Answer to Question #99605 in Electricity and Magnetism for Ali hamza

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 =



Mod V~


 is always the same,
although the direction of V~ changes in time.

(a) 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)Derive an expression for the magnitude of the magnetic field for y > 0 in terms of q, m, R
and V .
(c)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) 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)   the direction of the magnetic field for y > 0 is out of the page

(b) the magnitude of the magnetic force is given by formula

$F = qvB (1)$

We now apply Newton’s Second Law

$F=ma = \frac {mv×v} {R} (2)$

We put (1) in (2)

$B=\frac {mv×v} {qR}$

(c)   The direction of the magnetic field for y < 0 is into the page

We use Newton’s Second Law again noting that the radius of the circular orbit is R / 2 and so the magnitude of the magnetic field is

$B=\frac {2mv} {qR}$

(d) For each circular leg we use t = d / v and find that the time to reach point P is

$T=\frac {3πR} {2v}$