So far we’ve explored forces on permanent magnets & current-carrying conductors. But what about free-moving charged particles?
- Force on a charge moving in a magnetic field
- F = Bqv sin θ
- Determination of v & e/me for electrons
First, let’s introduce ourselves to a few charged particles:
- α-particle (Helium-4 nucleus)
- β-particle (fast-moving nuclear electron)
These particles exhibit interesting effects when they pass through a magnetic field.
What is the force on a moving charged particle in a magnetic field?
F = Bqv sin θ
Here’s a derivation:
Remember the motor effect:
F = BIL sin θ
Current I is given by the number of particles (n) of charge q passing through a point in time t:
I = nq/t
Subbing this in:
F = (Bnq/t)L sin θ
Since v = L/t,
F = Bnqv sin θ
If we are looking at the force acting on a single particle, n = 1, so
F = Bqv sin θ
For a charged particle moving perpendicular to a magnetic field,
F = Bqv
How does this effect the path of the particle?
It causes deflection.
The particle will travel across an arc of a circle. This is an example of Circular Motion (read up on it here).
We can calculate the radius of this arc as follows:
Here, the centripetal force is provided by the magnetic force:
Fc = FB
mv2/r = Bqv
The exact shape of the particle’s path depends on many factors.
Here are a few:
|Straight line||The charged particle enters a magnetic field parallel to the field lines.|
As all the velocity is along the field lines, there is NO force acting on the particle, so its motion is unchanged.
|Arc||The charged particle enters a magnetic field perpendicular to the field lines.|
It exits the magnetic field before it can complete 1 circular orbit.
Velocity selection of charged particles (see next post)
|Circular Orbit||The charged particle enters a magnetic field perpendicular to the field lines.|
It does not manage to exit the magnetic field before completing an orbit.
It will continue to move in this circular orbit unless the magnetic field strength changes.
Determining the mass-charge ratio of a particle (see below)
Original video: https://www.youtube.com/watch?v=a2_wUDBl-g8
|The charged particle enters a magnetic field at an angle to the field lines.|
There is a component of velocity along the field lines, which is unchanged.
There is also a component of velocity perpendicular to the field lines, which IS changed.
Thus, the particles follow a spring-like path.
This is useful in electron microscopes, where the helical beam can be focused onto a target.
One important application of this is learning about the properties of newly-discovered charged particles.
Let’s take the electron as an example.
The mass-charge ratio of an electron is known as its specific charge.
It is given by e/me
To do this calculation, we must use a fine-beam tube:
- electrons are accelerated from rest using a high voltage V
- Helmholtz coils provide a permanent magnetic field B perpendicular to the beam
- the circular orbit of the electron beam can be seen as electrons strike low-pressure gas inside the tube, making it glow
Calculating the mass-charge ratio of a charged particle (electron)
|We could rearrange this to |
then find the velocity, magnetic field strength, & radius.
|Unfortunately, it is difficult to measure the velocity v, so we must use another method.|
|Enter a new relation!|
Electrical energy supplied in accelerating a charged particle from rest = final kinetic energy of particle
In a fine-beam tube, the electrons are accelerated from rest via a voltage V.
Since electric potential energy = eV,
|We can now combine the 2 equations we have to eliminate v!|
|With the values of V & B (which we have set ourselves) along with r (which we can measure), we can calculate e/me|
In the next post, we will look at what happens when a charged particle enters a magnetic field AND an electric field.