Electricity and Magnetism (SL) — IB Physics SL SL Study Guide

For: IB Physics SL candidates sitting IB Physics SL.

Covers: Charge, current, voltage, resistance and Ohm’s law, series and parallel circuits, magnetic fields around current-carrying wires, forces on moving charges in magnetic fields, and forces on current-carrying conductors.

You should already know: IGCSE Physics, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Physics SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Electricity and Magnetism (SL)?

Electricity and magnetism (E&M) is the study of interactions between charged particles, both at rest (electrostatics) and in motion (current electricity and magnetism), and is one of the four fundamental forces of nature. In the IB Physics SL syllabus, this topic makes up 15% of your final exam score, with questions appearing in both multiple-choice Paper 1 and structured-response Paper 2. It links directly to thermal physics (for heating effects of current) and atomic physics (for charge properties of subatomic particles), and forms the foundation for optional topics like engineering physics.

2. Charge, current, voltage

Charge

Charge is a quantized physical property of matter that causes it to experience a force in an electromagnetic field. The SI unit of charge is the Coulomb (C), and the smallest possible unit of charge is the elementary charge $e=1.60\times 10^{-19}$ C, equal to the charge of a proton (or the negative of the charge of an electron). Total charge on an object is given by: $$Q=ne$$ where $n$ is an integer number of excess or deficit electrons. Like charges repel, opposite charges attract, and charge is always conserved in closed systems.

Current

Current is the rate of flow of charge past a fixed point in a circuit, defined as: $$I=\frac{\Delta Q}{\Delta t}$$ The SI unit of current is the Ampere (A), where 1 A = 1 C/s. Conventional current is defined as the flow of positive charge from the positive to negative terminal of a power source, which is the opposite direction to the actual flow of electrons in metal conductors. Examiners regularly test understanding of this distinction in direction-based questions.

Voltage (Potential Difference)

Voltage (or potential difference) is the work done per unit charge to move a charge between two points in a circuit, defined as: $$V=\frac{W}{Q}$$ The SI unit of voltage is the Volt (V), where 1 V = 1 J/C. When referring to the total energy supplied per unit charge by a battery or generator, the term electromotive force (emf, symbol $\epsilon$) is used, though it has the same units as potential difference.

Worked Example

A portable power bank transfers 1800 C of charge to a phone over 30 minutes, with an output voltage of 5 V. Calculate (a) the average current supplied, (b) the total work done by the power bank.

• (a) Convert 30 minutes to 1800 s: $I=\frac{1800}{1800}=1$ A
• (b) Rearrange $V=W/Q$: $W=VQ=5\times 1800=9000$ J = 9 kJ

3. Resistance and Ohm's law

Resistance is a measure of how much a circuit component opposes the flow of current, with the SI unit Ohm ($\Omega$), where 1 $\Omega$ = 1 V/A.

Ohm's Law

For ohmic conductors (e.g. metal resistors) at constant temperature, the current through the conductor is directly proportional to the potential difference across it: $$V=IR$$ Ohmic conductors have a linear V-I graph passing through the origin. Non-ohmic components (filament lamps, diodes, thermistors) have non-linear V-I graphs, as their resistance changes with temperature or current direction.

Factors affecting wire resistance

The resistance of a uniform wire depends on its material, length and cross-sectional area: $$R=\rho \frac{L}{A}$$ where $\rho$ is the resistivity of the material (constant for a given material at fixed temperature, unit $\Omega$m), $L$ is the length of the wire, and $A$ is its cross-sectional area.

Worked Example

A nickel wire with resistivity $7.0\times 10^{-8}$ $\Omega$m has a length of 1.5 m and diameter of 0.2 mm. Calculate its resistance, and the current flowing through it when connected to a 3 V battery.

• First calculate cross-sectional area: $r=0.1$ mm = $1\times 10^{-4}$ m, $A=\pi r^2=\pi (1\times 10^{-4}{)}^2\approx 3.14\times 10^{-8}$ m²
• $R=7.0\times 10^{-8}\times \frac{1.5}{3.14\times 10^{-8}}\approx 3.34$ $\Omega$
• $I=V/R=3/3.34\approx 0.90$ A

A common exam trap here is giving you diameter instead of radius for area calculations: always halve diameter first, and convert units to meters.

4. Series and parallel circuits

Series Circuits

Components connected end-to-end in a single loop are in series, with the following rules:

1. Current is the same through all components: $I_{total}=I_1=I_2=I_3...$
2. Total potential difference is the sum of pds across individual components: $V_{total}=V_1+V_2+V_3...$
3. Total resistance is the sum of individual resistances: $R_{total}=R_1+R_2+R_3...$ Ammeters are always connected in series to measure current, and have very low resistance to avoid changing the circuit current.

Parallel Circuits

Components connected across the same two points in a circuit are in parallel, with the following rules:

1. Potential difference is the same across all components: $V_{total}=V_1=V_2=V_3...$
2. Total current is the sum of currents through individual components: $I_{total}=I_1+I_2+I_3...$
3. Reciprocal of total resistance is the sum of reciprocals of individual resistances: $\frac{1}{R_{total}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}...$ Voltmeters are always connected in parallel to measure potential difference, and have very high resistance to avoid drawing current from the circuit.

Worked Example

Two resistors of 3 $\Omega$ and 6 $\Omega$ are connected (a) in series, (b) in parallel across a 9 V battery. Calculate total resistance and total current for both configurations.

• (a) Series: $R_{total}=3+6=9$ $\Omega$, $I=9/9=1$ A
• (b) Parallel: $\frac{1}{R_{total}}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$, so $R_{total}=2$ $\Omega$, $I=9/2=4.5$ A For two parallel resistors, you can use the shortcut $R_{total}=\frac{R_1{R}_2}{R_1+R_2}$ to save time in exams.

5. Magnetic field around a wire and on a moving charge

Magnetic fields are regions where a force is exerted on magnetic materials, moving charges, or current-carrying conductors, represented by field lines that travel from north to south pole outside a magnet.

Magnetic field around a current-carrying wire

A straight wire carrying current produces a magnetic field of concentric circles around the wire. The direction of the field is given by the right-hand grip rule: grip the wire with your right hand, point your thumb in the direction of conventional current, and your curled fingers point in the direction of the magnetic field lines. Field strength increases with higher current, and decreases with distance from the wire.

Force on a moving charge in a magnetic field

A charged particle moving through a magnetic field experiences a force given by: $$F=qvB\sin \theta$$ where $q$ is the charge of the particle, $v$ is its speed, $B$ is the magnetic flux density (unit Tesla, T), and $\theta$ is the angle between the velocity vector and the magnetic field lines. If the particle moves parallel to the field ($\theta =0$), force is zero; if it moves perpendicular ($\theta =90^{\circ }$), force is maximum ($F=qvB$).

The direction of the force is given by Fleming's Left Hand Rule: point your first finger in the direction of the magnetic field, second finger in the direction of conventional current (same as velocity for positive charges, opposite for negative charges), and your thumb points in the direction of the force.

Worked Example

An electron (charge $-1.6\times 10^{-19}$ C) travels at $2\times 10^7$ m/s perpendicular to a 0.4 T magnetic field. Calculate the magnitude of the force on the electron.

• $\theta =90^{\circ }$, so $\sin \theta =1$. Magnitude of charge is $1.6\times 10^{-19}$ C, so $F=1.6\times 10^{-19}\times 2\times 10^7\times 0.4=1.28\times 10^{-12}$ N

6. Force on a current-carrying conductor

Since current is a flow of moving charges, a current-carrying wire placed in a magnetic field also experiences a force, given by: $$F=BIL\sin \theta$$ where $I$ is the current in the wire, $L$ is the length of the wire inside the magnetic field, and $\theta$ is the angle between the current direction and the magnetic field. As with moving charges, force is zero if the current is parallel to the field, and maximum if perpendicular. Direction is again given by Fleming's Left Hand Rule, with the second finger pointing in the direction of conventional current.

This force is the operating principle of electric motors, where a loop of wire carrying current in a magnetic field experiences a rotational force.

Worked Example

A 0.3 m long wire carrying 4 A of current is placed at an angle of 60° to a 0.6 T magnetic field. Calculate the force on the wire.

• $\sin 60^{\circ }=\frac{\sqrt{3}}{2}\approx 0.866$, so $F=0.6\times 4\times 0.3\times 0.866\approx 0.62$ N Exam tip: Only use the length of wire that is actually inside the magnetic field, not the full length of the wire, when calculating this force.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Mixing up electron flow and conventional current for Fleming's rules. Why: Students first learn electron flow in IGCSE, so default to negative charge direction. Correct move: Always use conventional current (positive to negative) for Fleming's second finger; for electrons, point the second finger opposite to the direction of electron movement.
• Pitfall 2: Forgetting to take the final reciprocal when calculating parallel resistance. Why: Rushing through circuit calculations, stopping after summing reciprocals. Correct move: Always check that your total parallel resistance is smaller than the smallest individual resistor, a quick sanity check to catch errors.
• Pitfall 3: Ignoring the $\sin \theta$ term in magnetic force calculations. Why: Students memorize the formula but skip checking the angle between velocity/current and field. Correct move: If the question states the charge or wire is parallel to the field, immediately write $F=0$ with justification, no calculation needed.
• Pitfall 4: Using diameter instead of radius for cross-sectional area of wires. Why: Most questions give diameter instead of radius, leading to 4x errors in area. Correct move: Circle the word "diameter" in the question, halve it to get radius, and convert all units to meters before calculating area.
• Pitfall 5: Confusing ammeter and voltmeter connection rules. Why: Mixing up series and parallel requirements for measurement tools. Correct move: Remember: Ammeter goes in A series, Voltmeter goes in parallel: the first letters match the connection type.

8. Practice Questions (IB Physics SL Style)

Question 1 (3 marks)

A 12 V car headlight has a resistance of 3 $\Omega$ when operating. (a) Calculate the current through the headlight when switched on. (b) Calculate the total charge that flows through the headlight during a 20 minute drive.

Solution

• (a) Use Ohm's Law: $I=V/R=12/3=4$ A (1 mark for formula, 1 mark for correct answer)
• (b) Convert 20 minutes to 1200 s: $Q=I\Delta t=4\times 1200=4800$ C (1 mark for correct time conversion and answer, accept 4.8 × 10³ C)

Question 2 (4 marks)

A circuit has a 4 $\Omega$ resistor in series with a parallel combination of a 6 $\Omega$ and a 12 $\Omega$ resistor, connected across a 12 V battery. Calculate (a) total resistance of the circuit, (b) current through the 6 $\Omega$ resistor.

Solution

• (a) First calculate parallel combination resistance: $\frac{1}{R_{parallel}}=\frac{1}{6}+\frac{1}{12}=\frac{1}{4}$, so $R_{parallel}=4$ $\Omega$. Total resistance = 4 + 4 = 8 $\Omega$ (2 marks for correct parallel and series calculations)
• (b) Total circuit current = $12/8=1.5$ A. Potential difference across parallel combination = $IR=1.5\times 4=6$ V. Current through 6 $\Omega$ resistor = $V/R=6/6=1$ A (2 marks for correct pd calculation and current)

Question 3 (3 marks)

A 0.2 m long wire carrying 5 A of current is placed perpendicular to a uniform magnetic field, and experiences a force of 0.3 N. (a) Calculate the magnetic flux density of the field. (b) State two changes that would double the force on the wire.

Solution

• (a) Rearrange $F=BIL$: $B=F/(IL)=0.3/(5\times 0.2)=0.3$ T (2 marks for correct rearrangement and answer)
• (b) Any two of: double the current, double the length of wire in the field, double the magnetic flux density, change angle from 90° to a value where $\sin \theta$ doubles (1 mark for two correct answers)

9. Quick Reference Cheatsheet

10. What's Next

This topic forms the foundation for multiple core and optional topics in the IB Physics SL syllabus. It connects directly to subtopic 5.4, where you will use circuit rules and Ohm's law to calculate power dissipation and heating effects in resistors, as well as Topic 7 (atomic, nuclear and particle physics) where you will analyze the motion of charged particles in magnetic fields for mass spectrometry and particle detection experiments. If you choose the engineering physics optional topic, you will also build on this knowledge to study electromagnetic induction, alternating current, and transformers.

To reinforce your understanding, practice solving multi-loop circuit problems and magnetic force direction questions, paying close attention to unit conversions and mark scheme requirements for justifying answers. If you get stuck on any concept or problem, you can ask Ollie, our AI tutor, for personalized explanations, step-by-step walkthroughs, and additional practice questions tailored to your weak areas. You can also find more IB Physics SL study resources and past paper practice on the homepage.