# Magnetic Flux and Flux Linkage

## Magnetic Flux and Flux Linkage Revision

**Magnetic Flux and Flux Linkage**

To understand **electromagnetic induction**, we first must understand what **magnetic flux** and **flux linkage** are.

**Magnetic Flux**

When a wire passes through a **magnetic field**, the wire cuts through the **magnetic field lines**. The **magnetic flux** is the total amount of **magnetic field **that the wire passes through as it passes through the magnetic field.

As with the force in a magnetic field, if the conducting wire passes through the field at 90 degrees (perpendicular), the maximum amount of **magnetic flux** is measured. If the wire passes through **parallel **to the magnetic field, magnetic flux is zero.

The **magnetic flux** \left(\Phi\right) could also be defined as the number of **magnetic field lines** being cut through a given area by a conductor. As an equation, this can be written:

\Phi = BA

- \Phi is the
**magnetic flux**measured in Webers \left(\text{Wb}\right). - B is the
**magnetic flux density**in Tesla \left(\text{T}\right). - A is the
**cross-sectional area**in metres squared \left(\text{m}^2\right).

This equation can only be used when the **magnetic flu**x is **perpendicular **to the **cross-sectional area** of the conductor, as in the diagram above.

Alternatively, if the **cross-sectional area** and **magnetic flux** are not perpendicular, the equation below can be used:

\Phi = BA \cos \theta

where \theta is the **angle** between the cross-sectional area and the magnetic field.

**Example:** A wire of cross sectional 0.001 \text{m}^2 cuts through a magnetic field with density 2 \times 10^{-3} \: \text{T}. Calculate the magnetic flux density and give the appropriate units.

**[2 marks]**

\Phi = BA

\Phi = 2 \times 10^{-3} \times 0.001

\Phi = 2 \times 10^{-6} \: \text{Wb}

**Magnetic Flux Linkage**

The **magnetic flux density** and the force produced by a single wire is usually very low. Instead, using a coil of wire increases the **cross sectional area** and increases the **magnetic flux density** produced.

The **magnetic flux linkage** refers to the number of turns on a coil and multiples this by the magnetic flux of one wire. This gives the equation:

N\Phi = BAN

where N is the **number of turns** on a coil.

Again, if the **cross-sectional area** and **magnetic flux** are not perpendicular, the equation below can be used:

N \Phi = BAN \cos \theta

where \theta is the **angle** between the cross-sectional area and the magnetic field.

**Example:** A solenoid contains 250 turns of wire. Each piece of wire has a cross sectional area of 0.002 \text{m}^2. If the magnetic flux density is 8.0 \: \text{mT}, calculate the magnetic flux linkage.

**[2 marks]**

N\Phi = BAN

N\Phi = 8.0 \times 10^{-3} \times 0.002 \times 250

N\Phi = 4 \times 10^{-3} \: \text{Wb turns}

**Required Practical 11**

**Investigating Magnetic Flux Linkage.**

A **search coil** and **oscilloscope **can be used to investigate the effect of varying the angle between the search coil and a magnetic field on** magnetic flux linkage**. The equipment is shown below:

The circular coil of wire creates a **uniform magnetic field** of which the search coil is rotated within. The **oscilloscope** will display the varying EMF produced as the search coil rotates within the coil of wire.

When experimenting, the peak EMF will be determined when the search coil is **perpendicular **to the **magnetic field** of the coil of wire. A reading can then be taken every 10 \degree of rotation of the search coil.

The graph here may be plotted to show the relationship between **peak EMF** and the rotation of the search coil. \theta is the **angle of rotation**.

It is likely that the y-intercept of our experimental results will not be zero when the search coil and coil of wire are parallel. This is because of a degree of human error when trying to make sure the wires and search coil are **parallel**.

## Magnetic Flux and Flux Linkage Example Questions

**Question 1:** A wire of cross sectional area 0.005 \: \text{m}^2 cuts through a magnetic field with density 5 \times 10^{-3} \: \text{T}. Calculate the magnetic flux density and give the appropriate units.

**[2 marks]**

**Question 2:** Describe why a solenoid would produce a greater magnetic flux than a single loop of wire.

**[2 marks]**

A coil of wire (solenoid) would **increase the cross-sectional area compared to a single loop of wire**. We can see this in the equation \Phi = BA where \Phi and A are proportional.

**Question 3:** A solenoid contains 500 turns of wire. Each piece of wire has a cross sectional area of 0.010 \: \text{m}^2. If the magnetic flux density is 12.0 \: \text{mT}, calculate the magnetic flux linkage.

**[2 marks]**