A charged particle with initial velocity v 0 = 2.6 × 10 6 m/s in the positive x

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   m

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