What is magnetic flux?

If the blue surfaces shown in Figure 1 both have equal area and the angle $\theta$‍ is ${25}^{\circ }$‍, how much smaller is the flux through the area in Figure 1-left vs Figure 1-right?

Solution

Looking at the equation for magnetic flux through a surface, we can see that if the area and field remain the same then the only remaining factor is the angle. In this case the diagram shows the angle between the normal to the magnetic field and the blue surface. This is the same as the angle between the magnetic field and a vector normal to the blue surface. So as

$\cos {25}^{\circ }\simeq 0.91$‍

the magnetic flux through the tilted area is about 9% smaller than through the area normal to the field.

How do we measure magnetic flux?

Figure 2 shows a map of a non-uniform magnetic field measured near a sheet of magnetic material. If the green line represents a loop of wire, what is the magnetic flux through the loop?

Figure 2: A map of magnetic field measurements around a loop of wire (green)

Solution

To find the magnetic flux we need to know the area of the loop and the magnitude of the magnetic field intersecting it. In this case the field is not uniform so we need to take an average over the area enclosed by the coil.

$\begin{array}{rl}{B}_{avg}& =\frac{123\mathrm{mT}}{30}\\ & =4.1\mathrm{mT}\end{array}$‍

$\begin{array}{rl}\mathrm{\Phi }& =A{B}_{avg}\\ & =(0.05\mathrm{m}\cdot 0.06\mathrm{m})\cdot (4.1\mathrm{mT})\\ & =0.0123\mathrm{mWb}\end{array}$‍

Why is this useful?

1. When a coil of wire is moved through a magnetic field a voltage is generated which depends on the magnetic flux through the area of the coil. This is described by Faraday's law and is explored in our article on Faraday's law. Electric motors and generators apply Faraday's law to coils which rotate in a magnetic field as depicted in Figure 3. In this example the flux changes as the coil rotates. The description of magnetic flux allows engineers to easily calculate the voltage generated by an electric generator even when the magnetic field is complicated.

   Figure 3: Simplified diagram of a rotating coil in an electric generator (public domain).

2. Although we have thus-far only concerned ourselves with magnetic flux measured for a simple flat test-area, we can make our test-area a surface of any shape we like. In-fact, we can use a closed surface such as a sphere which encloses a region of interest. Closed surfaces are particularly interesting to physicists because of Gauss's law for magnetism. Because magnets always have two poles there is no possibility (as far as we know) that there is a magnetic monopole inside a closed surface. This means that the net magnetic flux through such a closed surface is always zero and therefore all the magnetic field lines going into the closed surface are exactly balanced by field lines coming out. This fact is useful for simplifying magnetic field problems.

Magnetic flux around a current-carrying wire

Figure 4 shows a square loop of wire placed near a current carrying wire. Using the dimensions shown in the figure, find the magnetic flux through a coil. If you don't know how to calculate the magnetic field around a wire, review our article on the magnetic field. Hint: it may be useful to plot the magnetic field vs vertical distance from the wire.

Figure 4: Magnetic flux through a loop near a straight current carrying wire.

Solution

Recall from our article on the magnetic field, the equation for magnetic field at a distance $r$‍ from a long straight wire carrying current $I$‍ is

$B=\frac{{\mu }_{0}I}{2\pi r}$‍.

Here ${\mu }_0$‍ is the permeability of free space. ${\mu }_0=4\pi \cdot {10}^{-7}\mathrm{T}\cdot \mathrm{m}/\mathrm{A}$‍.

If we plot the magnetic field as a function of vertical distance we can use the area under the curve between points on the $x$‍-axis that represent the area intersected by the coil to find the flux. There are many ways to evaluate this; the most accurate is to use calculus. However, a simple graphical evaluation is sufficient for many purposes. Figure 5 shows a plot with the gray area representing the area intersected by the coil.

Figure 5: Plot of magnetic field variation with distance in exercise 1.

Counting the gray squares in Figure 5, we have approximately 44.5 squares. So the total area is

$44.5\cdot (5\cdot {10}^{-3}\mathrm{m})\cdot (5\cdot {10}^{-6}\mathrm{T})=1.1125\cdot {10}^{-6}\mathrm{Tm}$‍

The area directly gives us magnetic field times distance but we need magnetic field times area. The field is not varying along the horizontal direction so we can simply multiply by the width of the coil.

$\begin{array}{rl}\mathrm{\Phi }& =(1.1125\cdot {10}^{-6}\mathrm{Tm})\cdot (75\cdot {10}^{-3}\mathrm{m})\\ & \simeq 8\cdot {10}^{-8}\mathrm{Wb}\end{array}$‍