * 10-31)
= 3.843 * 107 m/sec
=> magnetic field = 11000/3.843 * 107
= 2.862 * 10-4 T
Direction = The magnetic field must be perpendicular to both the electric field and the velocity vector of the electron.
b) Yes, if its velocity is the same as that of the electron.
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