H1. An electron moves with a speed of 8.0 x 106 m/s along the +x-axis. It enters
ID: 1786095 • Letter: H
Question
H1. An electron moves with a speed of 8.0 x 106 m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5×10-4 T, directed at an angle of 60° to the +x-axis and lying in the xy-plane. a) Find the magnitude of the magnetic force on the electron. b) Find the radius of curvature electron (as it travels in helical motion). H2. A beam of charged particles is directed along the axis of a velocity selector. An electric field, E, created by a capacitor with a plate separation of 2.0 mm. This electric field is perpendicular to the magnetic field. B = 0.60 T, as shown. a) what voltage across the plates will allow particles of speed 8.0 x 105 m/s to pass straight through without deflection? b) Which plate (upper or lower) is the positive plate? Tube B (into paper) (cutaway x x k x x view) 9 Capacitor platesExplanation / Answer
H1)Given,
v = 8 x 10^6 m/s ; B = 2.5 x 10^-4 T ; theta = 60 deg
a)We know that the magnetic force on an electron is given by:
F = q v B sin(theta)
F = 1.6 x 10^-19 x 8 x 10^6 x 2.5 x 10^-4 x sin60 = 2.8 x 10^-16 N
Hence, F = 2.8 x 10^-16 N
b)Let the radius be R
the magnetic force balances the centripital force, so the net force is:
we know that,
Fc - Fb = 0
mv^2/R - q v B sin(theta) = 0
mv^2/R = 2.8 x 10-16
R = 9.1 x 10^-31 x (8 x 10^6)^2/2.8 x 10^-16 = 0.208 m
Hence, R = 0.208 m = 20.8 cm
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