# Magnetic forces on moving charged particles

An **electric charge** moving through a **magnetic field** experiences a **force**.

The **magnitude of the force** can be increased by increasing:

- the
**strength**of the magnetic field; and - the
**speed**of the charge.

The **direction** of the force is always **perpendicular** to both the **direction of movement** of the charge and to the **magnetic field**. The direction of the magnetic field can be shown using field lines.

Reversing the direction of the magnetic field reverses the direction of the force on the particle.

The **magnitude of force on a moving charged particle** by a magnetic field is given by: $$$F=BQv \sin\text{ }\theta$$$ $$F$$ is the magnitude of force on the charged particle, $$B$$ is the magnetic flux density, $$v$$ is the speed of the charged particle and $$\theta$$ is the angle between the vectors for the magnetic field and the velocity.

$$v\sin\theta$$ is the component of the velocity of the charged particle that is perpendicular to the magnetic field.

The force is always perpendicular to the velocity and the magnetic field.

The force on a charged particle moving through a magnetic field can be determined using Fleming's left hand rule.

The force $$\Tred{F}$$ on the particle is indicated by the thumb.

The magnetic field $$\Tblue{B}$$ is indicated by the index finger. Remember that this points from north to south.

The index finger represents the component of the magnetic field that is perpendicular to the velocity. This is the same as the field when all the quantities are at right angles.

The conventional current $$\Tgreen{I}$$ is indicated by the second finger. This is the same as the direction of the motion (or velocity $$v$$) of a positive charge.

Remember this using the name of the US federal police - the ** FBI**! $$F$$ is the force, $$B$$ is the magnetic field and $$I$$ is the current.

Remember that for **negative charges**, the second finger needs to point in the direction of **conventional current**, so the direction of the force is reversed.

A moving charged particle in a static magnetic field (a field that does not vary over time) will undergo **circular motion** if it is projected into the field at right angles to the field.

This is because the force is always perpendicular to the velocity of the particle.

The magnetic force functions as the ** centripetal force**. This phenomenon is similar to the circular orbit of a satellite around a massive body due to gravitation.

The **radius** of circular motion can be derived by equating the magnetic force with the centripetal force as follows: $$$\begin{align*}& BQv=\frac{mv^{2}}{r}\\ \Rightarrow & BQ=\frac{mv}{r}\\ \Rightarrow & r=\frac{mv}{BQ}\end{align*}$$$ The **period** of the circular motion, $$T$$, can be obtained from the above by: $$$\begin{align*}& BQ=\frac{mv}{r}=m\omega=m(\frac{2\pi}{T})\\ \Rightarrow & T=\frac{2\pi m}{BQ}\end{align*}$$$ $$\omega$$ is the angular velocity.

The period of circular motion is **independent of the velocity of the charged particle.**

A **velocity selector** is a device used to separate charged particles based on their velocity. They are used mainly in laboratories for experiments which require particles of a certain velocity or energy.

It consists of electric and magnetic fields oriented perpendicularly to each other and to the direction of movement of the particles to be selected.

A moving charge $$Q$$ entering a velocity selector **experiences both electric and magnetic forces** whose magnitudes are given by $$\displaystyle{F_{\text{E}}=QE}$$ and $$\displaystyle{F_{\text{B}}=BQv}$$.

$$\vecphy{F_{\text{E}}}$$ and $$\vecphy{F_{\text{B}}}$$ are the electric and magnetic forces respectively. In this experiment, these forces have **opposite directions**. The magnitude of the total force $$F_{\text{T}}$$ is given by: $$$F_{\text{T}}=F_{\text{E}}-F_{\text{B}}$$$ To pass through the velocity selector **undeflected** the electric and magnetic forces acting on the particle must be equal (i.e. $$F_{\text{E}}=F_{\text{B}}$$). This means that the velocity of particle takes a very specific value given by:$$$\begin{align*}QE & =BQv \\ \Rightarrow v & =\frac{E}{B}\end{align*}$$$ Any particle with a different velocity will be deflected in either an upwards or downwards direction depending on its velocity. Only particles with a velocity of $$v=E/B$$ will be **"selected"** to pass through undeflected.