IB Physics SL D Fields Questions

Practise IB Physics SL fields through electric, gravitational and magnetic interactions, interpreting field diagrams and applying equations to data.

Exam points

Apply gravitational field strength, potential, orbital and escape relationships to fields and energy data.
Calculate electric and magnetic field strength, force, potential and charged-particle motion using field diagrams and right-hand rules.
Analyse charged-particle trajectories, velocity selection and energy or momentum changes in electromagnetic fields.

Covered syllabus points

D.1 Gravitational fields D.2 Electric and magnetic fields D.3 Motion in electromagnetic fields

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Study D. Fields conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

[Maximum number: 3]

A satellite powered by solar cells directed towards the Sun is in a polar orbit about the Earth.

The satellite is orbiting the Earth at a distance of 6600 km from the centre of the Earth.

Determine the orbital period for the satellite.

Mass of Earth $=6.0 \times 10^{24} \mathrm{~kg}$

$\begin{aligned} & \frac{m v^{2}}{r}=G \frac{M m}{r^{2}} & \text { leading to } T^{2}=\frac{4 \pi^{2} r^{3}}{G M} & T=5320 \text { «s» } \end{aligned}$

Alternative 2
$\begin{aligned} & \text { « } V=\sqrt{\frac{G m_{E}}{r}} \text { » }=\sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{6600 \times 10^{3}}} \text { OR } 7800 \text { « } \mathrm{ms}^{-1} \text { » } & \text { distance }=2 \pi r=2 \pi \times 6600 \times 10^{3} \text { « } \mathrm{m} \text { » or } 4.15 \times 10^{7} \text { « } \mathrm{m} \text { » } & \text { « } T=\frac{d}{v}=\frac{4.15 \times 10^{7}}{7800} \text { » }=5300 \text { « } \mathrm{S} \text { » } \end{aligned}$

Accept use of $\omega$ nistead of v

[Maximum number: 5]

Ion-thrust engines can power spacecraft. In this type of engine, ions are created in a chamber and expelled from the spacecraft. The spacecraft is in outer space when the propulsion system is turned on. The spacecraft starts from rest.

The mass of ions ejected each second is $6.6 \times 10^{-6} \mathrm{~kg}$ and the speed of each ion is $5.2 \times 10^{4} \mathrm{~m} \mathrm{~s}^{-1}$ . The initial total mass of the spacecraft and its fuel is 740 kg . Assume that the ions travel away from the spacecraft parallel to its direction of motion.

(a)

In practice, the ions leave the spacecraft at a range of angles as shown.

[ 2 ]

(i)

Outline why the ions are likely to spread out.

[ 2 ]

(b)

On arrival at the planet, the spacecraft goes into orbit as it comes into the gravitational field of the planet.

[ 3 ]

(i)

Outline what is meant by the gravitational field strength at a point.

[ 2 ]

(ii)

Newton's law of gravitation applies to point masses. Suggest why the law can be applied to a satellite orbiting a spherical planet of uniform density.

[ 1 ]

[Maximum number: 5]

Two oppositely charged parallel plates are a distance 8.0 cm apart. The potential difference between the plates is 120 V . An alpha particle is placed on the positively charged plate and released from rest. Gravity is ignored.

(a)

Calculate the electric field between the plates.

[ 1 ]

$E=\ll \frac{120}{8.0 \times 10^{-2}}=>1.5 \times 10^{3} N C^{-1} \boldsymbol{O} \boldsymbol{R} \mathrm{Vm}^{-1}$

[1]

(b)

Show that the acceleration of the alpha particle is about $7 \times 10^{10} \mathrm{~ms}^{-2}$ .

[ 2 ]

$\begin{aligned} F=<e E=>3.2 \times 10^{-19} \times 1.5 \times 10^{3} \text { OR } 4.8 \times 10^{-16} \mathrm{~N} a=<\frac{F}{m}=>\frac{4.8 \times 10^{-16}}{4 \times 1.67 \times 10^{-27}} \text { OR } 7.2 \times 10^{10} \mathrm{~ms}^{-2} \quad \end{aligned}$

Allow ECF from a).

Award [1] if they use a charge of e and a mass of 2 u obtaining the right result.

[2]

(c)

A magnetic field directed into the plane of the page is now established between the plates. An alpha particle enters the region between the plates with a horizontal speed of $5.0 \times 10^{5} \mathrm{~m} \mathrm{~s}^{-1}$ . The particle is not deflected.

Calculate the magnitude of the magnetic field.

[ 2 ]

$\begin{aligned} F_{\mathrm{m}} =F_{\mathrm{e}} \text { OR } q v B=q E \text { OR } B=F_{\mathrm{e}} / q v \quad B =\ll \frac{E}{v}=\frac{1.5 \times 10^{3}}{5.0 \times 10^{5}}=>3.0 \times 10^{-3} T \end{aligned}$

Award [2] if $3.0 \times 10^{-3}$ « T » is seen as the answer without working.

Allow ECF from a) and b) i)

[2]

[Maximum number: 12]

This question is in two parts. Part 1 is about simple harmonic motion and forced oscillations. Part 2 is about electric and magnetic force fields.

Part 1 Simple harmonic motion and forced oscillations
The graph shows the variation with time of the displacement of an object undergoing simple harmonic motion.
displacement / mm

time / ms

(a)

Part 2 Electric and magnetic force fields

[ 12 ]

(i)

Define electric field strength.

[ 2 ]

(ii)

The diagram shows a pair of horizontal metal plates. Electrons can be deflected vertically using an electric field between the plates.

Figure for Question (ii) — IB Physics SL

[ 5 ]

(i)

Label, on the diagram, the polarity of the metal plates which would cause an electron positioned between the plates to accelerate upwards.

[ 1 ]

(ii)

Draw the shape and direction of the electric field between the plates on the diagram.

[ 2 ]

(iii)

Calculate the force on an electron between the plates when the electric field strength has a value of $2.5 \times 10^{3} \mathrm{NC}^{-1}$ .

[ 2 ]

$F=2.5 \times 10^{3} \times 1.6 \times 10^{-19}$ ;
$4.0 \times 10^{-16} \mathrm{~N}$ ;

Marking guidance:

Award [2] for a bald correct answer.

(iii)

The diagram shows two isolated electrons, X and Y , initially at rest in a vacuum. The initial separation of the electrons is 5.0 mm . The electrons subsequently move apart in the directions shown.

Figure for Question (iii) — IB Physics SL

[ 5 ]

(i)

Show that the initial electric force acting on each electron due to the other electron is approximately $9 \times 10^{-24} \mathrm{~N}$ .

[ 2 ]

(c) (i) use of $F=\frac{\left(1.60 \times 10^{-19}\right)^{2}}{4 \pi \varepsilon_{0}\left(5.0 \times 10^{-3}\right)^{2}}$ or $F=\frac{\left(1.60 \times 10^{-19}\right)^{2}}{\left(5.0 \times 10^{-3}\right)^{2}} \times 8.99 \times 10^{9}$ ;
$9.2 \times 10^{-24} \mathrm{~N} ;$

(ii)

Discuss the motion of one electron after it begins to move.

[ 3 ]

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