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Electromagnetism review
A summary of all of Module 6 Electromagnetism, every formula with its units and when to use it, key words, how the ideas connect, and a mixed HSC-style test on charged particles in fields, the motor effect, induction, transformers, motors and generators.
Use this page to check the whole module. Read the summary, make sure you can use every formula and explain every key word, then do the test without looking back.
Module summary
| Topic | What you need to know |
|---|---|
| Charges in electric fields (6A) | Between parallel plates the field is uniform, E = V/d. The force qE is constant, so a charge accelerates uniformly. Work done W = qV = qEd becomes kinetic energy. A charge entering perpendicular to the field follows a parabola, like a projectile. |
| Charges in magnetic fields (6B) | F = qvB sin θ, perpendicular to both v and B, so it does no work and the speed is constant. Perpendicular entry gives a circle of radius r = mv/qB. Parallel entry: no force. At an angle: a helix. Crossed E and B fields select v = E/B. |
| Motor effect (6C) | F = lIB sin θ on a current-carrying conductor in a field. Direction by the right-hand palm rule. Measured with a current balance. |
| Parallel wires (6D) | Each wire is in the other's field: F/l = μ0I1I2/2πr. Same direction currents attract, opposite repel. Newton's third law: equal forces even for unequal currents. The ampere was defined from this until 2019; now it is defined by fixing the charge on the electron. |
| Flux and Faraday (6E) | Φ = BA cos θ (= B∥A), in webers. A changing flux induces an emf: ε = −NΔΦ/Δt. The flux can change by changing B, A or the angle. |
| Lenz's law (6F) | The induced current opposes the change in flux that produces it. This is required by conservation of energy. Eddy currents in conductors give magnetic braking and heating. |
| Transformers (6G) | Vp/Vs = Np/Ns; ideal: VpIp = VsIs. Real losses: resistance heating, eddy currents (reduced by laminations), hysteresis, flux leakage. Transmission at high voltage cuts I2R losses. |
| DC motors (6H) | τ = nIAB sin θ, maximum with the plane of the coil parallel to B. The split-ring commutator reverses the coil current each half turn. Back emf grows with speed: I = (V − εback)/R. |
| Generators and induction motors (6I) | Peak emf ε0 = NBAω. Slip rings give AC; a split ring gives DC. Loading a generator makes it harder to turn (Lenz). Induction motor: a rotating stator field induces currents in a squirrel-cage rotor, which follows the field but always slips behind it. |
Formulas
| Formula | Quantities and units | Use it for |
|---|---|---|
| $E = \dfrac{V}{d}$ | E (V m−1), V (V), d (m) | Uniform field between parallel plates |
| $F = qE$ | F (N), q (C) | Force on a charge in an electric field |
| $W = qV = qEd$ | W (J) | Work done on, or kinetic energy gained by, a charge |
| $\tfrac{1}{2}mv^2 = qV$ | m (kg), v (m s−1) | Speed after acceleration from rest |
| $F = qvB\sin\theta$ | B (T), θ between v and B | Force on a moving charge in a magnetic field |
| $r = \dfrac{mv}{qB}$ | r (m) | Radius of circular path (from qvB = mv2/r) |
| $F = lIB\sin\theta$ | l (m), I (A), θ between conductor and B | Motor effect on a conductor |
| $\dfrac{F}{l} = \dfrac{\mu_0 I_1 I_2}{2\pi r}$ | μ0 = 4π × 10−7 T m A−1, r (m) | Force per unit length between parallel wires |
| $\Phi = B_{\parallel}A = BA\cos\theta$ | Φ (Wb), θ between B and the normal | Magnetic flux |
| $\varepsilon = -N\dfrac{\Delta\Phi}{\Delta t}$ | ε (V), N turns, Δt (s) | Induced emf (Faraday, with Lenz's sign) |
| $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$ | p primary, s secondary | Transformer voltage ratio |
| $V_pI_p = V_sI_s$ | (ideal transformer) | Transformer current ratio |
| $P_{loss} = I^2R$ | R of the line (Ω) | Transmission line losses |
| $\tau = nIA_{\perp}B = nIAB\sin\theta$ | τ (N m), A (m2), θ between B and the normal | Torque on a motor coil |
| $I = \dfrac{V - \varepsilon_{back}}{R}$ | R of the coil | Current in a running DC motor |
| $\varepsilon_0 = NBA\omega$ | ω = 2πf (rad s−1) | Peak emf of a generator |
Putting it together
The whole module rests on two ideas and the connection between them.
- A magnetic field exerts a force on a moving charge. On a single charge it is qvB sin θ. On the charges drifting along a wire it adds up to lIB sin θ. On a coil it makes a torque, nIAB sin θ. On a wire in another wire's field it gives the force between parallel conductors. Every motor runs on this.
- A changing magnetic flux induces an emf. Faraday's law tells you how big; Lenz's law tells you which way. Every generator and transformer runs on this.
- The link is energy. Moving a conductor through a field pushes its charges along (motor force on the charges, so an emf). The induced current then feels a motor force that opposes the motion (Lenz). So a generator needs work to turn it, a motor generates a back emf, a magnet falls slowly through a copper pipe, and a transformer's primary draws more current when the secondary is loaded. In every case, the induced effects take energy from the thing causing the change, and energy is conserved.
Electric fields (6A) do work on charges and change their speed. Magnetic forces (6B) are always perpendicular to velocity, so they change direction but never speed. Instruments like the mass spectrometer and the cyclotron combine both: electric fields to accelerate, magnetic fields to steer.
Key words
| Word | Meaning |
|---|---|
| Electric field strength | Force per unit positive charge at a point, in N C−1 or V m−1 |
| Electron volt (eV) | The energy gained by an electron accelerated through 1 V: 1.602 × 10−19 J |
| Magnetic flux density (B) | The strength of a magnetic field, in tesla (T) |
| Motor effect | The force on a current-carrying conductor in a magnetic field |
| Ampere | SI unit of current; since 2019 defined by fixing e = 1.602176634 × 10−19 C exactly |
| Magnetic flux (Φ) | The amount of magnetic field passing through an area, in webers (Wb = T m2) |
| Electromagnetic induction | Producing an emf in a conductor by changing the flux through it |
| Lenz's law | The induced emf drives a current that opposes the change in flux causing it |
| Eddy currents | Circulating currents induced in a solid conductor by a changing flux |
| Step-up / step-down transformer | Increases / decreases the voltage (more / fewer secondary turns) |
| Laminations | Thin insulated sheets making up a transformer or motor core, to reduce eddy currents |
| Torque | The turning effect of a force, τ = r⊥F, in N m |
| Split-ring commutator | Reverses the current in a motor coil every half turn; gives DC from a generator |
| Slip rings | Complete rings giving continuous connection to a rotating coil; give AC from a generator |
| Back emf | The emf induced in a spinning motor coil, opposing the supply |
| Synchronous speed | The speed of the rotating field in an induction motor, 120f/p rpm |
| Slip | How far an induction motor's rotor lags behind the field, as a fraction of synchronous speed |
Test yourself
An electron is accelerated from rest through a potential difference of 500 V. Its final speed is closest to
v = √(2qV/m) = √(2 × 1.602 × 10−19 × 500 / 9.109 × 10−31) = 1.33 × 107 m s−1. The last option forgot the square root.
A proton and an alpha particle enter the same uniform magnetic field at right angles with the same speed. The ratio of the radius of the alpha particle's path to the proton's is about
r = mv/qB. The alpha particle has about 4 times the mass and 2 times the charge, so r is about twice as large.
Which statement about a charged particle moving in a uniform magnetic field is always true?
The force is always perpendicular to the velocity, so no work is done. A circle needs perpendicular entry.
Two long parallel wires 5.0 cm apart carry 20 A and 10 A in opposite directions. The force on a 2.0 m length of either wire is
F = μ0I1I2l/2πr = 2 × 10−7 × 20 × 10 × 2.0 / 0.050 = 1.6 × 10−3 N. Opposite currents repel.
A 200-turn coil of area 4.0 × 10−3 m2 lies perpendicular to a 0.50 T field. The field is reduced to zero in 0.10 s. The average induced emf is
ΔΦ = 0.50 × 4.0 × 10−3 = 2.0 × 10−3 Wb; ε = NΔΦ/Δt = 200 × 2.0 × 10−3 / 0.10 = 4.0 V.
An ideal transformer has 1200 primary turns and 60 secondary turns. It is connected to 230 V AC and its secondary supplies 3.0 A. The primary current is
Vs = 230 × 60/1200 = 11.5 V. VpIp = VsIs, so Ip = 11.5 × 3.0 / 230 = 0.15 A.
Electricity is transmitted at very high voltage mainly because
P = VI: raising V lowers I, and the loss depends on I2.
A DC motor is running at constant speed when its load is suddenly removed. Its back emf and current
The motor speeds up, so the back emf rises and the current I = (V − εback)/R falls.
Label the transformer.
Which principle does each device mainly rely on?
Put the steps from power station to home in order.
A charged particle moving perpendicular to a uniform magnetic field moves in a because the force is always perpendicular to its velocity. The magnetic flux through an area is measured in . The direction of an induced current is given by law, which follows from the conservation of . Transformer cores are to reduce eddy currents. An AC generator uses rings, and a DC motor uses a split-ring .
Electric cars use the same machine as a motor when accelerating and as a generator when braking ("regenerative braking"). Analyse how the principles of electromagnetism and the conservation of energy apply to this machine in both modes. Refer to the motor effect, Faraday's law, Lenz's law and back emf.
Motor mode. When the car accelerates, the battery drives current through the coils of the machine, which sit in a magnetic field. Each current-carrying side of a coil experiences a force (the motor effect, F = lIB sin θ), and the forces on opposite sides produce a torque, τ = nIAB sin θ, that turns the rotor and the wheels. As the coils rotate, the flux through them changes, so by Faraday's law an emf is induced. By Lenz's law this back emf opposes the supply, so the current is I = (V − εback)/R. The electrical power converted to mechanical power is εbackI, increasing the car's kinetic energy, while I²R is lost as heat.
Generator mode. When the driver brakes, the controller stops supplying current and connects the coils to charge the battery. The car's momentum keeps the rotor turning, so the flux through the coils still changes and an emf is induced (Faraday's law), driving a current into the battery. This induced current is in a magnetic field, so it also experiences forces. By Lenz's law the current flows in the direction that opposes the change causing it, the rotation, so the forces produce a torque against the motion, slowing the car.
Energy. In both modes Lenz's law guarantees conservation of energy. In motor mode, the back emf is how electrical energy is taken from the battery and converted to kinetic energy. In generator mode, the opposing torque means the car must do work against it, so its kinetic energy decreases by the amount converted to electrical energy and stored chemically in the battery. If the induced current aided the motion, the car would speed up while charging its battery, creating energy. Not all the kinetic energy is recovered: some is lost as heat through I²R in the windings, eddy currents in the core and friction, and conventional friction brakes are still needed for hard stops and at low speed, where the induced emf is too small.