A charged particle of mass m-4X108 kg, moving with constant velocity in the y di

Question

A charged particle of mass m-4X108 kg, moving with constant velocity in the y direction enters a region containing a constant magnetic field B 2.1T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.77 m, 0) and leaves the region at (x,y)-0, 0.77 m a time t-822 ps after it entered the region. 1) With what speed v did the particle enter the region containing the magnec field? 1471.4 m/s Submit 2) What is Fx, the x-component of the force on the particle at a time t1-274 s after it entered the region containing the magnetic field 0 N Submit 3) What is Fy, the y-component of the force on the particle at a time t 1-274 s after it entered the region containing the magnetic field 0 N Submit 4) What is q, the charge of the particle? Be sure to include the correct sign.

Explanation / Answer

Solution of first problem:

1) Since the magnetic force is perpendicular to the velocity of the particle, speed of the particle is constant.

So, speed of particle, |v| = (d / 2) / t = ( * 0.77 / 2) / (822 * 10-6) = 1471.4 m/s

2) Magnitude of force on the particle, F = mv2/d = (4 * 10-8) * 1471.42 / 0.77 = 0.1125 N

Angular position of the particle at t = 274 s from x-axis, = (274 / 822) * (/2) = 0.524 rad

Fx = Fcos (-i) = -0.1125 * cos(0.524) i = (-9.74 * 10-2 N) i

3) Fy = Fsin (-j) = -0.1125 * sin(0.524) i = (-5.63 * 10-2 N) j

4) At t = 0,

F = q(v x B) = mv2/d (-i)

=> q[(v j) x (B k)] = -(mv2/d) i

=> q(vB i) = -(mv2/d) i

=> q = -mv/BR = -(4 * 10-8) * 1471.4 / (2.1 * 0.77) = -3.64 * 10-2 C

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