Magnetic flux, $\Phi_B = BA\cos\theta$, counts how much field passes through a loop. Faraday's law, next, says that what matters for induction is how fast the flux changes. So before using it, we need to see every way flux can change, and how to find it when the field is not uniform.
Key idea
$$\Phi_B = \int \vec{B}\cdot d\vec{A}\qquad(\text{webers: } 1\ \text{Wb} = 1\ \text{T·m}^2)$$
For a uniform field and a flat loop, $\Phi_B = BA\cos\theta$ with $\theta$ from the normal. The flux changes when $B$ changes, when the area changes, or when the angle changes.
Why it works
The dot product picks out the part of $\vec{B}$ that crosses the surface, exactly as for electric flux. When $\vec{B}$ varies across the loop, the loop must be cut into strips where it is nearly constant, and the strips added with an integral. Each of the three factors can change in time: a magnet moving closer changes $B$, a sliding rod changes $A$, and a spinning coil changes $\theta$, even in a perfectly steady field.
Worked examples
Worked example: A tilted loop
A square loop $10\ \text{cm}$ on a side sits in a uniform $0.40\ \text{T}$ field, with its normal at $60^\circ$ to the field. Find the flux.
- $A = (0.10)^2 = 0.010\ \text{m}^2$.
- $\Phi = BA\cos 60^\circ = (0.40)(0.010)(0.50) = 2.00\ \text{mWb}$.
Answer: $\Phi = 2.00\ \text{mWb}$
Worked example: Turning in a steady field
The same loop is turned until its normal is at $90^\circ$ to the field. How much did the flux change?
- At $90^\circ$, $\cos 90^\circ = 0$: no field passes through.
- $\Delta\Phi = 0 - 2.00\ \text{mWb} = -2.00\ \text{mWb}$, even though $B$ never changed.
Answer: $\Delta\Phi = -2.00\ \text{mWb}$
Worked example: Beside a long wire
A long wire carries $20\ \text{A}$. A rectangular loop in the same plane has sides of $20\ \text{cm}$ parallel to the wire, at $2.0$ and $10\ \text{cm}$ from it. Find the flux.
- The field $\dfrac{\mu_0 I}{2\pi r}$ is stronger near the wire, so use strips parallel to the wire, of area $(0.20)\,dr$, each with one value of $B$.
- $$\Phi = \int_{0.02}^{0.10}\frac{\mu_0 I}{2\pi r}(0.20)\,dr = \frac{\mu_0 I(0.20)}{2\pi}\ln 5 = (2 \times 10^{-7})(20)(0.20)\ln 5$$
- $\Phi = 1.29 \times 10^{-6}\ \text{Wb}$, about $1.29\ \mu\text{Wb}$.
Answer: $\Phi \approx 1.29\ \mu\text{Wb}$
Common mistake
"A steady field cannot change the flux." It can, if the loop turns: $\cos\theta$ changes even though $B$ does not. That is exactly how a generator works, spinning coils in the steady field of a magnet. Check by listing all three factors in $BA\cos\theta$ and asking which of them is changing.