### Magnetic Field

In the presence of a magnetic field $\vec{B}$ (a vector field), a moving charge q experiences a magnetic force $\vec{F}$.

Note: $F \propto q$, $F \propto B$, $F \propto v$

The magnetic force $\vec{F}$ is always perpendicular to both $\vec{B}$ and $\vec{v}$. The magnetic force on a charge q moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by:

$$\vec{F} = q \vec{v} \times \vec{B}$$

The **SI unit of B**: tesla, T. $1 \, \text{T} = 1 \, \text{NA}^{-1}\text{m}^{-1}$

Tesla is usually too big for everyday usage. Hence, there is another unit for B: gauss, G. $1 \, \text{G} = 10^{-4} \, \text{T}$

### Motion of charged particles in magnetic field

When a charged particle moves through a region of space where both electric and magnetic fields are present, both fields exert forces on the particle. The total force is given by: (also called Lorentz force)

$$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$$

Motion of a charged particle under the action of a magnetic field alone is always motion with constant speed.

Consider a particle with positive charge q moving with velocity $\vec{v}$ on a horizontal plane in a uniform magnetic field $\vec{B}$ directed into the horizontal plane. The particle will undergo circular motion due to the magnetic force. We can calculate the radius of the circular motion:

$$\begin{aligned}\vec{F} &= q\vec{v} \times \vec{B} \\ qvB &= m \frac{v^{2}}{R} \\ R &= \frac{mv}{qB} \end{aligned}$$

We note that the period of the circular motion is given by: $T = \frac{2 \pi R}{v}$. We can calculate the frequency of the circular motion, more commonly known as the cyclotron frequency:

$$\begin{aligned}f &= \frac{1}{T} \\ &= \frac{v}{2 \pi R} \\ &= \frac{qB}{2 \pi m} \end{aligned}$$

Note that the above scenario is for velocity being perpendicular to the magnetic field.

If the direction of the initial velocity is not perpendicular to the magnetic field, the velocity component parallel to the field will be constant giving rise to a linear motion in the direction of the magnetic field while the velocity component perpendicular to the field will give rise to a circular motion. The final motion of the particle will be a spiral motion, spiraling out in the direction of the magnetic field.

In a non-uniform magnetic field, produced by two circular coils separated by some distance, a magnetic bottle is produced. Particles close to either end of the magnetic bottle will experience a magnetic force towards the center of the region. Particles with appropriate speeds will spiral repeatedly to and fro, from one end of the region to the other. Magnetic bottles are used to trap charged particles temporarily, which is exploited to confine very hot plasmas.

### Applications Of Motion Of Charged Particles

#### Velocity Selector

It is used in velocity selector, where only particles with velocity $v = \frac{E}{B}$ will be able to pass through without being deflected by the fields. Refer to the diagram above. The positive charge will experience a force upwards due to the electric field and a force downwards from the magnetic field. When the charge is not deflected, this means that both forces are equal.

How do you obtain that mysterious ratio? We start from the Lorentz force:

$$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$$

For the particle to pass through without being deflected, the Lorentz force acting on the particle should be 0. Hence,

$$\begin{aligned} q (-E + vB) &= 0 \\ v &= \frac{E}{B} \end{aligned}$$

#### Mass Spectrometers

In mass spectrometer, the material to be investigated is ionized and fed into a velocity selector. The velocity selector will ensure that the ions exiting will have the same velocity. These same-velocity ions are then fed into a region with the magnetic field lines perpendicular to the velocity of the ions. From the EM theory, the ions will experience a magnetic force and hence, undergo circular motion. The radius of the circular motion made by the ions will depend on the mass and charge. We can see this by:

$$\begin{aligned} F &= Bqv \\ \frac{mv^{2}}{R} &= Bqv \\ R &= \frac{mv}{qB} \end{aligned}$$

#### Thomson’s e/m experiment

J.J Thomson was the first scientist who measured charge to mass ratio (e/m) of an electron. The apparatus, as shown above, consists of a highly evacuated glass tube which is fitted with electrodes. Electrons are produced by thermionic emission from the electrical heating of a tungsten filament. The electrons are made to accelerate (due to a potential difference) and form a beam. The resulting beam is passed through a region with electric and magnetic field of known magnitudes. At the end, they hit on a zinc sulphide screen.

The kinetic energy gained by the electron due to the potential difference is given by:

$$\begin{aligned} K.E. = eV &= \frac{1}{2} m v^{2} \\ v &= \sqrt{\frac{2eV}{m}} \end{aligned}$$

Notice that the electric and magnetic fields essentially functions as a velocity selector. From velocity selector, we have: (E is electric field strength)

$$v = \frac{E}{B}$$

Hence,

$$v = \frac{E}{B} = \sqrt{\frac{2eV}{m}}$$

Making $\frac{e}{m}$ the subject, we obtain:

$$\frac{e}{m} = \frac{E^{2}}{2VB^{2}}$$

By knowing the E, V and B, you can calculate the electron’s charge to mass ratio.