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

Question

A proton (q = 1.6 X 10-19 C, m = 1.67 X 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 = 0.56 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (5 X 105 m/s, 1.9 X 105 m/s). what's the velocity of the proton? whats the radius of the curvature of motion? whats the y-coordinate as the proton leaves the magnetic field? what is Bz, the Z component of the magnetic field?

Explanation / Answer

1. velocity= sqrt((5*10^5)^2+(1.9*10^5)^2) = 5.35*10^5m/s

2) D=R*sin(theta) where tan(theta)=1.9/5==>sin(theta)=0.355

==> R=1.57 m

3) Y coordinate=R*(1-cos(theta))=0.102 m

4) Bz=mv^2/qvR

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