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IB Physics – Theme D: Fields

Gravitational, electric and magnetic fields, induction (HL).

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11 questions  ·  70 marks  ·  answers included  ·  IB Physics SL/HL (2023 syllabus)
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This worksheet covers:

  • Exam-style practice on Theme D
  • Gravitational fields and orbits (D.1)
  • Electric and magnetic fields (D.2)
  • Motion in electromagnetic fields (D.3)
  • Induction (D.4, HL only)

Sections A and B are accessible to SL and HL; Section C is HL. Use $G = 6.67\times10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$, $k = 8.99\times10^{9}\,\text{N m}^2\,\text{C}^{-2}$, $e = 1.60\times10^{-19}\,\text{C}$, $m_e = 9.11\times10^{-31}\,\text{kg}$, $m_p = 1.67\times10^{-27}\,\text{kg}$. Calculator required; allow about 75 minutes.

Questions

Section A – Core skills

SL and HL. Give answers to 2 or 3 significant figures.
1

Mars has mass $6.42\times10^{23}\,\text{kg}$ and radius $3.39\times10^{6}\,\text{m}$.(a) Calculate the gravitational field strength at the surface of Mars.(b) Calculate the gravitational potential at the surface of Mars.(c) Hence determine the escape speed from the surface of Mars.

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(a) $g = \dfrac{GM}{R^2} = 3.73\,\text{N kg}^{-1}$(b) $V = -\dfrac{GM}{R} = -1.26\times10^{7}\,\text{J kg}^{-1}$(c) $\tfrac{1}{2}v^2 = \dfrac{GM}{R} \Rightarrow v = 5.03\times10^{3}\,\text{m s}^{-1}$
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2

Two point charges are placed on the $x$-axis: $+4.0\,\mu\text{C}$ at $x = 0$ and $-2.0\,\mu\text{C}$ at $x = 0.30\,\text{m}$.(a) Calculate the magnitude and direction of the electric field at $x = 0.10\,\text{m}$.(b) Calculate the electric potential at $x = 0.10\,\text{m}$.

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(a) $E_1 = \dfrac{k(4.0\times10^{-6})}{0.10^2} = 3.60\times10^{6}$, $E_2 = \dfrac{k(2.0\times10^{-6})}{0.20^2} = 4.50\times10^{5}$, both in the $+x$ direction. $E = 4.05\times10^{6}\,\text{N C}^{-1}$ towards $+x$(b) $V = k\left(\dfrac{4.0\times10^{-6}}{0.10} – \dfrac{2.0\times10^{-6}}{0.20}\right) = 2.70\times10^{5}\,\text{V}$
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3

A straight wire of length $0.40\,\text{m}$ carries a current of $3.0\,\text{A}$ in a uniform magnetic field of flux density $0.050\,\text{T}$. The wire makes an angle of $30^\circ$ with the field.(a) Calculate the magnetic force on the wire.(b) State the angle between the wire and the field for which the force is a maximum, and explain how the direction of the force is found.

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(a) $F = BIL\sin\theta = 0.050 \times 3.0 \times 0.40 \times \sin 30^\circ = 0.030\,\text{N}$(b) $90^\circ$; direction is perpendicular to both the current and the field, given by the left-hand rule (or $\mathbf{F} = I\mathbf{L}\times\mathbf{B}$).
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4

Explain, in terms of field lines, why the gravitational field strength near the surface of the Earth can be treated as uniform, but the field around the Earth as a whole is radial. State how the gravitational field strength $g$ is related to the gravitational potential $V$.

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Field lines point towards the centre of the Earth, so over the whole planet the field is radial and $g \propto 1/r^2$. Over distances much smaller than the Earth’s radius the lines are effectively parallel and equally spaced, so the field is approximately uniform. $g = -\dfrac{\Delta V}{\Delta r}$ (field strength is the negative gradient of potential).
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Section B – Exam-style

SL and HL.
5

A satellite of mass $1200\,\text{kg}$ orbits the Earth (mass $5.97\times10^{24}\,\text{kg}$, radius $6.37\times10^{6}\,\text{m}$) in a circular orbit of radius $3R_E$.(a) Show that the orbital speed is about $4.6\,\text{km s}^{-1}$.(b) Calculate the orbital period in hours.(c) Calculate the total energy of the satellite in orbit.(d) Calculate the minimum energy needed to lift the satellite from the Earth’s surface to the orbit radius (ignore the kinetic energy needed for orbit).

Show answer
(a) $v = \sqrt{\dfrac{GM}{3R_E}} = \sqrt{\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{1.911\times10^{7}}} = 4.56\times10^{3}\,\text{m s}^{-1}$(b) $T = \dfrac{2\pi r}{v} = 2.63\times10^{4}\,\text{s} = 7.3\,\text{h}$(c) $E = -\dfrac{GMm}{2r} = -1.25\times10^{10}\,\text{J}$(d) $\Delta E_p = GMm\left(\dfrac{1}{R_E} – \dfrac{1}{3R_E}\right) = 5.0\times10^{10}\,\text{J}$
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6

Two parallel metal plates are $5.0\,\text{mm}$ apart with a potential difference of $400\,\text{V}$ between them. An electron is released from rest at the negative plate.(a) Calculate the electric field strength between the plates.(b) Calculate the force on the electron.(c) Calculate the speed of the electron when it reaches the positive plate.(d) Calculate the time taken to cross the gap.

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(a) $E = \dfrac{V}{d} = 8.0\times10^{4}\,\text{V m}^{-1}$(b) $F = eE = 1.28\times10^{-14}\,\text{N}$(c) $\tfrac{1}{2}mv^2 = eV \Rightarrow v = 1.19\times10^{7}\,\text{m s}^{-1}$(d) $a = F/m = 1.41\times10^{16}\,\text{m s}^{-2}$; $t = \sqrt{2d/a} = 8.4\times10^{-10}\,\text{s}$
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7

A proton moving at $2.0\times10^{6}\,\text{m s}^{-1}$ enters a uniform magnetic field of $0.25\,\text{T}$ at right angles to the field.(a) Explain why the proton follows a circular path and why its speed does not change.(b) Calculate the radius of the path.(c) Show that the period of the motion is independent of the speed, and calculate it.(d) A uniform electric field is now added so that the proton passes through undeflected. State the direction of the electric force relative to the magnetic force and calculate the electric field strength required.

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(a) The magnetic force is always perpendicular to the velocity, so it acts as a centripetal force; it does no work, so kinetic energy and speed are constant.(b) $r = \dfrac{mv}{qB} = \dfrac{1.67\times10^{-27}\times2.0\times10^{6}}{1.60\times10^{-19}\times0.25} = 8.4\times10^{-2}\,\text{m}$(c) $T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$, no $v$; $T = 2.6\times10^{-7}\,\text{s}$(d) Opposite to the magnetic force; $qE = qvB \Rightarrow E = vB = 5.0\times10^{5}\,\text{V m}^{-1}$
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8

A geostationary satellite has an orbital period of $86\,164\,\text{s}$.(a) State two features of a geostationary orbit other than its period.(b) Using Kepler’s third law in the form $T^2 = \dfrac{4\pi^2 r^3}{GM}$, calculate the height of the satellite above the Earth’s surface.

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(a) Any two: orbit is in the equatorial plane; satellite moves in the same sense as the Earth’s rotation; it remains above the same point on the equator.(b) $r = \left(\dfrac{GMT^2}{4\pi^2}\right)^{1/3} = 4.21\times10^{7}\,\text{m}$; height $= 4.21\times10^{7} – 6.37\times10^{6} = 3.58\times10^{7}\,\text{m}$ (about $36\,000\,\text{km}$)
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Section C – HL: Induction

HL only.
9

A flat coil of 200 turns and cross-sectional area $4.0\times10^{-4}\,\text{m}^2$ is placed with its plane perpendicular to a magnetic field. The flux density increases uniformly from $0$ to $0.30\,\text{T}$ in $0.050\,\text{s}$.(a) Define magnetic flux linkage.(b) Calculate the emf induced in the coil.(c) State Lenz’s law and explain how it is a consequence of conservation of energy.

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(a) Product of the number of turns and the magnetic flux through each turn, $N\Phi = NBA\cos\theta$.(b) $\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = \dfrac{200 \times 4.0\times10^{-4} \times 0.30}{0.050} = 0.48\,\text{V}$(c) The induced emf acts so as to oppose the change producing it. If it reinforced the change, flux and current would grow without an energy input; instead work must be done against the induced effect to produce the electrical energy.
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10

A metal rod of length $0.60\,\text{m}$ slides at a constant $4.0\,\text{m s}^{-1}$ along two parallel horizontal rails in a uniform vertical magnetic field of $0.20\,\text{T}$. The rails are connected by a resistor of $1.2\,\Omega$; the rod and rails have negligible resistance.(a) Show that the emf induced in the rod is $0.48\,\text{V}$.(b) Calculate the current in the circuit.(c) Calculate the magnetic force on the rod and state its direction relative to the motion.(d) Show that the mechanical power needed to keep the rod moving equals the electrical power dissipated in the resistor.

Show answer
(a) $\varepsilon = BLv = 0.20 \times 0.60 \times 4.0 = 0.48\,\text{V}$(b) $I = \dfrac{0.48}{1.2} = 0.40\,\text{A}$(c) $F = BIL = 0.20 \times 0.40 \times 0.60 = 0.048\,\text{N}$, opposing the motion (Lenz)(d) Mechanical power $= Fv = 0.048 \times 4.0 = 0.19\,\text{W}$; electrical power $= \varepsilon I = 0.48 \times 0.40 = 0.19\,\text{W}$
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11

A rectangular coil rotates at constant angular speed $\omega$ in a uniform magnetic field.(a) Explain why the induced emf is sinusoidal and state the position of the coil when the emf is a maximum.(b) The peak emf is $12\,\text{V}$ at $50\,\text{Hz}$. State the rms emf and the peak emf if the rotation rate is doubled.(c) Describe two ways in which the design of a transformer core reduces energy losses.

Show answer
(a) Flux linkage $= NBA\cos\omega t$, so $\varepsilon = NBA\omega\sin\omega t$, which is sinusoidal; maximum when the plane of the coil is parallel to the field (flux is zero but changing fastest).(b) $\varepsilon_{\text{rms}} = \dfrac{12}{\sqrt{2}} = 8.5\,\text{V}$; peak emf $\propto \omega$, so $24\,\text{V}$ at $100\,\text{Hz}$.(c) Laminated core (insulated layers) reduces eddy currents; soft iron core with low hysteresis loss (easily magnetised and demagnetised); core shaped so almost all flux links both coils.
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Related worksheets

More worksheets in IB Physics (SL & HL):

IB Physics – Theme A: Space, Time and MotionIB Physics – Theme B: The Particulate Nature of MatterIB Physics – Theme C: Wave BehaviourIB Physics – Theme E: Nuclear and Quantum PhysicsIB Physics – Data-Based Questions (Paper 1B)

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