Describe and quantitatively analyse electromagnetic induction using Faraday's law (induced EMF = - N dPhi/dt) and Lenz's law, including motional EMF, eddy currents and the induction coil

What this dot point is asking

NESA wants you to state Faraday's law in the form $\varepsilon = -N \, d\Phi / dt$, apply Lenz's law to determine the direction of an induced current, work with motional EMF ($\varepsilon = BLv$) as a special case, and explain qualitatively where eddy currents and the induction coil sit in the same framework. Faraday's law is the keystone of the rest of Module 6: transformers and motors all rely on it.

The answer

Faraday's law

For a single conducting loop linked by a magnetic flux $\Phi(t)$, the induced EMF around the loop is:

For a coil of $N$ turns, each turn intercepts the same flux, so the EMFs add in series and the total induced EMF is:

The flux through one turn is $\Phi = B A \cos \theta$. Anything that changes $B$, $A$ or $\theta$ produces an EMF:

• Changing $B$: a magnet moved toward or away from a stationary coil, or a changing current in a nearby circuit (the basis of the transformer).
• Changing $A$: a conducting rod sliding on rails to enlarge or shrink the circuit area (motional EMF).
• Changing $\theta$: a coil rotated in a steady field (the AC generator).

The minus sign encodes Lenz's law.

Lenz's law

The induced EMF and induced current always act in a direction that opposes the change in flux that produced them.

In practice, to find the direction of the induced current:

1. Identify the direction of the external flux through the coil and whether it is increasing or decreasing.
2. The induced current produces its own magnetic field inside the coil; it points opposite to the external field if external flux is increasing, and along the external field if external flux is decreasing.
3. Use the right-hand rule for a current loop (fingers curl with current, thumb along its magnetic field) to read off the direction of the induced current itself.

Lenz's law is a statement of energy conservation. If the induced current reinforced the change in flux, it would accelerate the motion or amplify the field that caused it, creating energy from nothing.

Motional EMF

A conducting rod of length $L$ moving with velocity $v$ perpendicular to a uniform field $B$ (and with $L$, $v$ and $B$ mutually perpendicular) has free charges in it experiencing a magnetic force $F = qvB$ along the rod. Charge separates until an electric field inside the rod balances the magnetic force. The resulting EMF between the ends of the rod is:

This is exactly the Faraday's law result for the case where the rod is part of a circuit and slides to change the enclosed area: $d\Phi / dt = B \, dA / dt = B L v$.

If the rod is part of a closed circuit of resistance $R$, the induced current is $I = \varepsilon / R$, and the magnetic force on this current-carrying rod opposes the motion (Lenz's law again).

Eddy currents

When a bulk conductor (a sheet, disc or block of metal) experiences a changing flux, induced EMFs drive circulating currents called eddy currents inside the conductor. They oppose the change in flux that produced them, so they exert a drag force on whatever is causing the flux change.

Examples:

• Magnetic braking. A conducting disc spinning between the poles of a magnet has currents induced in it. These currents experience a magnetic force opposing the rotation, slowing the disc. Used in train brakes and gym equipment.
• Induction cooktops. A high-frequency AC field induces eddy currents in the base of a ferromagnetic pot, dissipating energy as heat directly in the pot.
• Aluminium-tube demo. A magnet dropped down an aluminium tube falls slowly because eddy currents in the tube wall oppose the change in flux as the magnet moves.

Eddy currents are useful for braking and heating, but they are unwanted losses in transformer cores and motor armatures. To reduce them, the iron in those devices is laminated (thin sheets electrically insulated from each other), which breaks the eddy-current paths.

The induction coil

An induction coil is a transformer-like device used to produce high-voltage pulses from a low-voltage DC source. It has:

1. A primary coil of relatively few turns wound on a soft iron core.
2. A secondary coil of many more turns wound on the same core.
3. An interrupter (a mechanical or electronic switch) that rapidly opens and closes the primary circuit.

Each time the interrupter breaks the primary current, the primary's magnetic field collapses rapidly, producing a fast $d\Phi / dt$ through the secondary. With the large turns ratio, the induced secondary EMF can be tens of kilovolts, enough to produce a spark across an air gap.

Historically used for X-ray tubes and Tesla coils. Modern car ignition coils work on the same principle: the breaker (or transistor) interrupts the primary 12 V supply many times per second, producing tens of kilovolts at the spark plugs.

Worked example: rotating coil

A 100-turn rectangular coil of area $0.020$ m$^2$ rotates at $50$ Hz in a uniform field of $0.10$ T. Find the peak induced EMF.

The flux through one turn is $\Phi(t) = B A \cos(\omega t)$, where $\omega = 2 \pi f = 2 \pi \times 50 = 314$ rad/s.

$\varepsilon = -N \frac{d\Phi}{dt} = N B A \omega \sin(\omega t)$.

Peak EMF:

$\varepsilon_{\max} = N B A \omega = 100 \times 0.10 \times 0.020 \times 314 = 63$ V.

This is the operating principle of the AC generator: a constant-magnitude EMF that varies sinusoidally with the rotation angle.

Try it: Induced EMF calculator for change in flux, turns and time.

Examples in context

Example 1. EMF induced in a Snowy Hydro 2.0 stator coil. Each stator coil of a Snowy 2.0 generator has $N = 18$ turns and links a peak flux of $\Phi_{\max} = 0.20 \text{ Wb}$ per turn. The rotor spins at $f = 50/3 \text{ Hz}$ (with 6 pole-pairs producing 50 Hz output), so $\Phi(t) = \Phi_{\max} \cos(2 \pi f t)$. The induced EMF peaks at $\varepsilon_{\max} = N \Phi_{\max} 2 \pi f = 18 \times 0.20 \times 2 \pi \times 50 = 1131 \text{ V}$ per coil. Hundreds of coils in series produce the $20 \text{ kV}$ generator output that the step-up transformer raises to $330 \text{ kV}$ for transmission. Lenz's law ensures the induced current opposes the rotor's motion, supplying the mechanical "load" the turbine sees.

Example 2. Eddy-current braking on a Sydney Trains carriage. Sydney Metro trains use electromagnetic eddy-current brakes for high-speed retardation. A coil energised on the carriage produces $B = 0.80 \text{ T}$ over an area $A = 0.50 \text{ m} \times 0.20 \text{ m} = 0.10 \text{ m}^2$ near the steel rail. As the train moves at $v = 30 \text{ m/s}$, the rail experiences a $d\Phi/dt = B \cdot 0.20 \text{ m} \cdot v = 0.80 \times 0.20 \times 30 = 4.8 \text{ V/m}$ per metre of rail (motional EMF). Induced eddy currents in the rail dissipate the train's KE as heat. Braking force is independent of mechanical contact, eliminating brake-pad wear.

Try this

Q1. State Faraday's law and Lenz's law in one sentence each. [2 marks]

• Cue. Faraday: induced EMF equals minus rate of change of flux linkage. Lenz: the induced current opposes the flux change that caused it.

Q2. A coil of $200$ turns experiences a flux change from $0.040 \text{ Wb}$ to $0.010 \text{ Wb}$ over $0.50 \text{ s}$. Calculate the average induced EMF. [3 marks]

• Cue. $\varepsilon = -N \Delta\Phi / \Delta t = -200 \times (0.010 - 0.040)/0.50 = +12 \text{ V}$.

Q3. A conducting rod of length $L = 0.50 \text{ m}$ slides at $v = 4.0 \text{ m/s}$ on parallel rails in a field $B = 0.30 \text{ T}$ perpendicular to the plane of the rails. (a) Calculate the motional EMF. (b) If the circuit resistance is $0.20 \Omega$, find the induced current. (c) Determine the force required to keep the rod moving at constant $v$. [2+2+2 marks]

• Cue. (a) $\varepsilon = B L v = 0.30 \times 0.50 \times 4.0 = 0.60 \text{ V}$. (b) $I = 0.60/0.20 = 3.0 \text{ A}$. (c) $F = B I L = 0.30 \times 3.0 \times 0.50 = 0.45 \text{ N}$.