Q12. Magnetic force requires a charged particle to have velocity. Thus the magnetic force will always be zero unless the particle starts with some velocity. An electric field exerts a force on any charged particle of magnitude F = qE, and therefore, can make an electron at rest accelerate.
P5. Using the left hand rule - the force is directed South. The magnitude will be 7.45 x 10-13 N
P9. To produce a circular path, the magnetic field is perpendicular to the velocity. The magnetic force
provides the centripetal acceleration:
qvB = mv2/r, or
r = mv/qB;
0.25 m = (6.6 ´ 10–27 kg)(1.6 ´ 107 m/s)/2(1.60 ´ 10–19 C) B, which gives B = 1.3 T.
P14. The equation we derived yesterday comes up in this problem: qvB = mv^2/R. The difficult part is that we can write speed in terms of the radius: 2*pi*R/T = v. Making this substitution results in the expression .5mv^2 = 2*pi*qBR^2/T where T is the period of revolution.
17. skip this - we will do this in class Friday.
55. This is very similar to our result from class, but we don't know the speed of the particle when it enters the magnetic field. We DO know the particle is accelerated through a potential difference V...How can we relate a change in potential of a charge Q to the final speed it has if it starts at rest?
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