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Electricity and magnetism review
Module 4 review: summary of electrostatics, circuits and magnetism, every formula with its symbols and units, a comparison of gravitational, electric and magnetic fields, a multi-step worked problem, and a mixed HSC-style test.
Module 4 has three strands that turn out to be one story. Charges at rest make electric fields; charges moving through a potential difference carry energy round a circuit; and charges moving at all make magnetic fields. Before Year 12 Module 6 puts the strands together into motors and generators, make sure each idea here is something you can explain in words as well as calculate.
Summary
| Topic | The big ideas |
|---|---|
| 4A Charge | Charge is quantised (q = ne) and conserved. Electrons move. Friction gives equal and opposite charges; conduction gives the same sign; induction (with earthing) the opposite sign. Charged objects attract neutral ones by polarisation. |
| 4B Electric fields | E = F/q, a vector in the direction of the force on a + charge. Lines + to −, never cross, closer = stronger. Fields add as vectors. Uniform between plates: E = V/d. |
| 4C Coulomb's law | F = kq₁q₂/r², inverse square, equal and opposite on the two charges. Net force = vector sum. Point-charge field E = kq/r². |
| 4D Potential difference | V = ΔU/q, W = qV = qEd. Equipotentials ⟂ field lines; no work along them. From rest: qV = ½mv². |
| 4E Current | I = q/t, conventional current + to −, slow drift (I = nqvA) but fast field. V = W/q. Ammeter series, voltmeter parallel. |
| 4F Resistance | R = V/I at a point; Ohm's law (constant R) only for ohmic conductors. R = ρL/A; metals' R rises with temperature, semiconductors' falls. Filament, diode, LED, thermistor are non-ohmic. |
| 4G Kirchhoff | Junction rule = conservation of charge; loop rule = conservation of energy. Series: same I, R adds. Parallel: same V, 1/R adds. Voltage divider. |
| 4H Power | P = VI = I²R = V²/R; E = Pt; 1 kWh = 3.6 MJ. 230 V parallel household wiring; breakers protect wiring, earth wire and RCD protect people. |
| 4I Magnets | Poles in pairs; lines N to S outside, closed loops; B in teslas. Domains; soft vs hard materials; Curie temperature. |
| 4J Currents | Right-hand grip rule. Straight wire B = μ₀I/2πr (∝ 1/r). Loop like a short magnet. Solenoid B = μ₀NI/L, uniform inside. |
Formulae and constants
| Relationship | Formula | Units |
|---|---|---|
| Electric field | $E = \dfrac{F}{q}$, $E = \dfrac{V}{d}$ | N C−1 = V m−1 |
| Coulomb's law | $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$ | N |
| Work and potential difference | $W = qV$, $V = \dfrac{\Delta U}{q}$ | J, V = J C−1 |
| Current | $I = \dfrac{q}{t}$, $I = nqvA$ | A = C s−1 |
| Resistance | $V = IR$, $R = \dfrac{\rho L}{A}$ | Ω |
| Power and energy | $P = VI = I^2R = \dfrac{V^2}{R}$, $E = Pt$ | W, J (or kWh) |
| Resistors | $R_s = R_1 + R_2 + \dots$, $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$ | Ω |
| Kirchhoff | $\sum I_{in} = \sum I_{out}$, $\sum \varepsilon = \sum IR$ | |
| Field of a straight wire | $B = \dfrac{\mu_0 I}{2\pi r}$ | T |
| Field inside a solenoid | $B = \dfrac{\mu_0 NI}{L}$ | T |
Constants: $e = 1.602 \times 10^{-19}$ C; $\varepsilon_0 = 8.854 \times 10^{-12}$ C2 N−1 m−2 (so $k = 8.99 \times 10^9$ N m2 C−2); $\mu_0 = 4\pi \times 10^{-7}$ T m A−1; $m_e = 9.109 \times 10^{-31}$ kg; $m_p = 1.673 \times 10^{-27}$ kg.
Three fields compared
| Gravitational | Electric | Magnetic | |
|---|---|---|---|
| Source | Mass | Charge | Moving charge (currents, electron spin in magnets) |
| Field strength | g = F/m (N kg−1) | E = F/q (N C−1) | B (T) |
| Attract or repel? | Always attracts | Both | Both (between poles) |
| Field lines | End on masses | Start on +, end on − | Closed loops; no monopoles |
| Point source | g ∝ 1/r² | E ∝ 1/r² | Long straight wire: B ∝ 1/r |
| Uniform field | Near Earth's surface | Between parallel plates | Inside a solenoid; between flat poles |
| Can it be shielded? | No | Yes, by a closed conductor | Partly, by soft iron |
Putting it together
A 3.6 kW storage hot-water element runs from the 230 V mains for 2.5 hours each night. Its wire is nichrome (ρ = 1.1 × 10−6 Ω m) with a cross-sectional area of 1.0 × 10−6 m2. (a) Find the current and the element's resistance. (b) Find the length of wire needed. (c) Find the energy used per night in kWh and joules, and the charge that flows. (d) Find the magnetic field 2.0 cm from the supply cable's active wire.
$I = \dfrac{P}{V} = \dfrac{3600}{230} = 15.7$ A; $R = \dfrac{V}{I} = \dfrac{230}{15.7} = 14.7$ Ω.
$L = \dfrac{RA}{\rho} = \dfrac{(14.7)(1.0 \times 10^{-6})}{1.1 \times 10^{-6}} = 13$ m.
$E = Pt = (3.6)(2.5) = 9.0$ kWh $= 9.0 \times 3.6 \times 10^{6} = 3.2 \times 10^{7}$ J. $q = It = (15.7)(2.5 \times 3600) = 1.4 \times 10^{5}$ C.
$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi \times 10^{-7})(15.7)}{2\pi(0.020)} = 1.6 \times 10^{-4}$ T, about three times Earth's field. (The mains is AC, so this field reverses 100 times a second; and the neutral in the same cable carries the return current, so the fields mostly cancel a little further away.)
Test yourself
A small sphere with charge −5.0 nC is in a uniform field of 4.0 × 103 N C−1 directed east. The force on it is
F = qE = (5.0 × 10−9)(4.0 × 103) = 2.0 × 10−5 N, opposite to E because the charge is negative.
Two charges exert a force of 12 N on each other. One charge is tripled and the separation is doubled. The new force is
F' = 12 × 3/4 = 9.0 N.
An electron is accelerated from rest through 2.0 kV. Its final kinetic energy is
K = qV = (1.602 × 10−19)(2000) = 3.2 × 10−16 J = 2.0 keV.
A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel to a 12 V battery. The total power delivered is
P = V²/R for each: 144/6 + 144/3 = 24 + 48 = 72 W.
Which graph shape shows that a component is ohmic?
Ohmic means I ∝ V, which is a straight line through the origin.
A long straight wire carries 8.0 A. At what distance is its magnetic field equal to 4.0 × 10−5 T?
r = μ0I/(2πB) = (2 × 10−7)(8.0)/(4.0 × 10−5) = 0.040 m.
Moving a +2.0 μC charge along an equipotential line 3.0 cm long in a field of 500 V m−1 requires
Along an equipotential ΔV = 0, so W = qΔV = 0.
Which statement correctly compares electric and magnetic field lines?
There are no magnetic monopoles, so magnetic field lines have no start or end.
Label the field of a straight wire.
Which conservation law does each statement rely on?
Doubling the distance from a point charge its electric field, but doubling the distance from a long straight wire its magnetic field. One volt is one per coulomb. In parallel circuits each branch has the same . The field inside a solenoid depends on the turns per metre and the .
A student claims: "Electric, magnetic and gravitational fields all behave the same way: they all come from a point source and get weaker with the square of distance." Evaluate this claim, referring to the sources of the fields, the shapes of their field lines and how their strength depends on distance. Include at least one calculation to support your answer.
The claim is only partly correct. Gravitational and electric fields of point sources do behave similarly: both are radial, and both follow an inverse square law, g = GM/r² and E = kq/r². For example, 0.10 m from a 2.0 nC charge E = (8.99 × 109)(2.0 × 10−9)/(0.10)² = 1.8 × 103 N C−1, and at 0.20 m it is a quarter of that, 450 N C−1. They differ in that mass is always positive, so gravity only attracts, while charge comes in two signs, so electric fields can cause attraction or repulsion, and electric field lines start on positive and end on negative charges.
Magnetic fields are different in kind. They are produced by moving charges (currents, and the spin of electrons in magnetised materials), not by a magnetic "charge": no isolated magnetic pole exists, so magnetic field lines form closed loops rather than starting or ending anywhere. Their dependence on distance depends on the geometry. Around a long straight wire, B = μ0I/2πr, which is inverse, not inverse square: 10 A gives 4.0 × 10−5 T at 5.0 cm and 2.0 × 10−5 T at 10 cm, halving rather than quartering. Inside a long solenoid, B = μ0NI/L is uniform and does not depend on position at all, just as the electric field between parallel plates and the gravitational field near Earth's surface are uniform.
Overall, the claim fits gravitational and electric fields of point sources, but fails for magnetic fields and for any uniform field.
A 12 V battery is connected to a 2.0 Ω resistor in series with a parallel combination of a 6.0 Ω resistor and a filament globe. When the circuit is switched on, the globe has a resistance of 3.0 Ω. (a) Calculate the current from the battery and the power dissipated in the globe. (b) Explain how the current from the battery would change if the globe's filament heated up and its resistance rose.
(a) Parallel pair: R = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω. Total = 2.0 + 2.0 = 4.0 Ω. I = 12/4.0 = 3.0 A. Potential difference across the parallel pair = (3.0)(2.0) = 6.0 V, so the globe's power is P = V²/R = 36/3.0 = 12 W.
(b) As the filament heats, its resistance increases, so the resistance of the parallel combination increases and so does the total resistance of the circuit. With the same emf, the current from the battery decreases. Less potential difference is then dropped across the 2.0 Ω resistor, so the parallel section, including the globe, receives a slightly larger share of the 12 V.