**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/m
_{e}for electrons

**First, let’s introduce ourselves to a few charged particles:**

- electron
- proton
- 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:

Remember that centripetal force F_{c} is the NET FORCE on any moving object.

Here, the centripetal force is provided by the magnetic force:

F_{c} = F_{B}

mv^{2}/r = Bqv

Rearranging this,

__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. Applications: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. Applications:Determining the mass-charge ratio of a particle (see below) |

HelixOriginal 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.Applications: |

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/m_{e}

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)__

Remember, So, |

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/m_{e} |

*In the next post, we will look at what happens when a charged particle enters a magnetic field AND an electric field.*