Answer to Question #99707 in Electricity and Magnetism for ak

Question

A particle of mass m and charge q > 0 is at the origin at t = 0 and moving upward with velocity V~ = V yˆ.
Its subsequent trajectory is shown in the sketch. The magnitude of the velocity V =
V~
is always the same,

although the direction of V~ changes in time.



(a) (1 point) For y > 0, this positive charge is moving in a constant magnetic field which is either into the
page or out of the page. Is that magnetic field for y > 0 into or out of the page?
(b) (3 points) Derive an expression for the magnitude of the magnetic field for y > 0 in terms of q, m, R
and V .
(c) (3 points) For y < 0, the charge is moving in a different constant magnetic field. Is that field for y < 0
into or out of the page? What is the magnitude of that magnetic field in terms of q, m, R and V ?
(d) (3 points) How long does it take the charge to move from the origin to point P (see sketch) along the
x-axis. Give your answer in terms of R and V

Answer 1

A positive particle experiences Lorentz and centripetal force which are in balance:

$$qvB=m\frac{v^2}{R}.$$

The direction of the Lorentz force can be determined according to the left-hand rule: place your 4 fingers toward the velocity vector, the magnetic field vector must hit into bottom of your hand.

a) The thumb shows the direction of the Lorentz force:

b) From the equation express the field:

$$B=\frac{mv}{qR}.$$

c) The magnitude can be found according to the same expression, while the direction can be found according to the rule shown in the figure.

d) The time required to make 1 circle along the orbit is

$$T=L/v=2\pi R/v.$$

From the first equation express the velocity:

$$v=\frac{qBR}{m}.$$

Hence, the time for a circle is

$$T=\frac{2\pi m}{qB}.$$

The time for a half will be two times shorter, a quarter of a circle will take four time shorter than $T$ and so on.