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

Question

A proton (q = 1.6 times 10^-19C, 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 = 0.63 m in the x-direction. The proton leaves the field having a velocity vector (v_x, v_y) = (2.1 times 10^5 m/s, 0.9 times 10^5 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 containing the magnetic field? What is B_z, the z-component of the magnetic field? Note that B_z is a signed number. If the incident velocity v were increased, how would h and theta change, if at all? h and theta would both increase h and theta would both decrease h would increase and theta would decrease h would decrease and theta would increase Neither h nor theta would change

Explanation / Answer

1)

The proton is moving in x-direction initally

magnetic fied is along z- axis

The force on proton would be along y-axis perpendicular to x and y

hence the velocity of proton along x-axis will not change.

Velocity of proton after leaving the magnetic field

vx = 2.1e+5 m/s, vy = 0.9e+5 m/s

Initial velocity of proton as it enetrs the field

v = 2.1e+5 m/s ---- vy =0

2) Radius of curvature R :

along x-axis   v= 2.1e+5 m/s

displacement s= 0.63 m

time spent in the field   t= 0.63/2.1e+5 = 3.0e-6 s

along y-axis

intial vel vi =0     final vf = 0.9e+5 m/s

time t = 3.0e-6 s

acceleration a = 0.9e+5/3.0e-6 = 3.0e+10 m/s2

Force on proton ma = 1.67e-27*3.0e+10 = 5.01e-17 N

= mv2/R , centifugal force

radius of curvature   R = 1.67e-27*(2.1e+5)/5.01e-17

                                     = 1.47m

3) along y-axis a= 3.0e+10 m/s2

     time spent   t = 3.0e-6 s

      y-dispalcment   h = 0.5*3.0e+10*9.0e-12

                                    = 0.135 m

4) magnitude magnetic force = qvB = 5.01e-17 N

B = 5.01e-17/(1.6e-19*2.1e+5)

       = 1.49 mT

F is +y and v is +x hence B is -z dircetion

Bz = -1.49mT

5) if the intial velocity is increased the proton would spend less time in the magnetic field and the force would act for alesser time and the y displacement (h) would be less

h and theeta would decrease.

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