1) 2) 3) 4) A proton (q = 1.6 times 10-19 C, m = 1.67 times 10-27 kg) moving wit

Question

1)

2)

3)

4)

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 = 0.45 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (5.3 times 105 m/s, 2.8 times 105 m/s). What is v, the magnitude of the velocity of the proton as it entered the region containing the magnetic field? What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field? What is h, the y co-ordinate of the proton as it leaves the region conating the magnetic field? What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.

Explanation / Answer

q = 1.6 X 10^-19 C, m = 1.67 X 10^-27 kg, D = 0.45 m,

(vx, vy) = (5.3 X 10^5 m/s, 2.8 X 10^5 m/s).

1)

v = sqrt(vx^2 + vy^2) = 5.99 X 10^5 m/s

2)

R = D/sin(theta), where tan(theta) = vy/vx

R = 0.96 m

3)

R - h = D/tan(theta) = D*vx/vy = 0.85 m

so h = 0.11 m

4)

Bz = mv/(qR) =6.51*10^-3 T

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