A proton with a velocity v_i = (2i + 4j - 1k) Mm/s moves into a region with a ma
ID: 2077744 • Letter: A
Question
A proton with a velocity v_i = (2i + 4j - 1k) Mm/s moves into a region with a magnetic field B_0 = (-5i + 3j + 2k)G. (a) what is the magnetic force vector acting on the particle? After you find the vector force, calculate its magnitude. (b) Prove that the particle travels on a circle and not a helix. (c) What is the radius that the proton follows? (d) What is its frequency of revolution? (e) After going around that circle for exactly 2.5 cycles, what is the proton's new velocity v_f? (f) What is the change in kinetic energy of the particle between v_i and v_f?Explanation / Answer
a) vi = (2i +4j - 1k) mm/s = (2i +4j - 1k) * 10^-3 m/s
|vi| = 10^-3 * sqrt (2^2 + 4^2 + 1^2) = 4.58 * 10^-3 m/s
B0 = (-5i + 3j +2k) G = (-5i + 3j +2k) * 10^-4 T
q = 1.6 * 10^-19 C
F = q (v X B)
v X B = 2i X 3j + 2i X 2k + 4j X -5i + 4j X 2k -1k X -5i -1k X 3j
= 6k - 4j + 20k + 8i + 5j +3i
= (11i + 1j + 26k)
F = q (11i + 1j + 26k) * 10^-7 N
|F| = q* sqrt(11^2 + 1^2 + 26^2) = q * 28.25 * 10^-7 N = 4.52 * 10^-25 N
b) We take dot product of F and v
F.v = 22 + 4 - 26 = 0
=> Force is perpendicular to velocity. If F.v not equal to 0 then particle travels in helix else particle travels in circular path
c) F = mv^2/r where m = 1.67 * 10^-27 kg
=> r = mv^2/F = 1.67 * 10^-27 * 21 * 10^-6/ 4.52 * 10^-25 = 7.76 * 10^-8 m
d) f = 1/T = v/d = v/(2*pi*r) = 0.094 * 10^-3/10^-8 = 9.4 * 10^3 Hz
e) Velocity of particle is not change as it is orbiting with constant radius
vf = vi = 4.58 * 10^-3 m/s
But since it has travelled 2.5 revolutions, direction of vf is opposite that of vi (due to the half revolution)
=> vf = -(2i +4j - 1k) * 10^-3 m/s
f) Change in kinetic energy in circular motion is zero
So, in this case, change in kinetic energy is zero
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