Describe how magnetic flux can be sensed by the changing alignment of a magnet on a compass needle and quantitatively analyse the concept of magnetic flux density B and flux Phi = B A cos theta in a magnetic field

What this dot point is asking

NESA wants you to distinguish magnetic flux density $B$ (the field at a point, in tesla) from magnetic flux $\Phi$ (the total field through a surface, in weber), apply $\Phi = B A \cos \theta$ correctly, and connect this concept to the qualitative idea of a compass needle responding to field direction. Flux is the bridge to Faraday's law in the next dot point.

The answer

Magnetic flux density B

The magnetic flux density (often just called the magnetic field) $\vec{B}$ at a point is a vector describing the strength and direction of the magnetic field there. It is what a compass needle aligns with, and it determines the force on a moving charge ($F = qvB$) or on a current ($F = BIL \sin \theta$).

SI unit: the tesla (T). Equivalent forms:

Typical magnitudes:

• Earth's surface field: about $5 \times 10^{-5}$ T.
• Bar magnet near a pole: 0.01 to 0.1 T.
• MRI scanner: 1.5 to 3 T (some research machines reach 7 T).
• Strong laboratory electromagnet: up to 10 T.
• Neutron star: $10^8$ T (and rising).

Magnetic flux

For a flat surface of area $A$ placed in a uniform field $\vec{B}$, the magnetic flux through the surface is:

where $\theta$ is the angle between $\vec{B}$ and the normal to the surface (the vector perpendicular to the surface). SI unit: the weber (Wb), where 1 Wb = 1 T m$^2$.

Two ways to picture it:

1. Flux is the "amount of field passing through" the surface. More field, more area, or more alignment with the surface normal all increase flux.
2. Flux is the dot product $\Phi = \vec{B} \cdot \vec{A}$, where $\vec{A}$ is the area vector (magnitude $A$, direction along the normal).

Special angles:

• $\theta = 0°$ (field along the normal): $\Phi = BA$, maximum flux.
• $\theta = 90°$ (field in the plane of the surface): $\Phi = 0$, no flux through the surface.
• $\theta = 180°$: $\Phi = -BA$, maximum negative flux (the field passes through the surface in the opposite sense).

The angle convention (watch this)

The $\theta$ in $\Phi = B A \cos \theta$ is the angle between $\vec{B}$ and the normal to the surface, not between $\vec{B}$ and the surface itself. Questions sometimes give the angle between the field and the plane of a coil; you must take the complement.

"Plane at 30° to the field" $\Rightarrow$ normal at 60° to the field $\Rightarrow$ $\cos 60° = 0.5$.

"Normal at 30° to the field" $\Rightarrow$ $\cos 30° = 0.866$.

Flux through multiple turns

A coil of $N$ turns links flux $N$ times (each turn intercepts the same flux, in series). The flux linkage is:

Faraday's law uses flux linkage: $\varepsilon = - N \, d\Phi / dt$, not just $d\Phi / dt$. We treat this in the induction dot point.

Compass needles and flux qualitatively

A compass needle is a small magnetic dipole. It aligns with the local field direction so that its north pole points along $\vec{B}$. By placing compasses (or sprinkling iron filings) over a region you can map the direction of $\vec{B}$ at every point, hence the field line pattern. The density of the lines (lines per unit area perpendicular to them) is proportional to the flux density $B$, hence the name.

If you tilt a small loop of wire in a uniform field while watching the field lines, the number of lines threading the loop changes as $\cos \theta$. That is the geometric content of $\Phi = BA \cos \theta$.

Worked example: rotating coil

A square coil of side $0.20$ m and $50$ turns is rotated in a uniform field of $0.30$ T. Find the maximum flux linkage and the flux linkage when the coil normal is at $45°$ to the field.

Area: $A = 0.20^2 = 0.040$ m$^2$.

Maximum flux linkage (normal aligned with field, $\theta = 0°$):

$\Psi_{\max} = N B A = 50 \times 0.30 \times 0.040 = 0.60$ Wb.

At $\theta = 45°$:

$\Psi = N B A \cos 45° = 0.60 \times 0.707 = 0.42$ Wb.

As the coil rotates, the flux linkage oscillates between $+0.60$ Wb and $-0.60$ Wb, with the rate of change driving the induced EMF in a generator (Faraday's law).

Examples in context

Example 1. Flux through a Snowy 2.0 generator rotor pole face. Each salient pole of a Snowy 2.0 generator has a $0.40 \text{ m} \times 0.30 \text{ m}$ face producing $B = 1.6 \text{ T}$ at the air gap. With the pole face perpendicular to $\vec{B}$ (i.e. normal to the surface aligned with the field, $\theta = 0$), the flux through the pole face is $\Phi = B A \cos 0^{\circ} = 1.6 \times 0.40 \times 0.30 = 0.192 \text{ Wb}$. As the rotor spins past a stator coil, the flux linking that coil rises and falls sinusoidally between $\pm 0.192 \text{ Wb}$ at the rotation frequency, driving the induced AC EMF that feeds the NSW grid.

Example 2. Earth's flux through a Coffs Harbour orienteering compass. The Earth's magnetic field strength at Coffs Harbour is $B \approx 5.5 \times 10^{-5} \text{ T}$, inclined $60^{\circ}$ to horizontal. A horizontal compass needle ($A = 2.0 \times 10^{-4} \text{ m}^2$) has flux $\Phi = B A \cos 60^{\circ}$ (the vertical component of $\vec{B}$ passes through the horizontal area) $= 5.5 \times 10^{-5} \times 2.0 \times 10^{-4} \times 0.50 = 5.5 \times 10^{-9} \text{ Wb} = 5.5 \text{ nWb}$. This minuscule flux is enough to align the needle with Earth's field. A 3D field sensor (e.g. in a smartphone) reads all three components and combines them.

Try this

Q1. Define magnetic flux density and magnetic flux, giving SI units of each. [2 marks]

• Cue. Flux density $B$ in tesla (T = Wb/m$^2$); flux $\Phi = B A \cos\theta$ in weber (Wb = T m$^2$).

Q2. A circular coil of radius $5.0 \text{ cm}$ sits with its plane at $30^{\circ}$ to a uniform field of $0.40 \text{ T}$. Calculate the flux linking the coil. [3 marks]

• Cue. Area $A = \pi r^2 = 7.85 \times 10^{-3} \text{ m}^2$. Angle to normal is $60^{\circ}$, so $\Phi = 0.40 \times 7.85 \times 10^{-3} \times \cos 60^{\circ} = 1.57 \times 10^{-3} \text{ Wb}$.

Q3. A rectangular loop $0.20 \text{ m} \times 0.10 \text{ m}$ is placed in a field $B = 0.50 \text{ T}$. (a) Calculate the maximum flux through the loop. (b) Find the flux when the loop normal is at $45^{\circ}$ to $\vec{B}$. (c) State and explain when the flux is zero. [2+2+2 marks]

• Cue. (a) $\Phi_{\max} = 0.50 \times 0.02 = 0.010 \text{ Wb}$. (b) $\Phi = 0.010 \cos 45^{\circ} = 7.07 \times 10^{-3} \text{ Wb}$. (c) Normal at $90^{\circ}$ to $\vec{B}$ (loop plane parallel to $\vec{B}$).