Magnetic Fields Tutorial 3  Force
on a Charge
Interaction of Charged Particles with a Magnetic Field
We
know that a magnetic field and an electric current interact to produce a force.
Since a current is a flow of charge, it is reasonable to suppose that a
magnetic field exerts a force on individual charge carriers.
We
find that in a magnetic field, the force acts on a stream of electrons always at
90^{o} to the direction of
the movement. Therefore the path is
circular.
Consider
a charge q
moving through a magnetic field
B at a
constant velocity v.
The charge forms a current that moves a certain distance,
l, in a time
t.
We
know:
Velocity
= distance
¸
time
Current
= charge
¸
time
F = BIl
So
we can substitute into this relationship to give us:
F = B ´ (q/t) ´ vt = Bqv
So
the formula now becomes:
F = Bqv
[
F
force in N;
B
– field strength in
T;
q
–
charge in C;
v –
speed in m s^{1}]
The charge is usually the electronic charge, 1.6 ´ 10^{19} C.
If the magnetic field is at an angle q to the magnetic field, the equation is modified to:
F = Bqv sin q
If no angle is mentioned in the question, assume that the angle is 90^{o}.
An electron accelerated to 6.0 ´ 10^{6} m s^{1} is deflected by a magnetic field of strength 0.82 T. What is the force acting on the electron? Would it be any different for a proton? 
Remember
that the direction of the electrons’ movement is in the opposite
direction to the conventional current. So if the electrons are
going from left to right, the conventional current is going from right
to left. 
Path
of charged particles in a Magnetic Field
We have seen that the force always acts on the wire at 90^{o}, and that gives us the condition for circular motion.
We can combine the
relationship
a = v^{2}/r with
Newton II to give us:
Therefore:
The v on the left cancels to get rid of the v^{2} term on the right:
This
rearranges to give us:
Worked
Example An electron passes through a cathode ray tube with a velocity of 3.7 × 10^{7} m s^{1}. It enters a magnetic field of flux density 0.47 mT at a right angle. What is the radius of curvature of the path in the magnetic field? 
Combine F = Bqv and F = mv^{2}/r to give:

r
= 9.11 × 10^{31} kg × 3.7 × 10^{7} m
s^{1} = 0.39
m = 39 cm 0.47 × 10^{3} T × 1.6 × 10^{19} C

In a particle physics experiment, a detector is placed in a magnetic field of 0.920 T. A particle is found to produce a circular track of radius 0.500 m. Other experiments have shown that the particle carries a charge of +1.60 ´ 10^{19} C and that its speed was 3.00 ´ 10^{7} m s^{1}.
What is the mass of the particle? How does it compare to the mass of an electron (9.11 ´ 10^{31} kg)? 
The cyclotron is a particle accelerator that relies on this idea. The machine’s main components are two Dshaped electrodes ("Dees") in an evacuated chamber, placed between the poles of a large electromagnet.
From the top it looks like this:
Notice
that the beam of particles is not circular, but a spiral.
This is because the particles are being accelerated by the electric field
between each Dshaped electrode (called a dee).
As their speed increases, so does the radius of the curved path.
If
a particle of charge
q
enters one of the dees with a speed
v,
it will move in a semicircular path of radius
r.
Þ rearranging gives us
We
can work out from
t = s/v
what time it takes for the charge to travel:
For 1 revolution:
s = 2pr
Therefore:
The r terms cancel:
Since f = 1/t:
Rearranging gives us:
A cyclotron has magnets of flux density 1.50 T and the polarity changes with a frequency of 2.00 MHz. A large charged particle which has a charge of +2e is inserted into the machine and is accelerated. Calculate the mass of the particle. 
Click HERE to see an animation by Stephen Lucas, a past student of mine. I am most grateful for his permission to use it.