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 = 0.79 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (4.4 X 105 m/s, 2.2 X 105 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 conating the magnetic field?

m

4)What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.

Explanation / Answer

1) v = sqrt(vx2 + vy2) = 2.2 x 105 x sqrt(22 + 12) = 4.91935 x 105 m/s

2) R = D/sin = Dv/vy = 0.79 x sqrt(5) = 1.7665 m

3) h = R(1 - cos) = 1.7665(1 - vx/v) = 0.1865 m

4) now initial velocity is in x direction and force in y direction => Bz is in negative z direction

now Bz = -(mv/qR) z^

= -(1.67 X 10-27 x 4.91935 x 105)/(1.6 X 10-19 x 1.7665) z^ = -2.9066 x 10-3 T z^

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