Faraday's discovery, coming up next, depends on one quantity: how much magnetic field passes through a loop of wire. That amount is the magnetic flux. It uses the cosine from Sine, Cosine and Tangent and the field you met in Magnetic Fields and Their Sources.
Key idea
For a flat loop of area $A$ in a uniform field $B$, $$\Phi = BA\cos\theta,$$ where $\theta$ is the angle between $\vec{B}$ and the loop's normal (the line sticking straight out of the loop). The unit is the weber: $1\ \text{Wb} = 1\ \text{T·m}^2$.
Why it works
Think of field lines as rain and the loop as a hoop. Held flat under the rain, the hoop catches the most. Tilted, it catches less, because the area it shows to the rain shrinks to $A\cos\theta$. Held edge-on, it catches none. The flux counts field lines the same way: $B$ times the area facing the field.
Common mistake
Measuring $\theta$ from the plane of the loop instead of from its normal. If a problem says "the field makes $30^\circ$ with the plane of the loop", the angle from the normal is $90^\circ - 30^\circ = 60^\circ$. Quick check: a field lying in the loop's plane must give zero flux, and $\cos 90^\circ = 0$ does that.
Worked example: A tilted loop
A loop of area $0.050\ \text{m}^2$ sits in a $0.40\ \text{T}$ field. Its normal is $60^\circ$ from the field. Find the flux.
- $$\Phi = BA\cos\theta = (0.40)(0.050)\cos 60^\circ.$$
- $\cos 60^\circ = 0.50$, so $\Phi = 0.010\ \text{Wb}$.
Answer: $0.010\ \text{Wb}$
Worked example: Angle from the plane
A circular loop of radius $0.10\ \text{m}$ is in a $0.20\ \text{T}$ field that makes $30^\circ$ with the plane of the loop. Find the flux.
- Area: $A = \pi r^{2} = \pi (0.10)^{2} = 0.0314\ \text{m}^2$.
- Angle from the normal: $\theta = 90^\circ - 30^\circ = 60^\circ$.
- $$\Phi = (0.20)(0.0314)\cos 60^\circ = 3.1 \times 10^{-3}\ \text{Wb}.$$
Answer: $3.1 \times 10^{-3}\ \text{Wb}$
A coil of $N$ turns has the same flux through each turn, so the total, $N\Phi$, is $N$ times bigger. You will use that in Faraday's law.