A proton (q = 1.6 times 10^-19 C, m = 1.67 times 10^-27 kg) moving with constant

Question

A proton (q = 1.6 times 10^-19 C, m = 1.67 times 10^-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x, y) = (0,0) as shown. The magnetic field extends for a distance D = 0.43 m in the x-directlon. The proton leaves the field having a velocity vector (v_x, v_y) = (5.1 times 10^5 m/s, 2.7 times 10^5 m/s). 1) What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field? m/s 2) What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field? m 3) What is h, the y co-ordinate of the proton as it leaves the region containing the magnetic field? m 4) What it B_z, the z-component of the magnetic field? Note that B_2, is a signed number. T 5) If the incident velocity v were increased, how would h and theta charge, if at all?

Explanation / Answer

1) What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field?
v = sqrt(vx^2 + vy^2) = 5.77 X 10^5 m/s

2) What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field?
R = D/sin(theta), where tan(theta) = vy/vx

tan(theta)= 2.7x10^5 / 5.1x10^5 =0.52941

theta= 27.90

R= 0.43/sin (27.90) =0.919 m

3)

R-h= Rcos(theta)

h= R-Rcos(theta)

h= 0.919 - 0.919 cos (27.90)

h= 0.107 m

4) What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.
Bz = mv/(qR) = 6.55*10^-3 T

5) If the velocity were increased, the radius of curvature would increase by the same factor since R is proportional to v. If the radius of curvature increases, the angle through which the particle bends in a distance D also decreases. If the angle decreases, then the displacement h also decreases.

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