A charged particle of mass m = 7.8 times 10^-8 kg, moving with constant velocity

Question

A charged particle of mass m = 7.8 times 10^-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 3.5T aligned with the positive z-axis as shown. The particle enters the region at (x, y) = (0.64 m, 0) and leaves the region at (x, y) = 0, 0.64 m a time t = 653 mu s after it entered the region. With what speed v did the particle enter the region containing the magnetic field? m/s What is F_x, the x-component of the force on the particle at a time t_1 = 217.7 mu s after it entered the region containing the magnetic field. N What is F_y, the y component of the force on the particle at a time t_1 = 217.7 mu s after it entered the region containing the magnetic field. N What is q, the charge of the particle? Be sure to include the correct sign. Mu C 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? Increase B by a factor of 2 Increase B by less than a factor of 2 Decrease B by less than a factor of 2 Decrease B by a factor of 2 There is no change that can be made to B to keep the trajectory the same.

Explanation / Answer

m = 7.8 * 10-8 kg, B = 3.5 T, r = 0.64, t = 653 * 10-6 s

a) T = 2 pi r / v

v = (pi * 0.64) / (2 * 653 * 10-6) = 1.54 * 103 m/s

b) A = pi / 2 * t1 / t = pi / 2 * 217.7 / 653 = 0.524 rad = 30 deg

Fx = [-m v2 / r] cos A = [-7.8 * 10-8 * (1.54 * 103)2 / 0.64] * cos 30 = -0.250 N

c) Fy = [-m v2 / r] sin A = [-7.8 * 10-8 * (1.54 * 103)2 / 0.64] * sin 30 = -0.145 N

d) q = m v / B r = (7.8 * 10-8 * (1.54 * 103)) / (3.5 * 0.64) = -5.36 * 10-5C

e) There is no change that can be made to B to keep the trajectory the same

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