magnetic flux and Faraday induction law

Magnetic flux $\Phi$ is the flux of the magnetic induction vector $\vec{B}$ through a surface; for a small area element $d\vec{A}$ it is $d\Phi = \vec{B}\cdot d\vec{A}$, and for the whole surface $\Phi = \int \vec{B}\cdot d\vec{A}$. If the field is uniform and the area is flat, then $\Phi = BA\cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the normal to the area. The sign of the flux is set by the chosen normal direction, so reversing the normal reverses the sign of $\Phi$. The field $\vec{B}$ from several sources is found by superposition, so first add the fields and then compute the flux of the total field. Faraday's law states that the induced emf in a circuit equals the negative rate of change of magnetic flux: $\mathcal{E} = -d\Phi/dt$; for $N$ turns, $\mathcal{E} = -N\,d\Phi/dt$. Lenz's law fixes the sign: the induced current flows so that its magnetic field opposes the change in flux.

A flat square loop with side $a = 0.20$ m is in a uniform magnetic field with induction $B = 0.50$ T. The normal to the loop makes an angle $\theta = 60^\circ$ with $\vec{B}$. Find the magnetic flux through the loop.

A coil has $N = 200$ turns of area $A = 1.5\times 10^{-3}$ m$^2$. A uniform magnetic field is perpendicular to the plane of the turns and decreases from $B_1 = 0.40$ T to $B_2 = 0.10$ T in $\Delta t = 0.050$ s. Find the average induced emf and state the direction of the induced current if the initial field was directed away from the observer.

A long straight wire carrying current $I$ lies in the plane of a rectangular loop. The loop has length $L$ (parallel to the wire) and width $w$ (perpendicular to the wire); the nearest side of the loop is at distance $d$ from the wire. The magnetic field of the wire at distance $r$ is $B = \mu_0 I/(2\pi r)$ and is perpendicular to the plane of the loop. The current decreases as $I(t) = I_0 - kt$, where $k>0$. Find the magnetic flux through the loop at time $t$ and the induced emf.

Self-study with a simulation. Open an electromagnetic induction simulation (for example, PhET "Electromagnetic Induction" or a similar one) and complete the steps:

1. Take a pickup coil and a bar magnet. Slowly move the magnet into and out of the coil; record when the galvanometer deflects.
2. Repeat the motion at a higher speed. Compare the maximum deflection; explain it using $\mathcal{E} = -d\Phi/dt$.
3. Leave the magnet stationary inside the coil. Check whether there is current; explain using $d\Phi/dt=0$.
4. Rotate the coil by $90^\circ$ so that the coil axis becomes perpendicular to the field. Compare the deflections for the same magnet motion.
5. Draw a conclusion: what exactly must change for an emf to appear?

Show that magnetic flux depends on the angle between the field and the normal, and that induced emf is determined by the rate of change of flux.

Diagram: a flat loop, normal vector $\vec{n}$, field lines $\vec{B}$, and angle $\theta$; next to it, graphs of $\Phi(t)$ and $\mathcal{E}(t) = -d\Phi/dt$ for a uniformly changing flux.

A bottle dynamo is pressed against the tyre. Its coil has $N = 2000$ turns, each of area $A = 3.0\times10^{-4}\ \mathrm{m^2}$, and the field in the gap is $B = 0.15\ \mathrm{T}$. At $40\ \mathrm{km/h}$ the wheel turns $5$ times per second, and the flux through one turn reverses every $0.05\ \mathrm{s}$: during that time $\Delta\Phi = 2BA = 9.0\times10^{-5}\ \mathrm{Wb}$.

Faraday's law gives the average emf over half a period:

$$\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 2000\cdot\frac{9.0\times10^{-5}\ \mathrm{Wb}}{0.05\ \mathrm{s}} \approx 3.6\ \mathrm{V}.$$

That is enough to make a $6\ \mathrm{V}$, $3\ \mathrm{W}$ lamp glow dimly, and nearly normally when you speed up to $50\ \mathrm{km/h}$.

The example exercises both halves of the theory: first $\Phi = BA\cos\alpha$ fixes the flux through one turn, then $\varepsilon = -N\,\Delta\Phi/\Delta t$ turns the rate of change of flux into an emf, and the factor $N$ shows that turns in series add their emfs.

Inside the back of a phone there is a receiver coil: $N = 10$ turns, each of area $A = 3.0\ \mathrm{cm^2} = 3.0\times10^{-4}\ \mathrm{m^2}$. The pad's coil drives an alternating field of amplitude $B = 5.0\ \mathrm{mT}$ at a frequency $150\ \mathrm{kHz}$.

In half a period, $\Delta t = 1/(2f) = 3.3\times10^{-6}\ \mathrm{s}$, the flux through one turn changes by $\Delta\Phi = 2BA = 3.0\times10^{-6}\ \mathrm{Wb}$, so the induced emf is

$$\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 10\cdot\frac{3.0\times10^{-6}}{3.3\times10^{-6}} \approx 9\ \mathrm{V}.$$

A rectifier converts this $9\ \mathrm{V}$ of alternating voltage into $5\ \mathrm{V}$ of direct voltage and charges the battery at about $5\ \mathrm{W}$. Hold a static magnet against the coil instead: the flux stops changing, $\Delta\Phi = 0$, and the voltage disappears.

What this case exercises is the requirement that the flux actually change: a steady field gives $\Delta\Phi = 0$ and zero emf, so the charger must drive its coil at the high frequency $150\ \mathrm{kHz}$. The formula $\Phi = BA$ with the turn's area explains why the receiver coil is made flat and wide.

Under the glass of an induction hob sits a coil of $N = 25$ turns with a total area $A = 2.0\times10^{-2}\ \mathrm{m^2}$. The electronics drive it with a current at $20\ \mathrm{kHz}$, producing a field of amplitude $B = 20\ \mathrm{mT}$ above the glass.

In half a period, $\Delta t = 25\ \mu\mathrm{s}$, the flux through the coil changes by $\Delta\Phi = 2BA = 8.0\times10^{-4}\ \mathrm{Wb}$, and the induced emf reaches

$$\varepsilon = N\frac{\Delta\Phi}{\Delta t} = 25\cdot\frac{8.0\times10^{-4}}{25\times10^{-6}} \approx 800\ \mathrm{V}.$$

This voltage drives eddy currents in the steel base of the pan, releasing about $2000\ \mathrm{W}$ — enough to boil $2\ \mathrm{L}$ of water in a few minutes. The glass and aluminium foil stay cool: the field inside them is too weak and their resistance too large.

The example exercises the role of area and turn count in $\Phi = BA\cos\alpha$ and in $\varepsilon = -N\,\Delta\Phi/\Delta t$: the same $20\ \mathrm{mT}$ field over an area of $2.0\times10^{-2}\ \mathrm{m^2}$ sets the flux, and $N = 25$ multiplies the result. The $20\ \mathrm{kHz}$ frequency matters precisely because the emf is proportional to $\Delta\Phi/\Delta t$.