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Topic D.2 · SL and HL

Electric and Magnetic Fields: notes and practice questions

Summary
  • This topic covers the fundamental properties of electric charges and magnetic fields, and their interactions.
  • Electric charges exert forces described by Coulomb's law: F=kq1q2r2F = k\frac{q_1q_2}{r^2}.
  • Electric field strength EE is defined as the force per unit charge: E=FqE = \frac{F}{q}.
  • The uniform electric field strength between parallel plates is given by E=VdE = \frac{V}{d}.
  • Electric charge is conserved and quantized, with Millikan's experiment providing evidence.
  • Charge transfer can occur through friction, electrostatic induction, and contact, including grounding.
  • Electric field lines represent the direction and relative strength (density) of an electric field.
  • Magnetic field patterns are studied for bar magnets, current-carrying straight wires, circular coils, and solenoids.
  • The direction of the magnetic field around a current-carrying wire can be determined.

How it is examined

Both papers. Field-line sketching is examined directly, and the guidance list above is effectively the list of diagrams that can be asked for. Equipotential questions are HL only. The parallel-plate result E=V/dE = V/d is the bridge into D.3, and most Paper 2 questions use it as the first step of a longer electron-deflection problem. An SL question must not ask about potential or equipotentials in either the gravitational or the electric case.

Given in the booklet

SL: Coulomb's law with k=1/4πε0k = 1/4\pi\varepsilon_0, E=F/qE = F/q, E=V/dE = V/d, plus the elementary charge, ε0\varepsilon_0 and k in the constants table, and the electronvolt conversion. HL adds Ep=kq1q2/rE_\text{p} = kq_1q_2/r, Ve=kQ/rV_\text{e} = kQ/r, the potential gradient and W=qΔVeW = q\Delta V_\text{e}. Permittivity values for materials other than free space are supplied in the question.

Key ideas
  • the direction of forces between the two types of electric charge
  • Coulomb's law as given by F=kq1q2r2F = k\dfrac{q_1 q_2}{r^2} for charged bodies treated as point charges, where k=14πε0k = \dfrac{1}{4\pi\varepsilon_0}
  • the conservation of electric charge
  • Millikan's experiment as evidence for quantization of electric charge
At HL
  • the electric potential energy EpE_\text{p} in terms of work done to assemble the system from infinite separation
  • the electric potential energy for a system of two charged bodies as given by Ep=kq1q2rE_\text{p} = k\dfrac{q_1 q_2}{r}
  • that the electric potential is a scalar quantity with zero defined at infinity
  • that the electric potential VeV_\text{e} at a point is the work done per unit charge to bring a test charge from infinity to that point, as given by Ve=kQrV_\text{e} = \dfrac{kQ}{r}

Guiding questions

  • Which experiments provided evidence to determine the nature of the electron?
  • How can the properties of fields be understood using both an algebraic approach and a visual representation?
  • What are the consequences of interactions between electric and magnetic fields?

Linking questions

  • How are electric and magnetic fields like gravitational fields?
  • What are the relative strengths of the four fundamental forces?
  • How can moving charges in magnetic fields help probe the fundamental nature of matter?
  • Charge is quantized. Which other physical quantities are quantized? (NOS)

Practice questions

22 questions · 4 easy · 17 medium · 1 hard
Showing 20 of 20

Question 1

EasyPaper 1A · calculator1 mark

The work done per unit positive charge to move a point charge from infinity to a point X in an electric field is the

A. electric field strength at X.

B. electric force at X.

C. electric potential energy at X.

D. electric potential at X.

Question 2

MediumPaper 2 · calculator5 marks
(a)

The fine-structure constant, α\alpha, is a fundamental physical constant that characterizes the strength of the electromagnetic interaction. It is given by the expression α=e24πϵ0ℏc\alpha = \frac{e^2}{4\pi\epsilon_0 \hbar c}.

(a) Show that the quantity e24πϵ0\frac{e^2}{4\pi\epsilon_0} has units of energy multiplied by distance.

[2]
(b)

(b) Hence, show that the fine-structure constant α\alpha is dimensionless.

[3]

Question 3

HardPaper 2 · calculator14 marks
(a)(i)

In a controlled environment, a technician is studying the behavior of microscopic charged dust particles. One such particle, with two excess electrons, is observed to be held perfectly stationary between two horizontal parallel metal plates. The uniform electric field between the plates is 2.5×105 V m−12.5 \times 10^5\,\text{V}\,\text{m}^{-1}. The density of the dust particle is ρ=1200 kg m−3\rho = 1200\,\text{kg}\,\text{m}^{-3}.

(a) (i) Calculate the radius of the dust particle.

[3]
(a)(ii)

(a) (ii) State one significant assumption made in your calculation in (a)(i).

[1]
(b)(i)

(b) (i) The electric field is suddenly switched off, and the same dust particle begins to fall, reaching a constant terminal speed of 1.5×10−4 m s−11.5 \times 10^{-4}\,\text{m}\,\text{s}^{-1}. Explain why the particle reaches a constant terminal speed.

[2]
(b)(ii)

(b) (ii) Using the data, estimate the dynamic viscosity of the air in the controlled environment.

[4]
(c)

(c) If the particle were to acquire only one excess electron and the original electric field was re-established, determine the new terminal velocity of the particle, stating its direction.

[4]

Question 4

EasyPaper 1A · calculator1 mark

Two point charges, +q+q and +Q+Q, are held a distance rr apart. The magnitude of the electrostatic force on charge +q+q is FF. The charge +q+q is replaced by a charge of +2q+2q and the separation is increased to 2r2r. What is the magnitude of the electrostatic force on charge +Q+Q?

A. F4\frac{F}{4}

B. F2\frac{F}{2}

C. FF

D. 2F2F

Question 5

MediumPaper 1A · calculator1 mark

A uniform copper wire of length 2.0 m2.0 \text{ m} has a resistance of 10Ω10 \Omega.

The wire is then drawn out, maintaining a constant volume, until its new length is 6.0 m6.0 \text{ m}.

What is the resistance of the new wire?

A. 10Ω10 \Omega

B. 30Ω30 \Omega

C. 90Ω90 \Omega

D. 270Ω270 \Omega

Question 6

EasyPaper 1A · calculator1 mark

Two identical positive point charges, each of magnitude +Q+Q, are separated by a distance 2d2d.

What is the magnitude of the electric field strength at the midpoint between the charges?

A. Zero

B. kQd2\frac{kQ}{d^2}

C. 2kQd2\frac{2kQ}{d^2}

D. kQ4d2\frac{kQ}{4d^2}

Question 7

MediumPaper 1A · calculator1 mark

A charged liquid flows through a rectangular channel of width 0.40m0.40 \text{m} at a constant speed of 1.5ms−11.5 \text{ms}^{-1}. The liquid carries a uniform surface charge density of 2.0mCm−22.0 \text{mCm}^{-2} on its exposed surface.

As the liquid passes a collection point, all the charge is continuously transferred to a conductor.

What is the current in the conductor?

A. 0.80mA0.80 \text{mA}

B. 1.2mA1.2 \text{mA}

C. 2.4mA2.4 \text{mA}

D. 3.0mA3.0 \text{mA}

Question 8

EasyPaper 1A · calculator1 mark

A hollow, positively charged conducting sphere has radius RR. Point P is at the center of the sphere, and point Q is on its outer surface. What is the relationship between the electric potential at P, VPV_P, and the electric potential at Q, VQV_Q?

A. VP=0V_P = 0 and VQ>0V_Q > 0

B. VP=VQV_P = V_Q

C. VP<VQV_P < V_Q

D. VP>VQV_P > V_Q

Question 9

MediumPaper 1A · calculator1 mark

In an experimental electromagnetic levitation system, a horizontal conductor of length 0.45m0.45 \text{m} is placed in a uniform vertical magnetic field. When the current in the conductor is increased by 3.0A3.0 \text{A}, the upward magnetic force acting on it increases by 6.75mN6.75 \text{mN}.

What is the strength of the magnetic field?

A. 0.50mT0.50 \text{mT}

B. 5.0mT5.0 \text{mT}

C. 50mT50 \text{mT}

D. 5.0T5.0 \text{T}

Question 10

MediumPaper 1A · calculator1 mark

A heating element of length LL is used in a laboratory oven. When the potential difference across the element is 120V120\text{V}, the power transferred by the element is 600W600\text{W}. A second heating element is made from the same material and has the same cross-sectional area. When a potential difference of 240V240\text{V} is applied across the second element, the power transferred is 1200W1200\text{W}.

What is the length of the second heating element?

A. L4\frac{L}{4}

B. L2\frac{L}{2}

C. LL

D. 2L2L

Question 11

MediumPaper 2 · calculator4 marks

A microscopic pollen grain, with a mass of 5.0×10−155.0 \times 10^{-15} kg, acquires a net negative charge due to the presence of three excess electrons. This pollen grain is introduced into a region between two horizontal parallel metal plates separated by a distance of 1515 mm.

(a) Calculate the potential difference that must be applied between the plates to keep the pollen grain suspended in equilibrium.

Question 12

MediumPaper 1A · calculator1 mark

A researcher is testing a new type of electromagnetic actuator. A straight segment of wire, part of the actuator, is placed perpendicularly within a uniform magnetic field.

The length of the wire segment within the field is 0.25 m0.25 \text{ m}.

When the current flowing through this segment is increased by 3.0 A3.0 \text{ A}, the magnetic force acting on the wire increases by 2.4 mN2.4 \text{ mN}.

What is the strength of the magnetic field?

A. 3.2 mT3.2 \text{ mT}

B. 32 mT32 \text{ mT}

C. 0.32 mT0.32 \text{ mT}

D. 3.2 T3.2 \text{ T}

Question 13

MediumPaper 1A · calculator1 mark

Two long, straight, parallel conductors P and Q initially carry currents in the same direction. This results in an attractive force per unit length of magnitude FF. The current in conductor P is II and the current in conductor Q is 3I3I. The following changes are then made:

  • The direction of the current in P is reversed and its magnitude is halved.
  • The separation between the conductors is tripled.
  • The current in Q remains unchanged.

What is the new magnitude and direction of the force per unit length on each conductor?

MagnitudeDirection
A.F6\frac{F}{6}repulsive
B.F6\frac{F}{6}attractive
C.F3\frac{F}{3}repulsive
D.F3\frac{F}{3}attractive

Question 14

MediumPaper 1A · calculator1 mark

In a particle scattering experiment, a proton is accelerated from rest through a potential difference of 2.5 MV2.5\text{ MV}. The proton is aimed directly at a stationary lithium nucleus. The proton number of lithium is 33.

What is the distance of closest approach between the proton and the lithium nucleus?

A. 3ke2.5×106\frac{3ke}{2.5 \times 10^6}

B. 3ke22.5×106\frac{3ke^2}{2.5 \times 10^6}

C. ke2.5×106\frac{ke}{2.5 \times 10^6}

D. 4ke2.5×106\frac{4ke}{2.5 \times 10^6}

Question 15

MediumPaper 1A · calculator1 mark

Two long, straight, parallel conductors P and Q are separated by a distance dd. Conductor P carries a current II and conductor Q carries a current 2I2I in the same direction. The force per unit length on each conductor is attractive and has a magnitude FF.

The direction of the current in Q is reversed. The current in P is tripled and the current in Q is halved. The separation between the conductors is also halved.

What is the new magnitude and direction of the force per unit length on each conductor?

A. 3F3F repulsive

B. 3F3F attractive

C. 6F6F repulsive

D. 6F6F attractive

Question 16

MediumPaper 1A · calculator1 mark

A small, charged spherical pollen grain is held stationary in a uniform vertical electric field. A second pollen grain of the same material has half the radius and double the charge of the first grain. This second grain is placed in the same electric field.

What is the initial motion of the second pollen grain?

A. It is stationary.

B. It moves with constant velocity upwards.

C. It accelerates upwards.

D. It accelerates downwards.

Question 17

MediumPaper 1A · calculator1 mark

A proton and a deuteron are accelerated from rest through the same potential difference. The mass of a deuteron is approximately twice the mass of a proton, and they have the same magnitude of charge.

What are the values for the ratio of the final kinetic energies Ek,protonEk,deuteron\frac{E_{k, \text{proton}}}{E_{k, \text{deuteron}}} and the ratio of the final de Broglie wavelengths λprotonλdeuteron\frac{\lambda_{\text{proton}}}{\lambda_{\text{deuteron}}}?

A. Kinetic energy ratio = 1; Wavelength ratio = 2\sqrt{2}

B. Kinetic energy ratio = 1; Wavelength ratio = 12\frac{1}{\sqrt{2}}

C. Kinetic energy ratio = 2; Wavelength ratio = 2

D. Kinetic energy ratio = 1; Wavelength ratio = 1

Question 18

MediumPaper 1A · calculator1 mark

Light passes from a type of optical glass into air. The critical angle for this interface is 42∘42^\circ.

What is the approximate value for the speed of light in the optical glass?

A. 1.8×108 m s−11.8 \times 10^8 \text{ m s}^{-1}

B. 2.0×108 m s−12.0 \times 10^8 \text{ m s}^{-1}

C. 2.2×108 m s−12.2 \times 10^8 \text{ m s}^{-1}

D. 2.4×108 m s−12.4 \times 10^8 \text{ m s}^{-1}

Question 19

MediumPaper 1A · calculator1 mark

Two long, parallel conductors, X and Y, are separated by a distance rr. They carry currents IXI_X and IYI_Y respectively. The magnetic force per unit length exerted on conductor X due to the current in conductor Y is FF.

The currents are changed to 2IX2I_X and 3IY3I_Y. The separation between the conductors is increased to 6r6r.

What is the new magnetic force per unit length on conductor X?

A. F6\frac{F}{6}

B. FF

C. 3F2\frac{3F}{2}

D. 6F6F

Question 20

MediumPaper 1A · calculator1 mark

Two long, parallel power cables, X and Y, are separated by a distance dd. They carry currents IXI_X and IYI_Y respectively. The magnetic force per unit length on cable X due to cable Y is FF.

During maintenance, the separation between the cables is reduced to d3\frac{d}{3} and the current in cable X is halved.

What is the new magnetic force per unit length on cable X?

A. F6\frac{F}{6}

B. 2F3\frac{2F}{3}

C. 3F2\frac{3F}{2}

D. 6F6F

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  • Answering a procedure question with a platitude.
  • Losing precision in Paper 1B. Uniquely to this paper, quoting the right number badly loses marks.
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What does Electric and Magnetic Fields cover in IB Physics?

This topic covers the fundamental properties of electric charges and magnetic fields, and their interactions. Electric charges exert forces described by Coulomb's law: F = k(q_1q_2)/(r^2). Electric field strength E is defined as the force per unit charge: E = (F)/(q).

Is Electric and Magnetic Fields SL or HL?

Both. SL and HL students study Electric and Magnetic Fields, and HL goes further: the electric potential energy E_p in terms of work done to assemble the system from infinite separation.

How do I revise Electric and Magnetic Fields for IB Physics?

Start from the core idea: this topic covers the fundamental properties of electric charges and magnetic fields, and their interactions. In the exam: both papers. Field-line sketching is examined directly, and the guidance list above is effectively the list of diagrams that can be asked for. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Electric and Magnetic Fields?

FourtyFive has 22 Electric and Magnetic Fields questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

Is FourtyFive free for Electric and Magnetic Fields practice?

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Can I handwrite Electric and Magnetic Fields answers on an iPad?

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