A charged particle with initial velocity v 0 = 2.6 × 10 6 m/s in the positive x
ID: 1672106 • Letter: A
Question
A charged particle with initial velocity v0= 2.6 × 106 m/s in the positivex-direction enters a region of depthd1 = 2 m that has a uniform magnetic fieldB = 0.017 T in the positive z-direction (out ofthe page). The magnetic field is zero elsewhere. The particleleaves the magnetic field region with a velocity vector at an angleq = 8o with respectto the x-axis.
(a) What is the magnitude v0' of theparticle's velocity when it exits the magnetic field region?
v0' = m/s
(b) What is the radius of curvature R of the particle'strajectory when in the region of the magnetic field?
R = m
(c) Calculate the ratio q/m of the charge tothe mass of the particle. Be sure to include the correct algebraicsign in your answer.
q/m = C/kg
(d) Calculate the displacement d2 in they-direction of the particle from its original trajectoryat the point where the particle exits the magnetic fieldregion.
d2 = m
Explanation / Answer
a. Since magnetic forceworks normal to the direction of motion of the particle, themagnitude of velocity is not altered, hence v0' = v0 = 2.6* 106 m/s = 2.6* 106 m/s b. The angle is verysmall, hence the length of arc from entering point to exiting pointcan be approximated to the length of magnetic field itself.then ( inradians) = arc length /radius 80 */180 = 2 m / r radius r = 2* 180 / 8 * 3.14 = 14.33 m c. radius of path ofchargedparticle r = m* v / B * q charge to massratio q /m = v / B * r = 2.6* 106 / 0.017 * 14.33 = 1.067* 107 C/kg d. tan = d2 /d1 tan80 = d2 /2 verticaldisplacement d2 = 0.281 m tan80 = d2 /2 verticaldisplacement d2 = 0.281 mRelated Questions
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