A particle with charge 2.15µC and mass 3.20 X 10-11 kg is initially travelling i

Question

A particle with charge 2.15µC and mass 3.20 X 10-11 kg is initially travelling in the +y-direc1ion with a speed U0 = 1.45 X 105 m/s. It then enters a region containing a uniform magnetic field that is directed into and perpendicular to, the page in Fig. The magnitude of the field is 0.420 T. The region extends a distance of 25.0 cm along the initial direc1ion of travel; 75.0 cm from the point of entry into the magnetic field region is a wall. The length of the field-free region is thus 50.0 cm. When the charged particle enters the magnetic field, it follows a curved path whose radius of curvature is R. It then leaves the magnetic field after a time t1 having been deflected a distance ?x1 The particle then travels in the field-free region and strikes the wall after undergoing a total deflection ?x.
(a) Determine the radius R of the curved part of the path.
(b) Determine t1 the lime the particle spends in the magnetic field.
(c) Determine ?x1 the horizontal deflec1ion at the point of exit from the field.
(d) Determine ?x, the total horizontal deflection.
this is the picture of teh problem. Thanks in Advance
http://www.solutioninn.com/physics/electricity-and-magnetism/magnetic-fields/particle-with-charge-2.15uc-and-mass-3.20-x-10-11-kg-is

Explanation / Answer

a) The magnetic force on the charge is F = Bqv = 0.42T*2.15e-6C*1.45e5m/s = 0.13 N. Since the magnetic force is always perpendicular to the velocity of the charge, it creates a centripetal acceleration on the particle which is a = F/m = 0.13N/3.2e-11kg = 4.06*10^9 m/s^2. And we have the equation a = v^2/R, so R = v^2/a = (1.45e5m/s)^2/4.06e9m/s^2 = 5.18m. b) The angle the particle rotates in the magnetic field is theta = arcsin(d/R) = arcsin(0.25m/5.18m) = 0.0483(rad). And the angular velocity of the particle is w = v/R = 1.45e5 / 5.18 = 2.8*10^4 rad/s. So t1 = theta/w = 0.0483/2.8e4 = 1.73e-6 s. c) x1 = R - Rcos(theta) = 6.04e-3m. d) x = (D-d)*tan(theta) + x1 = (0.75m - 0.25m)*tan0.0483 + 6.04e-3m = 3.02e-2m.

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