# Magnetic Flux Density

## Magnetic Flux Density Revision

**Magnetic Flux Density**

Motors and generators are extremely useful in everyday appliances. These work based on the interaction of **magnetic fields**. To fully explain how motors and generators work, we need to begin by understanding the concept of **magnetic flux density**.

**Magnetic Fields**

Any **current-carrying conductor** produces a **magnetic field **around it. If the **magnetic field** of the wire interacts with another **magnetic field**, a **force **will be exerted on the wire.

In the diagram on the right, a current carrying wire is being passed through a **magnetic field**. As the **magnetic field **of the wire interacts with the **magnetic field **of the magnets, a **force **is exerted on the wire. This **force **causes the wire to move if it’s not held in place.

The **force** applied to the **current-carrying wire** is determined by the equation:

F = BIL

- F is the
**force**on the wire in Newtons \left(\text{N}\right). - B is the
**magnetic flux density**in tesla \left(\text{T}\right). - I is the
**current**in amps \left(\text{A}\right). - L is the
**length**of the wire in metres\left(\text{m}\right).

It is important for the above equation to work that the current carrying wire and the **magnetic field **of the magnet are **perpendicular **to each-other.

If not, the equation can be adapted to:

F = BIL \sin \theta

- \theta is the
**angle**between the wire and the magnetic field lines of the magnets.

Therefore, if the current-carrying wire and the **magnetic field** are **parallel** \left(\theta = 0 \degree\right), \sin \theta would equal zero and **no force** would be produced.

**Example:** A current of 2 \: \text{A} flows through a wire perpendicular to a magnetic field of flux density \left(B\right) 100 \: \text{mT}. The length of the wire is 0.5 \: \text{m}. What is the magnitude of force on the wire?

**[2 marks]**

Because the wire is perpendicular to the field, \sin \theta = \sin 90 = 1 .

F = BIL

F = 100 \times 10^{-3} \times 2 \times 0.5

F = 100 \times 10^{-3} \: \text{N}

**Fleming’s Left Hand Rule**

Let’s take another look at the previous diagram:

We can see that the direction of the **force \left(F\right)**, **current \left(I\right)** and **Magnetic Flux \left(B\right)** have been labelled. **Fleming’s left hand rule** can be used to determine the direction of these variables. As the name suggests, it is important that the** left hand** is used and never the right.

Next to the diagram is a representation of **Fleming’s left hand rule**. The **thumb** represents the **force** or motion, the **first finger** is the direction of the **magnetic flux** and the** second finger** represents the **current **in the wire. As long as two of the directions are known, the third can be found by rotating the left hand to match the directions of the two variables known.

**Magnetic Flux Density**

The **magnetic flux density** is the strength of the **magnetic field** and is measured in Tesla’s. It can be defined as the magnitude of **force** acting on a current-carrying wire **per unit of current per unit of length** when the wire is placed within a magnetic field. This gives the following equation, which is just a rearranged version of the equation above:

B = \dfrac{F}{IL}

For this equation to be applicable, the magnetic field must be perpendicular to the wire. Magnetic flux density is given the unit Tesla \left(\text{T}\right) where 1 Tesla is equal to 1 Newton of force acting on a wire of 1 \: \text{m} in length with 1 \: \text{A} passing through it.

**Example:** A 32 \: \text{cm} length of wire is placed perpendicular to a magnetic field. Calculate the flux density when a current of 5 \: \text{A} flows through the wire, producing a force of 0.1 \: \text{N}.

**[2 marks]**

B = \dfrac{F}{IL}

B = \dfrac{0.1}{\left(5 \times 32 \times 10^{-2}\right)}

B = 0.063 \: \text{T}

**Required Practical 10**

**Investigating how force varies with current through a wire in a magnetic field.**

This experiment will use the following set up to observe how changing the current through a wire in a **magnetic field** will effect the force on this wire.

A similar experiment can be conducted where instead of the current through the wire being changed, the length of wire or strength of magnetic field is changed.

**Doing the experiment:**

- Set up the experiment as shown above, with the power supply initially switched off. Place the magnet on a
**top pan balance**. Reset the top pan balance to 0.0 \: \text{g}. Also measure the**length of wire**. - Ensure the wire is completely
**perpendicular to the magnets**as this would influence the experiments result if not. - Measure the
**length of the magnets**as this will give us the length of wire \left(L\right) inside the magnetic field. - Switch the power supply on and adjust the
**current**to 0.5 \: \text{A} using the**variable resistor**. - As a result of the current now flowing through the wire, an
**upwards force**is exerted on the wire. Due to**Newton’s 3rd Law**, there is equal and opposite force exerted on the magnet by the current carrying wire. Therefore there is a reading on the top pan balance. Record this reading on the top pan balance. - Repeat this method but each time increase the current in intervals of 0.5 \: \text{A}, up until an appropriate maximum current.
- Repeat the entire experiment for each current 3 times and take a mean average mass for each current.

**Analysing the results:**

As the force on a** current carrying wire** is given by the equation;

F = BIL

The top pan balance measures the mass pushing down upon it. We know that the corresponding force is the mass multiplied by the gravitational acceleration: F = mg.

Therefore we can rewrite the equation as:

mg = BIL

Plotting a graph of m against I would give a linear graph through the origin with a gradient equal to \dfrac{BL}{g}:

\begin{aligned} \textcolor{7cb447}{y} &= \textcolor{d11149}{m}\textcolor{2730e9}{x} \\ \textcolor{7cb447}{m} &= \textcolor{d11149}{\dfrac{BL}{g}}\textcolor{2730e9}{I}\end{aligned}

So in order to calculate the magnetic flux density we can use the equation:

\text{gradient} = \dfrac{BL}{g}

B = \dfrac{\text{gradient} \times g}{L}

## Magnetic Flux Density Example Questions

**Question 1:** What is magnetic flux density?

**[1 mark]**

The **strength of a magnetic field,** often measured in tesla.

**Question 2:** What would the force on a current-carrying wire be if the wire and the magnetic fields are acting parallel to each other? Explain your answer.

**[1 mark]**

The force would be \boldsymbol{0} \: \textbf{N}, **because** \boldsymbol{\sin \theta = \sin 0 = 0} \: \textbf{N}. Therefore F = BIL \sin \theta = 0 \: \text{N}.

**Question 3:** A magnet produces a magnetic flux density of 5 \: \text{mT}. If 2 \: \text{m} of wire passes through the magnetic field carrying 3 \: \text{A}, what is the magnitude of force applied?

**[2 marks]**