A charged particle of maw m = 5.8times10^-8 kg, moving with constant velocity in

Question

A charged particle of maw m = 5.8times10^-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.2T aligned with the positive z-axis as shown. The particle enters the region at (x, y) = (0.7 m, 0) and leaves the region at (x, y) = 0, 0.7 m a time t = 476 mus after it entered the region. 1) With what speed v did the particle enter the region containing the magnetic field? m/s 2) What is F_x, the x-component of the force on the particle at a time t_1 = 158.7 mus after it entered the region containing the magnetic field. N 3) What is F_y, the y-component of the force on the panicle at a time t_1 = 158.7 mus after it entered the region containing the magnetic field. N 4) What is q, the charge of the particle? Be sure to include the correct sign. muC 5) If the velocity of the incident charged particle were doubled, how would B have to change (keeping all other parameters constant) to keep the trajectory of the particle the same?

Explanation / Answer

1) v = (pi/2)d/t = 1.57*0.7/(476*10^-6) = 2309 m/s

2) Fx = Fo cos theta

= (-mv^2/R) cos [(pi/2)*158.7/476] = (-mv^2/R) cos(pi/6)

= -(5.8*10^-8*2309^2/0.7)* 0.866

= -0.383 N

3) Fy = Fo sin theta

= (-mv^2/R) sin[(pi/2)*158.7/476] = (-mv^2/R) sin(pi/6)

= -(5.8*10^-8*2309^2/0.7)* 0.5

= -0.221 N

4)W know,

qvB = mv^2/R

qB = mv/R

q = mv/RB

= 5.8*10^-8*2309/[0.7*1.2] =0.0001594 C = 159.4 uC

5) qB = mv/R

so B will be doubled

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