# Moving Charges

## Moving Charges Revision

**Moving Charges**

Instead of considering the **force** acting on a current carrying wire, we may also need to consider the force acting on a **single moving charged particle**.

**Force on a Charged Particle**

If a **charged particle** (such as a proton or electron) is passed through a **magnetic field**, the charged particle will experience a **force**. The direction of the force can be calculated using **Flemingâ€™s left hand rule**. However, for an electron the direction of current would be opposite to the direction of electron flow as current is defined as the rate of flow of positive charge.

The magnitude of the **force** may be calculated using:

F = BQv

- F is the
**force**on the charged particle in newtons \left(\text{N}\right). - B is the
**magnetic flux density**in tesla \left(\text{T}\right). - Q is the
**charge**of the particle in coulombs \left(\text{C}\right). - v is the
**speed**of the moving charged particle in metres per second \left(\text{ms}^{-1}\right).

As with the previous equation, all three variables must be at **right angles** (perpendicular) to each other to achieve maximum force. No force will be experienced if current and magnetic flux density are parallel.

**Example:** An electron passes perpendicular to a magnetic field. Given that the magnetic flux density is 0.5 \: \text{T} and the electron moves at a velocity of 6 \times 10^7 \: \text{ms}^{-1}, calculate the force on the electron.

**[3 marks]**

F = BQv

F = 0.5 \times 1.6 \times 10^{-19} \times 6 \times 10^7

F = 4.8 \times 10^{-12} \: \text{N}

**Cyclotrons**

A **cyclotron** is a type of** particle accelerator** that makes use of **magnetic fields** and** electric fields** to accelerate charged particles.

A source of **charged particles** is released at the centre of the **cyclotron** into one of the Dees. The **magnetic field** within the Dees always acts towards the centre and **perpendicular **to the motion of the **charged particle**. The magnetic field **accelerates** the particle but it **does not change the speed**. Instead, it only changes the direction of the particle.

When the particle leaves the Dee, it is **accelerated across the gap by an electric field**. As velocity has increased, the particle now takes a path with a **greater radius** than before and leaves the electrode again. The process repeats, repeatedly accelerating the particle.

The radius of the path of the charged particle is determined by its **velocity**. The **centripetal force** is caused by the force of the** magnetic field** in the charged particle. These forces must be balanced in order for the charged particle’s radius to remain constant.

Therefore we can equate the **centripetal force** equation with the **force of a magnetic field** equation to give:

\dfrac{mv^2}{r} = BQv

As there is a v on both sides of the equation we can cancel these out to give:

\dfrac{mv}{r} = BQ

Rearranging to make r the subject gives:

r = \dfrac{mv}{BQ}

This shows that the **radius** of the charged particle is **proportional** to the **mass **of the particle (which would be constant) and the **velocity** the particle is travelling at. It also shows the radius is inversely proportional to the charge of the charged particle and the magnetic flux density.

## Moving Charges Example Questions

**Question 1:** A proton passes perpendicular to a magnetic field. Given that the magnetic flux density is 0.2 \: \text{T} and the proton moves at a velocity of 1.1 \times 10^7, calculate the force on the electron.

**[3 marks]**

**Question 2:** Describe how a cyclotron uses magnetic fields.

**[2 marks]**

The **magnetic fields in a cyclotron are used to keep the charged particle in circular motion.** As the magnetic field acts perpendicular to motion, it **acts as a centripetal force** keeping the radius inside the Dee constant for each time it enters the Dee.

**Question 3:** A charged particle is accelerated so that its velocity doubles every time it passes the gap between the Dees. After crossing the gap 5 times, how much has the radius of the circular motion increased?

**[3 marks]**

As r =\dfrac{mv}{BQ}, radius and velocity are proportional. Therefore, the radius also doubles each time it crosses between the Dees. So after 5 times, the radius will increase by 2^5 times = 32 times.

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