4. Electricity and magnetism

Repulsion is the decisive test for two magnets: an unmagnetised magnetic material can be attracted by either pole, but it cannot repel a pole.

The nearest induced pole is opposite to the approaching magnet pole, so the magnetic material is attracted. Soft iron usually loses most of this induced magnetism when the magnet is removed.

Choose soft iron when magnetism should appear and disappear readily. Choose steel when the object must remain magnetised after the magnetising field is removed.

Bring a known magnet close without touching. Attraction shows that the sample is magnetic; no magnetic attraction shows that it is non-magnetic under the test conditions.

A permanent magnet and an electromagnet both produce magnetic fields around them. The field describes how they can exert forces without direct contact.

A complete sketch uses several smooth loops rather than isolated straight arrows. Between close unlike poles, central field lines run approximately straight from N to S.

Outside a bar magnet, follow the arrow from N towards S. At any point on a curved line, the field direction is along the tangent to that line.

For the pattern method, place paper over the magnet, sprinkle iron filings evenly and tap gently. The filings align with the local field and show the curved line pattern, but they do not by themselves show arrow direction.

For an electromagnet, switching current on creates the useful magnetic field and switching it off removes most of the field when a soft-iron core is used.

The interaction gives forces on both objects in opposite directions. If one magnet is held fixed, the force on the free magnet is still caused by the interaction of their fields.

Around a bar magnet, lines are usually closest near the poles, so the field is strongest there. To rank labelled points, compare local spacing at each point.

Always state the sign as well as saying that an object is charged. Positive and negative are two types of electric charge, not descriptions of whether an answer is good or bad.

The force is mutual: if two free like-charged objects repel, each moves away from the other; if two free unlike-charged objects attract, each moves towards the other.

Begin with discharged objects, keep the surfaces dry, avoid touching the rubbed area and use an insulating support. Repeat from the same starting distance so that a movement can be compared with an uncharged control.

If electrons move from a cloth to a rod, the rod becomes negative and the cloth becomes positive. If they move from the rod to the cloth, the rod becomes positive and the cloth becomes negative. Charge is transferred, not created.

Use the same cell, lamp, lead positions and sample dimensions where possible. Clean, firm contacts matter: an open contact can make a conductor look like an insulator.

A metal object can lose excess charge through a conducting path to Earth, so it is supported on an insulator when charge must be retained. Plastic can retain charge where it was rubbed because electrons cannot move freely through it.

The numerical sign and the unit answer different questions: + or - gives the type of net charge, while C identifies the measured quantity as electric charge.

The field explains how a charged object can exert force without direct contact. The field exists around the source charge even before a test charge is placed there.

At a point on a field line, read the arrow or tangent direction. A positive charge accelerates with that direction if electric force is the only unbalanced force; a negative charge accelerates oppositely.

Draw several lines distributed symmetrically and place consistent arrows on them. For the parallel-plate pattern, ignore end effects as required by the syllabus, so the central field is represented as uniform.

In a complete series path, charge is not used up by a component: the same continuing flow passes successive points, while the component transfers energy.

A reading off-scale needs a higher range; a very small reading on a high range is less precise and may need a lower range. Record the value with A or mA and convert units when required.

The drift of many free electrons produces an electric current even though each electron's drift is slow. The metal ions remain in their lattice positions.

A d.c. magnitude may be steady or may vary while remaining in the same direction. An a.c. waveform can have different shapes, but it must reverse direction.

Example: 7.0 C passes in 5.0 min = 300 s, so I = 7.0 / 300 = 0.023 A. Convert time to seconds and current prefixes to amperes before substitution.

Circuit arrows labelled I normally show conventional current. To add an electron-flow arrow, reverse the conventional-current direction along the same wire.

A source with a larger e.m.f. supplies more electrical energy to each coulomb of charge, not necessarily more total charge or current.

One volt is one joule per coulomb: 1 V = 1 J/C. Include V after a numerical e.m.f. value.

A larger p.d. across a component means that more work is done, or more electrical energy is transferred, for each coulomb passing through it.

Both e.m.f. and p.d. are measured in volts because both compare electrical work with charge. Their energy roles differ: a source supplies energy, while a component transfers it from the electrical pathway.

A negative digital reading or reversed analogue deflection indicates reversed d.c. lead polarity. A high-resistance voltmeter draws very little current so it minimally changes the circuit.

A 9.0 V battery moves 30 C around the circuit: W = E Q = 9.0 × 30 = 270 J. Label the result as energy supplied by the source.

If a resistor transfers 100 J when 10 C passes, V = W / Q = 100 / 10 = 10 V. If 5.0 C passes through a 2.0 V component, W = 2.0 × 5.0 = 10 J.

For a resistor with 6.0 V across it and 0.30 A through it: $R = 6.0 / 0.30 = 20\,\Omega$. The same relationship rearranges to $V = IR$ and $I = V/R$.

1. Build the circuit with the switch open and select meter ranges above the expected readings.
2. Close the switch briefly; record $I$ and $V$ together.
3. Adjust the variable resistor and repeat for several pairs.
4. For each pair, calculate $R = V/I$ with V, A and $\Omega$.

If one reading pair is 2.4 V and 0.20 A, then $R = 2.4/0.20 = 12\,\Omega$. Similar values across repeated low-current readings support a constant resistance.

A long, thin wire therefore tends to have a larger resistance than a short, thick wire of the same metal. A larger diameter means a larger cross-sectional area, so it lowers resistance when length is unchanged.

For the straight resistor line, gradient $\Delta I/\Delta V = 1/R$: a steeper line represents a smaller constant resistance. For a curved device, compare $V/I$ at the stated operating point instead of treating the whole graph as one constant gradient.

A wire has $R_1=0.14\,\Omega$, $L_1=2.0$ m and $A_1=0.40$ mm². For the same material with $L_2=3.0$ m and $A_2=0.90$ mm²: $R_2=0.14\times(3.0/2.0)\times(0.40/0.90)=0.093\,\Omega$.

For a circular wire, $A = \pi d^2/4$. Doubling diameter makes the area four times larger, so resistance becomes one quarter when length is unchanged. Combine simultaneous changes by multiplying the length factor by the inverse area factor.

A mains supply is also an electrical energy source for the circuit. The appliance does not store all the received energy permanently: its useful output and any heating are eventually transferred to the surroundings.

A 12 V supply delivers 0.35 A to a circuit: $P = 0.35 \times 12 = 4.2$ W. Rearrange the same equation as $I=P/V$ or $V=P/I$ when current or p.d. is required.

A 3.0 V torch lamp carries 20 mA for 5.0 minutes. Convert first: $I=0.020$ A and $t=300$ s. Then $E=3.0\times0.020\times300=18$ J.

1. Convert appliance power from W to kW.
2. Convert operating time to hours.
3. Calculate energy: $E(\text{kWh})=P(\text{kW})\times t(\text{h})$.
4. Calculate cost: energy in kWh × price per kWh.

A 2200 W heater runs for 48 minutes at $0.25 per kWh.$P=2.2$kW and$t=0.80$h, so$E=2.2\times0.80=1.76$kWh. Cost$=1.76\times0.25=$0.44$.

To convert a physical circuit into a diagram: identify every component, replace it with the correct standard symbol, trace each conducting path, and reproduce every branch and junction. A crossing is a connection only when the diagram marks it as a junction; label source polarity and meter positions when they matter.

An LED is drawn as a diode symbol with two small arrows pointing outwards to represent emitted light. It is normally connected in series with a resistor that limits the current. Reversing the LED reverses its bias, so a complete-looking circuit may still carry no current through that branch.

When interpreting a diode circuit, first mark the source polarity, then compare it with the diode direction, and only then decide which branches can conduct. A single diode in an alternating-current circuit conducts during one half-cycle and blocks during the other, producing current in only one direction through the load.

Current is the rate of flow of charge. Charge does not get used up by a lamp or resistor, and there is no junction where the flow can split, so the same amount of charge passes every point each second.

Before switching on, trace every complete path from one supply terminal to the other, check for an accidental zero-resistance path across the supply, put ammeters in series, and put voltmeters in parallel across the component being measured.

A 1.5 V cell in series with a 6.0 V battery, both aiding, gives $E_{ ext{total}}=1.5+6.0=7.5$ V. If the 1.5 V cell is reversed, the combined e.m.f. is $6.0-1.5=4.5$ V.

For 12 Ω, 8 Ω and 5 Ω in series, $R_{ ext{series}}=12+8+5=25$ Ω. The result must be greater than every individual resistance because another series resistor adds opposition.

If two branch currents are 0.30 A and 0.20 A, the source current is $0.30+0.20=0.50$ A. The currents need not divide equally unless the branches have the same resistance.

Adding a parallel resistor creates another path for charge. At the same p.d., more total current can flow from the source; since $R=V/I$, a larger total current means a smaller combined resistance.

A 12 V supply feeds two parallel branches. Both branches have 12 V across them. If one branch takes 0.40 A and the other 0.25 A, the source current is 0.65 A. Within a branch containing two series components with p.d.s 7 V and 5 V, the p.d.s add to 12 V.

Current measures charge passing per second, and electric charge is conserved. In a steady circuit, charge cannot be created, destroyed or continually accumulate at the junction. Therefore every coulomb arriving each second must leave through one of the outgoing branches.

If 1.2 A enters and currents of 0.45 A and 0.30 A leave along two branches, the remaining outgoing current is $1.2-0.45-0.30=0.45$ A.

For 3.0 Ω and 6.0 Ω in parallel, $R=(3.0 imes6.0)/(3.0+6.0)=18/9.0=2.0$ Ω. The answer passes the check because 2.0 Ω is below both 3.0 Ω and 6.0 Ω.

Reduce a mixed circuit in stages: first find each clear series or parallel group, redraw it as one equivalent resistor, then combine the next group. Keep units in ohms throughout.

The same 0.40 A flows through a 10 Ω conductor and a 20 Ω conductor in series. Their p.d.s are $0.40 imes10=4.0$ V and $0.40 imes20=8.0$ V, so the conductor with twice the resistance has twice the p.d.

To predict a change: identify the component across which $V_{out}$ is measured, decide how that component's resistance changes, then compare its share of the total series resistance. Moving a slider towards one end shortens one resistive section while lengthening the other, so the two output shares change in opposite directions.

A 1.2 kΩ resistor and a 2.4 kΩ thermistor are in series across 6.0 V, with the output across the thermistor. $V_{out}=6.0 imes2.4/(1.2+2.4)=4.0$ V. Check: the thermistor is two-thirds of the total resistance and receives two-thirds of the supply p.d.

Choose the smallest available rating above the appliance's normal operating current but below the maximum safe current for the cable. This prevents operation during normal use while disconnecting before the cable overheats. A 10 A kettle with choices 3 A, 5 A and 13 A needs 13 A.

A small overload may take longer to operate a protective device; a much larger fault current should disconnect much faster. A fuse or trip switch disconnects the circuit—it does not reduce and hold the current at its rating.

If a fault or overload makes the current too large, the fuse in the live wire melts and disconnects the supply before the appliance wiring or flexible cable overheats. The fuse therefore protects the circuit and cabling even though there is no earth wire.

An induced e.m.f. can exist across an open circuit. An induced current flows only when the conducting path is complete.

The effect can also be demonstrated by moving a straight wire across the field between magnet poles while it is connected to a sensitive meter. Reversing the motion reverses the deflection.

For a fair comparison, change one factor at a time and compare the size of the peak meter deflection. Reversing motion or field direction reverses polarity; it does not by itself make the e.m.f. larger.

Because the induced force opposes the motion, work must be done to keep the magnet, wire or coil moving. That mechanical energy is transferred to electrical energy in the circuit.

Reversing either the motion or the field reverses the induced current. Reversing both leaves the current direction unchanged because the two reversals cancel.

During the next half-turn, each side of the coil moves across the field in the opposite direction, so the induced e.m.f. and current reverse. Repetition produces an alternating output, with one complete a.c. cycle per complete coil revolution.

A generator may instead rotate a magnet beside a fixed coil. The field linking the fixed coil then changes; because the coil and its output wires are stationary, slip rings are not needed for that arrangement.

In the common end-view diagram with a horizontal N-to-S field, a horizontal coil line is parallel to the field and corresponds to a peak; a vertical coil line is perpendicular and corresponds to zero. Always use the field direction rather than relying on the words horizontal and vertical alone.

Faster rotation makes the linkage change faster, so the peaks are larger and the cycles are closer together: both peak e.m.f. and frequency increase.

For a solenoid, curl the right-hand fingers in the conventional-current direction around its turns. The thumb points along the field inside the solenoid and towards its north pole. Field lines continue outside from north to south and return inside from south to north.

Reverse the supply: the pattern keeps the same shape but every compass direction reverses. Move a compass farther from a straight wire, or from inside to outside a solenoid: the current's effect becomes weaker and Earth's field may influence the reading more.

When the relay-coil current stops, the soft iron rapidly loses its magnetism and a spring returns the armature and contacts. The control circuit and load circuit are electrically separate but mechanically linked by the moving contacts.

An alternating audio current repeatedly reverses the magnetic effect and therefore the force on the voice coil. The cone moves backwards and forwards, making surrounding air vibrate; the changing current waveform sets the sound vibration.

With alternating current, the current magnitude and direction change repeatedly, so the magnetic field strength varies and its direction repeatedly reverses.

Keep the rod position, current magnitude and magnet spacing unchanged when comparing directions. With the switch open there is no current and no motor-effect force; outside the field the effect should disappear or become much smaller.

If the current repeatedly reverses while the field stays fixed, the force repeatedly reverses and the conductor vibrates.

Read the field direction first, then the conventional current. Orient the two fingers before reading the thumb. The force is perpendicular to both field and current.

Identify the field direction from N to S or from dot/cross symbols. Convert the beam motion to conventional-current direction, then use Fleming's left-hand rule. For a negative particle, this gives the force opposite to that on a positive particle moving the same way.

When the beam crosses the field at right angles, the force deflects it sideways and continually changes its direction, producing a curved path while it remains in the field.

A twofold increase in one factor gives a twofold increase in the turning effect when the other conditions are unchanged. A decrease in a factor decreases the turning effect in the same qualitative way.

The turning effect also changes as the coil rotates. It is greatest when the forces have the largest perpendicular distance from the axis and becomes zero at the position where their lines of action pass through the axis.

After each half-turn, the two halves of the split ring contact the opposite brushes. The current in the coil reverses just as the sides exchange positions. Each side's force therefore also reverses, so the turning effect continues in the same rotational direction.

At the position where the turning effect is momentarily zero, the coil's motion carries it through. Commutation then restores a turning effect in the same rotational direction, allowing continuous rotation.

Reversing the supply polarity reverses the current relative to the fixed field, so the motor rotates in the opposite direction.

Copper is used for the coils because its low resistance limits heating. Soft iron is used instead of permanently magnetised material because its magnetisation follows the changing primary current readily.

First identify which coil is connected to the source: that is the primary. Then compare the output voltage with the input voltage to name the transformer. The physical left or right side of a drawing does not decide the terms.

Example: a primary has 800 turns at 240 V and the secondary has 40 turns. V_s = 240 × 40 / 800 = 12 V, so it is a step-down transformer.

Check the direction before accepting a number: fewer secondary turns must give a smaller secondary voltage, while more secondary turns must give a larger secondary voltage.

Follow the energy route: generator → step-up transformer → transmission cables → step-down transformer or transformers → consumers. The transformer beside the power station is step-up; the transformer before consumers is step-down.

An examination answer should link the electrical advantage to an economic result: less energy wasted lowers operating cost, while thinner cables and fewer supports can lower construction cost.

Soft iron magnetises and demagnetises readily, so the field follows the primary a.c. and provides a strong changing magnetic link between the two coils.

A steady d.c. current produces a steady magnetic field after switching. With no continuing change of field, there is no continuing induced e.m.f. in the secondary. A brief effect may occur only while the d.c. is switched on or off.

Example: an ideal transformer takes 0.20 A at 240 V and gives 12 V. I_s = 0.20 × 240 / 12 = 4.0 A. The lower output voltage is paired with a higher output current.

For the same ideal power, current changes inversely with voltage: a step-up transformer steps current down, while a step-down transformer steps current up.

The square makes current especially important. If current is halved while cable resistance is unchanged, P_loss becomes (1/2)² = 1/4 of its original value. If current is one tenth, the loss is one hundredth.

A complete explanation states the fixed transmitted power, uses high V → low I from P = VI, and then uses low I → much smaller I²R heating loss.