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

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 = 0.74 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (3.6 X 105 m/s, 1.7 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

40What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.T

5)If the incident velocity v were increased, how would h and change, if at all?

h and would both increase

h and would both decrease

h would increase and would decrease

h would decrease and would increase

Neither h nor would change

Explanation / Answer

Here , as the speed of proton will not change in magetic field

A)

initial speed , v = sqrt(vx^2 + vy^2)

v = sqrt(3.6^2 + 1.7^2) *10^5

v = 3.98 *10^5 m/s

the magnitude of velocity is 3.98 *10^5 m/s

B)

for the radius of curvature

theta = arctan(vy/vx)

theta = arctan(1.7/3.6)

theta = 25.3 degree

as D = R * sin(theta)

0.74 = R * sin(25.3)

R = 1.73 m

radius is 1.73 m

c)

h = R * (1 - cos(theta))

h = 1.73 * (1 - cos(25.3))

h = 0.166 m

the y co-ordinate of the proton as it leaves the region conating the magnetic field is 0.166 m

d)

for

R = m * V /(B * q)

1.73 = - 2.67 *10^-27 * 3.98 *10^5 /(B * 1.602 *10^-19)

B = -2.402 *10^-3 T

the z-component of the magnetic field is -2.402 *10^-3 T

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