Learn what magnetic flux means and how to calculate it.

What is magnetic flux?

Magnetic flux is a measurement of the total magnetic field which passes through a given area. It is a useful tool for helping describe the effects of the magnetic force on something occupying a given area. The measurement of magnetic flux is tied to the particular area chosen. We can choose to make the area any size we want and orient it in any way relative to the magnetic field.
If we use the field-line picture of a magnetic field then every field line passing through the given area contributes some magnetic flux. The angle at which the field line intersects the area is also important. A field line passing through at a glancing angle will only contribute a small component of the field to the magnetic flux. When calculating the magnetic flux we include only the component of the magnetic field vector which is normal to our test area.
If we choose a simple flat surface with area AA as our test area and there is an angle θ\theta between the normal to the surface and a magnetic field vector (magnitude BB) then the magnetic flux is,
More generally, the magnetic flux can be found using the vector dot product. If B\vec{B} is a magnetic field vector and A\vec{A} is the surface-normal vector to the test area then Φ=BA\Phi = \vec{B}\cdot\vec{A}.
Φ=BAcosθ\Phi = B A \cos{\theta}
In the case that the surface is perpendicular to the field then the angle is zero and the magnetic flux is simply BAB A. Figure 1 shows an example of a flat test area at two different angles to a magnetic field and the resulting magnetic flux.
Figure 1: Magnetic flux through given areas (blue) oriented at an angle (left) and normal to (right) the magnetic field.
Figure 1: Magnetic flux through given areas (blue) oriented at an angle (left) and normal to (right) the magnetic field.
Exercise 1:
If the blue surfaces shown in Figure 1 both have equal area and the angle θ\theta is 2525^\circ, how much smaller is the flux through the area in Figure 1-left vs Figure 1-right?
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
cos250.91\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?

The SI unit of magnetic flux is the Weber (named after German physicist and co-inventor of the telegraph Wilhelm Weber) and the unit has the symbol Wb\mathrm{Wb}.
Because the magnetic flux is just a way of expressing the magnetic field in a given area, it can be measured with a magnetometer in the same way as the magnetic field. For example, suppose a small magnetometer probe is moved around (without rotating) inside a 0.5 m20.5~\mathrm{m^2} area near a large sheet of magnetic material and indicates a constant reading of 5 mT5~\mathrm{mT}. The magnetic flux through the area is then (5103 T)(0.5 m2)=0.0025 Wb(5\cdot 10^{-3}~\mathrm{T})\cdot(0.5~\mathrm{m^2}) = 0.0025~\mathrm{Wb}. In the event that the magnetic field reading changes with position, it would be necessary to find the average reading.
A related term that you may come across is the magnetic flux density. This is measured in Wb/m2\mathrm{Wb/m^2}. Because we are dividing flux by area we could also directly state the units of flux density in Tesla. In fact, the term magnetic flux density is often used synonymously with the magnitude of the magnetic field.
Exercise 2:
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)
Figure 2: A map of magnetic field measurements around a loop of wire (green).
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.
Bavg=123 mT30=4.1 mT\begin{aligned}B_{avg} &= \frac{123~\mathrm{mT}}{30} \\ &= 4.1~\mathrm{mT}\end{aligned}
Φ=ABavg=(0.05 m0.06 m)(4.1 mT)=0.0123 mWb\begin{aligned} \Phi &= AB_{avg} \\&= (0.05~\mathrm{m}\cdot 0.06~\mathrm{m}) \cdot (4.1~\mathrm{mT}) \\ &= 0.0123~\mathrm{mWb}\end{aligned}

Why is this useful?

There are a couple of reasons why the description of magnetic flux can be more useful than that of a magnetic field directly.
  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).
    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

Exercise 1:
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.
Figure 4: Magnetic flux through a loop near a straight current carrying wire.
Recall from our article on the magnetic field, the equation for magnetic field at a distance rr from a long straight wire carrying current II is
B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}.
Here μ0\mu_0 is the permeability of free space. μ0=4π107 Tm/A\mu_0 = 4\pi\cdot 10^{-7}~\mathrm{T\cdot m / 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 xx-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.
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(5103 m)(5106 T)=1.1125106 Tm44.5 \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.
Φ=(1.1125106 Tm)(75103 m)8108 Wb\begin{aligned} \Phi &= (1.1125\cdot 10^{-6}~\mathrm{Tm})\cdot (75\cdot 10^{-3}~\mathrm{m}) \\ &\simeq 8\cdot 10^{-8}~\mathrm{Wb}\end{aligned}
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