D.3 Motion in electromagnetic fields: IB Physics SL exam-style questions
D.3 analyses how charged particles move in uniform fields. In an electric field the force is constant and the path is like a projectile's; in a magnetic field the force F = qvB sin θ is perpendicular to the velocity, so a particle moves in a circle at constant speed and its kinetic energy does not change.
You also need crossed electric and magnetic fields, the charge-to-mass ratio from a particle's path, the force on a current-carrying conductor F = BIL sin θ, and the force per unit length between parallel wires, which attract when the currents are in the same direction.
36 questions
267 marks
Paper 1A: 14
Paper 1B: 8
Paper 2: 14
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36 practice questions on D.3 Motion in electromagnetic fields
1D-1A-19
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksCalculate
A straight wire of length 0.40 m carries a current of 3.0 A at right angles to a uniform magnetic field of flux density 0.25 T.
What is the force on the wire?
Diagram NOT accurately drawnShow mark scheme
Marking point
Mark
Notes
Step 1F = BIL sin θ with θ = 90°, so F = BIL = 0.25 × 3.0 × 0.40 = 0.30 N.
✓ 1
Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThis leaves out the current: BL = 0.25 × 0.40 = 0.10.
BCorrect: F = BIL = 0.25 × 3.0 × 0.40 = 0.30 N.
CThis divides by the length instead of multiplying by it.
DThis divides by B instead of multiplying: IL/B = 3.0 × 0.40/0.25 = 4.8.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Calculate
2D-1A-20
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A straight wire carrying a current lies at 30° to a uniform magnetic field. The magnetic force on the wire is F.
The wire is turned until it is at right angles to the field. The current and the length of wire in the field do not change. What is the magnetic force on the wire now?
Show mark scheme
Marking point
Mark
Notes
Step 1F = BIL sin θ, where θ is the angle between the wire and the field: at 30°, F = BIL × 0.5.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2At 90° the force is BIL = F/sin 30° = 2F.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis uses cos θ, which is zero at 90°. The force is zero when the wire is parallel to the field, not perpendicular.
BThis multiplies by sin 30° again instead of dividing by it.
CThis uses F = BIL cos θ for the first position, giving F/cos 30° = 2F/√3.
DCorrect: BIL = F/sin 30° = 2F, the maximum force.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Determine
3D-1A-21
Force on a moving charge·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A proton and an alpha particle move at right angles to the same uniform magnetic field. The speed of the alpha particle is one quarter of the speed of the proton.
What is (magnetic force on the alpha particle)/(magnetic force on the proton)?
Show mark scheme
Marking point
Mark
Notes
Step 1F = qvB, so F ∝ qv in the same field; the alpha particle has charge 2e.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ratio = 2 × 1/4 = 0.5.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis allows for the lower speed but takes the charge of the alpha particle as e instead of 2e.
BCorrect: twice the charge and a quarter of the speed give 2 × 0.25 = 0.5.
CThis uses the mass ratio, 4, instead of the charge ratio, 2. The magnetic force does not depend on mass.
DThis allows for the doubled charge but not for the lower speed.
Syllabus understandingD.3 — the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θCommand term: Determine
4D-1A-22
Charge-to-mass ratio·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
A particle of charge q and mass m moves with speed v at right angles to a uniform magnetic field of flux density B. It follows a circular path of radius r.
Which expression gives the charge-to-mass ratio q/m of the particle?
Show mark scheme
Marking point
Mark
Notes
Step 1The magnetic force provides the centripetal force: qvB = mv²/r.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2q/m = v/Br.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: qvB = mv²/r rearranges to q/m = v/Br.
BThis is m/q, the inverted ratio.
CThis leaves v out of the magnetic force, writing qB = mv²/r.
DThis writes the centripetal force as mv²r instead of mv²/r.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; determination of the charge-to-mass ratio from the path in a uniform magnetic field (guidance) Command term: Deduce
5D-1A-23
Direction of the magnetic force·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksDeduce
A straight wire in the plane of the page carries a current towards the bottom of the page. A uniform magnetic field is directed out of the page.
What is the direction of the magnetic force on the wire?
Diagram NOT accurately drawnShow mark scheme
Marking point
Mark
Notes
Step 1Fleming's left-hand rule: first finger out of the page (field), second finger towards the bottom of the page (current), so the thumb (force) points towards the left of the page.
✓ 1
Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: the force is perpendicular to both the current and the field, towards the left of the page.
BThis reverses the rule (for example by using the right hand).
CThe force is perpendicular to the field, so it cannot point along the field.
DThe force is perpendicular to the current, so it cannot act along the wire.
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Deduce
6D-1A-24
Parallel current-carrying wires·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
Two long parallel wires 0.10 m apart in a vacuum carry currents of 5.0 A and 3.0 A in the same direction.
Which row gives the magnitude of the force per unit length on each wire and describes the force?
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Currents in the same direction attract.
✓ 1
Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis is the value at 0.20 m; the separation has been doubled.
BCorrect: 3.0 × 10−5 N m−1, and currents in the same direction attract.
CThe magnitude is right, but parallel currents in the same direction attract.
DThe 2π has been left out of the denominator.
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/2πr; the force is attractive when the currents flow in the same direction (guidance) Command term: Determine
7D-1A-25
Charge in a uniform electric field·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksIdentify
A proton is released from rest in a uniform electric field in a vacuum.
Which graph shows the variation of the kinetic energy Ek of the proton with the distance s it has moved?
Sketch graphs, not to scaleShow mark scheme
Marking point
Mark
Notes
Step 1The force eE on the proton is constant, so the work done after a distance s is eEs.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Ek = eEs, a straight line through the origin.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AEk ∝ s² would follow from v ∝ s; this confuses distance with time (Ek ∝ t² for constant acceleration).
BLevelling off would need a resistive force, as for terminal speed. In a vacuum the force eE stays constant.
CCorrect: work done = force × distance = eEs, so Ek is proportional to s.
DThis is the shape of the speed against distance (v ∝ √s), not the kinetic energy.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field Command term: Identify
8D-1B-05
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 1BEasy9 marks
Data-based question9 steps to full marksDetermine
A student investigates the force on a current-carrying conductor. Two flat magnets are fixed to a steel yoke that rests on a top-pan balance of resolution 0.01 g. Seen from above, the north pole is on the student's left and the south pole on the student's right. A stiff copper rod is clamped horizontally so that it runs directly away from the student through the gap between the poles, without touching the magnets. The length of rod in the field is equal to the length of the pole faces, (5.0 ± 0.1) cm.
When there is a current I in the rod, the balance reading decreases by Δm. The uncertainty in each value of Δm is ±0.02 g.
I / A
Δm / g
0.50
0.20
1.00
0.37
1.50
0.58
2.00
0.76
2.50
0.94
3.00
1.15
Graph drawn to scale
(a)
Show that the force on the rod when I = 3.00 A is about 1.1 × 10−2 N.
(1)
(b)
Draw the line of best fit for the data.
(1)
(c)
Determine, using the gradient of your line, the magnetic flux density B between the poles.
(2)
(d)
Deduce the direction of the current in the rod.
(2)
(e)
A calibrated magnetic field sensor placed between the poles reads 0.075 T. Ignoring the uncertainty in the gradient, determine the absolute uncertainty in your value of B and deduce whether the two values agree.
(2)
(f)
Suggest one change to the apparatus that would reduce the percentage uncertainty in Δm for the smallest currents.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = Δm g = 1.15 × 10−3 × 9.81 = 1.13 × 10−2 N
✓ 1
Must see full substitution with Δm in kg OR answer to at least 3 s.f.
Part (b)
Single straight line through all the error bars, passing through or very close to the origin
✓ 1
Do not accept a line joining the points.
Part (c)
gradient = 0.379 g A−1 (= 3.8 × 10−4 kg A−1)
✓ 1
Accept 0.37–0.39 g A−1. Gradient from a large triangle.
B = gradient × g/L = 3.79 × 10−4 × 9.81/0.050 = 0.0744 T
✓ 1
Allow ECF from the gradient. Accept 0.073–0.077 T. Award [0] for MP2 if L is not in metres.
Part (d)
The reading decreases, so the force on the magnets is upwards; by Newton's third law the force on the rod is downwards
✓ 1
The field is from left to right (N to S), so by the left-hand rule / F = IL × B the current is directed away from the student
✓ 1
Award MP2 only if a downward force on the rod is used. OWTTE
Part (e)
ΔB/B = ΔL/L = 0.1/5.0 = 2 %, so ΔB = ±0.0015 T
✓ 1
Allow ECF from (c). Accept ±0.001 T or ±0.002 T.
B = (0.0744 ± 0.0015) T, i.e. 0.0730–0.0759 T, which includes 0.075 T, so the values agree
✓ 1
The conclusion must be consistent with the candidate's range.
Part (f)
Use magnets with longer pole faces / several magnets in a row (larger L) OR stronger magnets, so that Δm is larger compared with the 0.01 g resolution
✓ 1
Accept a balance of higher resolution (e.g. 0.001 g). Do not accept “repeat the readings” alone.
Answers: (a) 1.13 × 10−2 N · (c) B = 0.0744 T · (e) B = (0.0744 ± 0.0015) T (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ Command term: Determine
9D-1B-06
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 1BMedium10 marks
Data-based question10 steps to full marksDetermine
A horizontal copper rod of mass (4.80 ± 0.01) g hangs from two light vertical conducting wires, forming a swing. A pair of large coils produces a uniform vertical magnetic field of flux density B in the region of the rod. The length of the rod between the two suspension wires is (0.120 ± 0.001) m.
When there is a current I in the rod it swings sideways and comes to rest with the suspension wires at an angle φ to the vertical, measured with a protractor to ±1°. In equilibrium, tan φ = BIL/(mg), where m is the mass and L the length of the rod. The graph shows tan φ against I with error bars.
I / A
φ / °
tan φ
1.0
5
0.087
2.0
10
0.176
3.0
15
0.268
4.0
20
0.364
5.0
24
0.445
6.0
28
0.532
Graph drawn to scale
(a)
Outline why the magnetic forces on the two suspension wires can be ignored.
(1)
(b)
Show that the absolute uncertainty in tan φ for I = 3.0 A is about 0.02.
(1)
(c)
Draw the line of best fit and the lines of maximum and minimum gradient for the data.
(2)
(d)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(e)
Determine B and its absolute uncertainty.
(2)
(f)
A calibrated magnetic field sensor gives B = 0.034 T. Comment on the student's result.
(1)
(g)
Suggest how the measurement of the deflection could be changed to reduce the uncertainty in the result.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The currents in the two suspension wires are in opposite directions, so the magnetic forces on them are equal and opposite (along the line of the rod) and cancel OR for small deflections the wires are almost parallel to the vertical field, so the force on them is very small
✓ 1
OWTTE
Part (b)
tan 16° − tan 15° = 0.2867 − 0.2679 = 0.0188
✓ 1
Accept ½(tan 16° − tan 14°) = 0.0187. Must see the calculation OR an answer to at least 2 s.f. (0.019).
Part (c)
Line of best fit passing through all the error bars, with points on both sides
✓ 1
Steepest and least steep lines that pass through all (or nearly all) the error bars
Accept ±0.005 to ±0.010 A−1. Allow ECF from the candidate's lines.
Part (e)
B = gradient × mg/L = 0.0893 × 4.80 × 10−3 × 9.81/0.120 = 3.50 × 10−2 T
✓ 1
Allow ECF from (d).
fractional uncertainty = 9.0 % + 0.2 % + 0.8 % ≈ 10 %, so B = (3.5 ± 0.4) × 10−2 T
✓ 1
MP2 is for adding fractional uncertainties AND matching the precision of value and uncertainty. Accept ignoring the (negligible) mass and length terms.
Part (f)
0.034 T lies within the range 3.1–3.9 × 10−2 T, so the result is consistent with the sensor value
✓ 1
Allow ECF from (e).
Part (g)
Measure the horizontal displacement of the rod with a ruler (or from a photograph against a grid) and calculate the angle from it, because ±1° is a large fractional uncertainty for small angles
✓ 1
Accept: use larger currents/larger angles so the fractional uncertainty in φ is smaller. Do not accept “use a more accurate protractor” without detail.
Answers: (b) 0.0188 · (d) (0.089 ± 0.008) A−1 · (e) B = (3.5 ± 0.4) × 10−2 T (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ Command term: Determine
10D-1B-07
Charge-to-mass ratio·D.3 Motion in electromagnetic fields
Paper 1BHard11 marks
Data-based question11 steps to full marksDetermine
A student uses a fine-beam tube to determine the charge-to-mass ratio e/m of the electron. Electrons are accelerated from rest through a potential difference V and then move at right angles to a uniform magnetic field produced by a pair of coils. The flux density is (8.0 ± 0.2) × 10−4 T and is kept constant. The electrons follow a circular path, made visible by the gas in the tube.
The student holds a ruler against the front of the glass bulb and, looking from the front, reads the diameter D of the circle to ±0.2 cm. The table gives r = D/2 and r²; the graph shows r² against V with error bars.
V / V
D / cm
r / cm
r² / 10−3 m²
150
10.1
5.05
2.55
180
11.2
5.60
3.14
210
12.0
6.00
3.60
240
12.9
6.45
4.16
270
13.6
6.80
4.62
300
14.4
7.20
5.18
Graph drawn to scale
(a)
Deduce that r² = (2m/(eB²)) V.
(2)
(b)
Show that the absolute uncertainty in r² for V = 300 V is about 1.4 × 10−4 m².
(1)
(c)
Draw the line of best fit for the data.
(1)
(d)
Determine e/m using the gradient of your line.
(2)
(e)
The uncertainty in the gradient is ±4 %. Determine the absolute uncertainty in your value of e/m and compare your result with the accepted value of 1.76 × 1011 C kg−1.
(3)
(f)
The ruler is about 1 cm in front of the plane of the beam. Explain whether this makes the value of e/m obtained too large or too small.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The magnetic force provides the centripetal force: evB = mv²/r, so r = mv/(eB)
✓ 1
eV = ½mv², so v² = 2eV/m; substituting gives r² = m²v²/(e²B²) = 2mV/(eB²)
✓ 1
Must see both equations combined; the answer is given.
Part (b)
Δr/r = 0.1/7.20 and Δ(r²) = 2 × (0.1/7.20) × 5.184 × 10−3 = 1.44 × 10−4 m²
✓ 1
Must see the doubling of the fractional uncertainty in r (Δr = 0.1 cm) OR an answer to 3 s.f.
Part (c)
Single straight line passing through all the error bars and (close to) the origin
Δ(e/m) = 0.09 × 1.80 × 1011 ≈ 0.2 × 1011 C kg−1, so e/m = (1.8 ± 0.2) × 1011 C kg−1
✓ 1
MP2 is for matching the precision of value and uncertainty. Allow ECF from (d).
The accepted value lies within the range 1.6–2.0 × 1011 C kg−1, so the result is consistent with it
✓ 1
Allow ECF.
Part (f)
Seen from the front, the lines of sight to the two sides of the circle converge, so they cross the ruler (nearer the eye) closer together: every diameter/radius is read too small
✓ 1
OWTTE
e/m = 2V/(B²r²) / the gradient is too small, so e/m is too large
✓ 1
Award MP2 only if the direction of the error in r is correct. Consistent with the student's high value.
Answers: (b) 1.44 × 10−4 m² · (d) e/m = 1.80 × 1011 C kg−1 · (e) (1.8 ± 0.2) × 1011 C kg−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field (the determination of the charge-to-mass ratio for a charged particle by investigating its path in a uniform magnetic field) Command term: Determine
11D-1B-08
Parallel current-carrying wires·D.3 Motion in electromagnetic fields
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
A student uses a current balance to determine the magnetic constant μ0. A straight horizontal brass rod rests on an insulating block on a top-pan balance of resolution 0.001 g. An identical rod is clamped directly above it, parallel to it. The two rods are connected in series so that the same current I flows through each, in opposite directions. The length of each rod is (0.300 ± 0.002) m and the separation of their centres is (5.0 ± 0.2) mm.
For each current the student records the increase Δm in the balance reading with the current in one direction, then reverses the current and records it again. The table gives the mean of the two values, with an uncertainty of ±0.002 g. The graph shows Δm against I² with error bars.
I / A
I² / A²
Δm / g
4.0
16
0.019
5.0
25
0.032
6.0
36
0.043
7.0
49
0.061
8.0
64
0.077
9.0
81
0.100
10.0
100
0.121
Graph drawn to scale
(a)
Explain why the balance reading increases when there is a current in the rods.
(2)
(b)
Draw the line of best fit and the lines of maximum and minimum gradient for the data.
(2)
(c)
Determine the gradient of the line of best fit and its absolute uncertainty.
(2)
(d)
Determine μ0 and its absolute uncertainty.
(3)
(e)
The accepted value of μ0 is 1.26 × 10−6 T m A−1. Comment on the student's result.
(1)
(f)
The rods are also in the Earth's magnetic field. Explain how reversing the current and averaging the readings removes the effect of the Earth's field on the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The currents in the rods are in opposite directions, so the rods repel each other
✓ 1
The force on the lower rod is downwards, so the force on the balance (and its reading) increases
✓ 1
Award MP2 only if repulsion is stated or implied.
Part (b)
Line of best fit through all the error bars, passing through or close to the origin
✓ 1
Steepest and least steep lines that pass through all the error bars
✓ 1
Part (c)
gradient = 1.21 × 10−3 g A−2
✓ 1
Accept 1.18 × 10−3 to 1.25 × 10−3 g A−2. Unit required.
Accept ±1 % to ±5 %. Allow ECF from the candidate's lines.
Part (d)
Force = Δm g and F = μ0I²L/(2πr), so μ0 = 2πr × gradient × g/L (gradient in kg A−2)
✓ 1
μ0 = 2π × 5.0 × 10−3 × 1.21 × 10−6 × 9.81/0.300 = 1.24 × 10−6 T m A−1
✓ 1
Allow ECF from (c).
percentage uncertainty = 2.5 % + 4 % + 0.7 % ≈ 7 %, so μ0 = (1.24 ± 0.09) × 10−6 T m A−1
✓ 1
MP3 is for adding the fractional uncertainties AND matching precision. Accept (1.2 ± 0.1) × 10−6.
Part (e)
1.26 × 10−6 lies within the range 1.15–1.33 × 10−6 T m A−1, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d).
Part (f)
The Earth's field exerts a force BIL on the lower rod that reverses direction when the current is reversed (it is proportional to I)
✓ 1
Reversing the current reverses both currents, so the force between the rods (proportional to I²) is unchanged; the mean of the two readings therefore contains only the force between the rods
✓ 1
OWTTE
Answers: (c) (1.21 ± 0.03) × 10−3 g A−2 · (d) μ0 = (1.24 ± 0.09) × 10−6 T m A−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/(2πr) where r is the separation between the two wires Command term: Determine
12D-1B-09
Charge in a uniform electric field·D.3 Motion in electromagnetic fields
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
In an electron deflection tube, electrons accelerated through a potential difference Va = (3.00 ± 0.05) kV enter the region between two horizontal parallel plates, travelling horizontally midway between them. The plates are (50.0 ± 0.5) mm apart and the upper plate is at a potential difference Vp = (1.20 ± 0.02) kV above the lower plate.
The path of the beam is shown on a fluorescent screen marked with a grid. A student reads the upward deflection y of the beam at horizontal distances x from the point where the beam enters the field. Each value of y has an uncertainty of ±1 mm. Theory predicts y = kx², where k = Vp/(4dVa) and d is the plate separation.
x / cm
y / mm
x² / cm²
4
3
16
5
5
25
6
7
7
10
49
8
13
64
9
16
10
20
100
Graph drawn to scale
(a)
Explain why the beam is deflected upwards, towards the upper plate.
(2)
(b)
Complete the table.
(1)
(c)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(d)
Determine the gradient of the line, in mm cm−2.
(2)
(e)
Calculate the predicted value of k, in mm cm−2, and its absolute uncertainty.
(2)
(f)
Deduce whether the data support the theory.
(1)
(g)
Suggest one change to the settings that would reduce the percentage uncertainty in the values of y.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the upper plate is at the higher potential, so the electric field between the plates is directed downwards, from the upper plate to the lower plate
✓ 1
electrons are negatively charged, so the electric force on them is opposite to the field, upwards
✓ 1
MP2 only scores if the direction of the field is correct or not stated. OWTTE
Part (b)
36 AND 81
✓ 1
Both needed; integers expected.
Part (c)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (b).
Single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (d)
Gradient calculated from a large triangle (at least half the line)
✓ 1
gradient = 0.201 mm cm−2
✓ 1
Accept 0.190–0.210 mm cm−2.
Part (e)
k = 1.20 × 103/(4 × 0.0500 × 3.00 × 103) = 2.00 m−1 = 0.200 mm cm−2
✓ 1
Unit conversion required: 1 m−1 = 0.1 mm cm−2.
percentage uncertainty = 1.7 % + 1.7 % + 1.0 % = 4.3 %, so k = (0.200 ± 0.009) mm cm−2
✓ 1
Accept ±0.008 to ±0.009 mm cm−2. Allow ECF for the value of k.
Part (f)
The measured gradient lies within the predicted range 0.191–0.209 mm cm−2 and the line passes through the origin, so the data support the theory
✓ 1
Allow ECF from (d) and (e).
Part (g)
Increase Vp (within the limit of the plates) or decrease Va, so that the deflections y are larger compared with the ±1 mm reading uncertainty
✓ 1
Do not accept “use a finer grid” as a change to the settings.
Answers: (b) 36, 81 · (d) 0.201 mm cm−2 · (e) k = (0.200 ± 0.009) mm cm−2(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field Command term: Determine
13D-2-08
Circular motion in a magnetic field·D.3 Motion in electromagnetic fields
Paper 2Hard15 marks
Short answer & extended response15 steps to full marksDetermine
Polonium-210 (21084Po) decays by emitting alpha particles, all with a kinetic energy of 5.30 MeV.
In an evacuated chamber, a narrow beam of these alpha particles moving to the right enters a large region of uniform magnetic field of flux density 1.50 T, directed into the page.
Mass of an alpha particle = 6.64 × 10−27 kg. Charge of an alpha particle = +3.20 × 10−19 C.
Diagram NOT to scale
(a)
(i)
State the nuclear equation for this decay. The nucleus formed is an isotope of lead (Pb).
(2)
(ii)
Suggest why the alpha particles must travel in a vacuum in this experiment.
(1)
(b)
(i)
Show that the speed of the alpha particles is about 1.6 × 107 m s−1.
(2)
(ii)
Draw, on the diagram, the path of the alpha particles in the magnetic field.
(2)
(iii)
Calculate the radius of the path.
(2)
(c)
Explain why the kinetic energy of the alpha particles does not change while they are in the magnetic field.
(2)
(d)
A student suggests that protons with the same kinetic energy, 5.30 MeV, could be distinguished from these alpha particles by the radius of their path in the same field. Determine whether the student is correct.
(4)
Show mark scheme
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Notes
Part (a)(i)
42α «or 42He» as a product
✓ 1
Nucleon and proton numbers both needed.
20682Pb as the other product
✓ 1
Nucleon and proton numbers both needed.
Part (a)(ii)
alpha particles are strongly ionizing, so in air they would lose their energy within a few centimetres «and not complete the path»
✓ 1
OWTTE
Part (b)(i)
Ek = 5.30 × 106 × 1.60 × 10−19 = 8.48 × 10−13 J
✓ 1
v = √(2 × 8.48 × 10−13/6.64 × 10−27) = 1.60 × 107 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
path curves towards the top of the page as soon as the particles enter the field
✓ 1
Fleming's left-hand rule: field into the page, positive particles moving to the right, force up the page.
path is an arc of a circle «constant curvature»: a semicircle leaving the field through its left-hand edge, moving to the left
Award [2] for CNA. Accept 0.220–0.222 m. Allow ECF from (b)(i).
Part (c)
the magnetic force is always perpendicular to the velocity
✓ 1
so the force does no work «W = Fs cos 90° = 0» and the kinetic energy stays constant
✓ 1
Part (d)
r = mv/qB = √(2mEk)/(qB)
✓ 1
ALTERNATIVE: r ∝ √m/q; for an alpha particle √(4mp)/(2e) = √mp/e, the same as for a proton.
for a proton r = √(2 × 1.67 × 10−27 × 8.48 × 10−13)/(1.60 × 10−19 × 1.50) = 0.222 m
✓ 1
the radii differ by only about 0.3 % «effectively equal»
✓ 1
both particles are positive, so they also curve the same way: they cannot be told apart by their paths and the student is not correct
✓ 1
MP4 only scores if MP3 scores or the proportionality argument is complete.
Answers: (b)(i) v = 1.60 × 107 m s−1 · (b)(iii) r = 0.221 m · (d) rp = 0.222 m (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; the kinetic energy of a charged particle stays constant in a magnetic field (guidance) (linked to E.3 — alpha decay equations and ionizing ability) Command term: Determine
14D-2-09
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDetermine
A model electromagnetic launcher has two horizontal parallel metal rails 0.080 m apart. A metal rod rests across the rails and is fixed to a light trolley. The total mass of the rod and trolley is 0.150 kg. A uniform vertical magnetic field of flux density 0.25 T acts over the whole length of the rails.
A supply drives a constant current of 12 A through the rails and the rod. The diagram shows the arrangement seen from above.
Diagram NOT to scale
(a)
(i)
Draw an arrow on the diagram to show the direction of the magnetic force on the rod.
(1)
(ii)
Calculate the magnitude of the magnetic force on the rod.
(1)
(b)
A constant frictional force of 0.030 N opposes the motion. The rod starts from rest and travels 1.20 m along the rails.
(i)
Calculate the acceleration of the rod and trolley.
(2)
(ii)
Show that the speed of the rod at the end of the rails is about 1.8 m s−1.
(1)
(c)
The potential difference across the terminals of the supply is 3.0 V throughout the launch.
(i)
Determine the efficiency with which energy from the supply is transferred to kinetic energy of the rod and trolley.
(4)
(ii)
Suggest what happens to most of the energy from the supply.
(1)
Show mark scheme
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Notes
Part (a)(i)
arrow along the rails, to the right «away from the supply»
✓ 1
Fleming's left-hand rule: field upwards «out of the page», current up the page, force to the right.
Part (a)(ii)
F = BIL = 0.25 × 12 × 0.080 = 0.24 N
✓ 1
Part (b)(i)
resultant force = 0.24 − 0.030 = 0.21 N
✓ 1
Allow ECF from (a)(ii).
a = 0.21/0.150 = 1.4 m s−2
✓ 1
Award [1 max] for 1.6 m s−2 «friction ignored».
Part (b)(ii)
v = √(2 × 1.4 × 1.20) = 1.83 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f. ALTERNATIVE: 0.21 × 1.20 = ½ × 0.150 × v².
Part (c)(i)
time of launch t = v/a = 1.83/1.4 = 1.31 s
✓ 1
ALTERNATIVE: t from s = ½at².
energy from the supply = VIt = 3.0 × 12 × 1.31 = 47 J
✓ 1
Ek = ½ × 0.150 × 1.83² = 0.25 J
✓ 1
Allow ECF from (b)(ii).
efficiency = 0.25/47 = 0.0053 «0.53 %»
✓ 1
Allow ECF. Accept 0.5–0.6 %.
Part (c)(ii)
it becomes internal energy «heating» of the rails, the rod and the connections because of their resistance
✓ 1
Do not accept «friction» alone «only 0.036 J».
Answers: (a)(ii) 0.24 N · (b)(i) 1.4 m s−2 · (b)(ii) 1.83 m s−1 · (c)(i) 0.53 % (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ (linked to A.2 — Newton's second law, A.3 — efficiency, and B.5 — P = IV) Command term: Determine
15D-2-10
Charge-to-mass ratio·D.3 Motion in electromagnetic fields
Paper 2Medium14 marks
Short answer & extended response14 steps to full marksDetermine
In an ion-beam laboratory, positive ions from an ion source are accelerated from rest through a potential difference of 46.7 kV. They then travel along an evacuated beam line, as shown. Two detectors, D₁ and D₂, 2.40 m apart, record each ion as it passes. The time between the two signals is 1.60 μs.
At X the ions enter a bending magnet, a region of uniform magnetic field perpendicular to the page. Inside the magnet the ions follow a quarter circle of radius 0.250 m and leave towards the target.
Diagram NOT to scale
(a)
(i)
Calculate the speed of the ions.
(1)
(ii)
Show that the charge-to-mass ratio q/m of the ions is about 2.4 × 107 C kg−1.
(2)
(b)
(i)
Draw, on the diagram, an arrow to show the direction of the magnetic force on an ion at X.
(1)
(ii)
State the direction of the magnetic field in the bending magnet.
(1)
(iii)
Explain why the path of an ion inside the magnet is an arc of a circle.
(2)
(iv)
Determine the magnetic flux density in the bending magnet.
(2)
(c)
The ion source can produce helium ions He⁺ (charge +e, mass 6.64 × 10−27 kg) and carbon ions C³⁺ (charge +3e, mass 1.99 × 10−26 kg).
(i)
Determine whether the beam could consist of He⁺ ions, of C³⁺ ions, or of either.
(2)
(ii)
Explain why no arrangement of electric and magnetic fields can separate these two kinds of ion if they enter it with the same velocity.
(2)
(iii)
Suggest a measurement that would identify the ions in the beam.
Must see full substitution OR answer to at least 3 s.f.
Part (b)(i)
arrow at X towards the top of the page «towards the centre of the arc»
✓ 1
Arrow must be perpendicular to the beam at X.
Part (b)(ii)
into the page
✓ 1
F = qv × B with q > 0, v to the right and F towards the top of the page. Allow ECF from (b)(i).
Part (b)(iii)
the magnetic force is always perpendicular to the velocity, so it does no work and the speed «and so the force qvB» is constant
✓ 1
a force of constant magnitude always perpendicular to the velocity is a centripetal force, giving uniform circular motion
✓ 1
OWTTE
Part (b)(iv)
qvB = mv²/r, so B = v/((q/m)r)
✓ 1
B = 1.50 × 106/(2.41 × 107 × 0.250) = 0.249 T
✓ 1
Accept 0.24–0.25 T. Award [2] for CNA. Allow ECF from (a).
Part (c)(i)
q/m for He⁺ = 1.60 × 10−19/6.64 × 10−27 = 2.41 × 107 C kg−1 AND q/m for C³⁺ = 4.80 × 10−19/1.99 × 10−26 = 2.41 × 107 C kg−1
✓ 1
Both values needed.
both equal the measured q/m, so the beam could be either «the data cannot decide»
✓ 1
MP2 only scores if MP1 scores. Do not accept a choice of one ion.
Part (c)(ii)
the electric force qE and the magnetic force qvB are both proportional to q, so the acceleration a = F/m depends only on q/m and the velocity
✓ 1
ions with the same q/m and the same velocity have the same acceleration at every point, so they follow identical paths
✓ 1
OWTTE
Part (c)(iii)
measure the kinetic energy of the ions, e.g. with a detector that stops them: a C³⁺ ion has three times the kinetic energy «140 keV, not 46.7 keV» of a He⁺ ion
✓ 1
Accept measuring the momentum or the mass of the ions, e.g. from a collision. Do not accept another E- or B-field measurement.
Answers: (a)(i) 1.50 × 106 m s−1 · (a)(ii) 2.41 × 107 C kg−1 · (b)(iv) 0.249 T · (c)(i) 2.41 × 107 C kg−1 for both ions (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field (the determination of the charge-to-mass ratio for a charged particle by investigating its path in a uniform magnetic field — guidance); the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; the motion of a charged particle in a uniform electric field (linked to A.1 — speed from a time of flight; A.2 — Newton's second law; A.3 — kinetic energy) Command term: Determine
16D-2-11
The velocity selector·D.3 Motion in electromagnetic fields
Paper 2Hard12 marks
Short answer & extended response12 steps to full marksDetermine
In an ion implanter used in making computer chips, singly charged boron ions, each of charge +1.60 × 10−19 C, are accelerated from rest through a potential difference of 20.0 kV. They then enter a velocity filter: two horizontal parallel plates 20.0 mm apart, with a uniform magnetic field of flux density 0.0500 T directed into the page between them.
The filter is set so that boron-11 ions, of mass 1.83 × 10−26 kg, pass straight through. Boron-10 ions have a mass of 1.66 × 10−26 kg.
Diagram NOT to scale
(a)
(i)
Draw and label, on the ion shown in the diagram, the electric force and the magnetic force on a boron-11 ion that passes straight through the filter.
(2)
(ii)
Deduce which plate is positive.
(1)
(b)
(i)
Show that the speed of a boron-11 ion entering the filter is about 5.9 × 105 m s−1.
(1)
(ii)
Calculate the potential difference between the plates.
(2)
(c)
Boron-10 ions, accelerated through the same potential difference, also enter the filter. Determine whether they pass straight through. If they do not, state the direction in which they are deflected.
(3)
(d)
A few boron-11 ions lose two electrons instead of one before they are accelerated through the same potential difference. Determine whether these ions pass straight through the filter.
(3)
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Notes
Part (a)(i)
magnetic force towards the top of the page AND electric force towards the bottom of the page, both labelled
✓ 1
Fleming's left-hand rule: positive ions moving to the right, field into the page, magnetic force up the page.
arrows of equal length
✓ 1
Part (a)(ii)
the upper plate, because the electric force on a positive ion must point down, so the field points from the upper plate to the lower plate
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
qE = qvB, so E = vB = 5.91 × 105 × 0.0500 = 2.96 × 104 V m−1
✓ 1
V = Ed = 2.96 × 104 × 0.0200 = 591 V
✓ 1
Award [2] for CNA. Accept 590–592 V.
Part (c)
boron-10 ions gain the same kinetic energy but have less mass, so they are faster: v = 5.91 × 105 × √(1.83/1.66) = 6.21 × 105 m s−1
✓ 1
the magnetic force qvB increases while the electric force qE is unchanged
✓ 1
so they do not pass straight through; they are deflected towards the top of the page «towards the positive plate»
✓ 1
MP3 only scores if MP2 scores.
Part (d)
the kinetic energy qV doubles, so v = √2 × 5.91 × 105 = 8.36 × 105 m s−1
✓ 1
doubling q doubles both the electric and the magnetic force, so whether they balance depends only on whether v = E/B
✓ 1
v > E/B, so the magnetic force is larger and the ions do not pass straight through «they are deflected towards the top of the page»
✓ 1
MP3 only scores if a larger speed is found or argued.
Answers: (b)(i) 5.91 × 105 m s−1 · (b)(ii) 591 V · (c) 6.21 × 105 m s−1 · (d) 8.36 × 105 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ Command term: Determine
17D-2-12
Parallel current-carrying wires·D.3 Motion in electromagnetic fields
Paper 2Easy10 marks
Short answer & extended response10 steps to full marksCalculate
In an aluminium smelter, very large direct currents are carried by long, straight, parallel metal bars called bus bars. Two bus bars P and Q are 0.12 m apart. P carries a current of 3.0 kA into the page and Q carries a current of 3.0 kA out of the page, as shown in the cross-section.
Diagram NOT to scale
(a)
(i)
Draw, on the diagram, three magnetic field lines around P due to the current in P only. Show the direction of the field.
(2)
(ii)
State the direction of the magnetic field of P at the position of Q.
(1)
(iii)
Hence, deduce whether the bus bars attract or repel each other.
(2)
(b)
The bus bars are clamped to supports every 2.5 m. Calculate the magnetic force on a 2.5 m section of Q.
(3)
(c)
During a fault, the current in both bus bars rises briefly to 40 kA.
(i)
Calculate the factor by which the force per unit length increases.
(1)
(ii)
Suggest why the clamps must be designed for this fault and not only for the normal current.
(1)
Show mark scheme
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Notes
Part (a)(i)
concentric circles centred on P, further apart with increasing distance from P
✓ 1
At least three lines.
arrows showing a clockwise direction
✓ 1
Right-hand grip rule: thumb into the page, fingers curl clockwise.
Part (a)(ii)
towards the bottom of the page «downwards»
✓ 1
Allow ECF from (a)(i).
Part (a)(iii)
the force on Q is to the right, away from P «Fleming's left-hand rule: current out of the page, field down the page»
✓ 1
Allow ECF from (a)(ii).
so the bus bars repel
✓ 1
Accept a conclusion based on «currents in opposite directions repel».
the force becomes about 180 times larger «about 6.7 kN on each section», so clamps designed only for the normal force could break and the bars could be pushed apart
✓ 1
OWTTE
Answers: (b) F/L = 15 N m−1; F = 38 N · (c)(i) 178 (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/2πr where r is the separation between the two wires; D.2 — magnetic field lines; the determination of the direction of the magnetic field based on the current direction in a current-carrying straight wire (guidance) Command term: Calculate
18D-2-13
Charge in a uniform electric field·D.3 Motion in electromagnetic fields
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
A space probe is driven by an ion engine. Xenon ions, each of mass 2.18 × 10−25 kg and charge +1.60 × 10−19 C, start almost at rest at grid P. A uniform electric field accelerates them to grid Q, which is 1.20 mm from P. The potential difference between the grids is 1100 V.
The ions leave the engine as a beam. The current in the beam is 1.40 A.
Diagram NOT to scale
(a)
(i)
Show that the speed of an ion at Q is about 4.0 × 104 m s−1.
(1)
(ii)
Determine the time an ion takes to travel from P to Q.
(2)
(b)
(i)
Show that about 9 × 1018 ions leave the engine each second.
(1)
(ii)
Determine the force exerted on the probe by the engine.
(3)
(iii)
The probe has a mass of 650 kg. Determine whether the engine could increase the speed of the probe by 1.0 km s−1 in 30 days. Assume that the mass of the probe does not change.
(3)
(c)
Electrons are injected into the ion beam after it passes grid Q. Explain why this is necessary.
(3)
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Notes
Part (a)(i)
Ek = qV = 1.60 × 10−19 × 1100 = 1.76 × 10−16 J, so v = √(2 × 1.76 × 10−16/2.18 × 10−25) = 4.02 × 104 m s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
the field is uniform, so the acceleration is uniform and the mean speed is v/2 «= 2.0 × 104 m s−1»
✓ 1
ALTERNATIVE: E = V/d = 9.2 × 105 V m−1, a = qE/m = 6.7 × 1011 m s−2, then t = v/a.
t = 1.20 × 10−3/2.0 × 104 = 6.0 × 10−8 s
✓ 1
Award [2] for CNA. Award [1 max] for 3.0 × 10−8 s «t = d/v».
Part (b)(i)
n = I/e = 1.40/1.60 × 10−19 = 8.75 × 1018 s−1
✓ 1
Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
momentum of one ion = mv = 2.18 × 10−25 × 4.02 × 104 = 8.76 × 10−21 kg m s−1
✓ 1
force = rate of change of momentum of the ions = n × mv «by Newton's third law, equal to the force on the probe»
✓ 1
F = 8.75 × 1018 × 8.76 × 10−21 = 0.077 N
✓ 1
Allow ECF. Accept 0.076–0.077 N.
Part (b)(iii)
a = F/m = 0.077/650 = 1.18 × 10−4 m s−2
✓ 1
Allow ECF from (b)(ii).
in 30 days Δv = 1.18 × 10−4 × 30 × 86 400 = 3.1 × 102 m s−1
✓ 1
ALTERNATIVE: time needed = 1000/a = 8.5 × 106 s ≈ 98 days.
this is less than 1.0 km s−1, so it could not
✓ 1
The conclusion must be consistent with the candidate's value.
Part (c)
positive charge continually leaves the probe in the beam
✓ 1
by conservation of charge the probe would build up an equal negative charge
✓ 1
the negatively charged probe would attract the positive ions back, reducing the force «the injected electrons neutralize the beam, so the probe stays uncharged»
✓ 1
OWTTE
Answers: (a)(i) 4.02 × 104 m s−1 · (a)(ii) 6.0 × 10−8 s · (b)(ii) 0.077 N · (b)(iii) 3.1 × 102 m s−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the conservation of electric charge (linked to A.2 — F = Δp/Δt, and B.5 — I = Δq/Δt) Command term: Determine
19D-1A-34
Charge in a uniform electric field·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark3 steps to full marksDetermine
An electron enters the uniform electric field between two parallel plates, moving parallel to the plates. It leaves the plates with a deflection y.
The potential difference between the plates is doubled and the entry speed of the electron is doubled. The electron still leaves the plates. What is (new deflection)/y?
Diagram NOT accurately drawnShow mark scheme
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Notes
Step 1The time between the plates is t = L/v: doubling v halves t.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration a = eV/md doubles when V doubles.
—
Step 3y = ½at², so the factor is 2 × (½)² = 0.5.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis allows for the shorter time but not for the doubled p.d.
BCorrect: y ∝ V/v², so the factor is 2/2² = 0.5.
CThis takes y ∝ V/v, so that the two changes cancel; the time enters squared.
DThis allows for the doubled p.d. but not for the shorter time between the plates.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field Command term: Determine
20D-1A-39
Circular motion in a magnetic field·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDetermine
A proton and an electron travel at the same speed at right angles to the same uniform magnetic field.
What is (radius of the path of the proton)/(radius of the path of the electron)?
Show mark scheme
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Notes
Step 1qvB = mv²/r gives r = mv/qB; v, B and the magnitude of q are the same, so r ∝ m.
—
All 2 steps must be completed — there is no mark for a part-answer.
AThe radius depends on the mass: at the same speed the heavier particle is harder to deflect.
BThis is √(mp/me), which would apply if the particles had the same kinetic energy, not the same speed.
CCorrect: r = mv/qB, so the ratio is mp/me = 1.8 × 103.
DThis squares the mass ratio.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field Command term: Determine
21D-1A-40
The velocity selector·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark2 steps to full marksCalculate
In a velocity selector the electric field strength is 4.5 × 104 V m−1 and the magnetic flux density is 0.030 T. The diagram shows the electric and magnetic forces on a negatively charged particle moving through the selector.
What speed must the particle have to pass through undeflected?
Diagram NOT accurately drawnShow mark scheme
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Notes
Step 1Undeflected: qE = qvB, so the charge cancels and v = E/B.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2v = 4.5 × 104/0.030 = 1.5 × 106 m s−1.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: v = E/B = 4.5 × 104/0.030 = 1.5 × 106 m s−1.
BThis multiplies E by B instead of dividing.
CThis is B/E, the inverted ratio.
DThe charge cancels from qE = qvB, which is what makes a velocity selector useful.
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields Command term: Calculate
22D-1A-41
Paths in electric and magnetic fields·D.3 Motion in electromagnetic fields
Paper 1AEasy1 mark
Multiple choice · 1 mark1 step to full marksIdentify
A positive ion enters a region of uniform field moving at right angles to the field lines. In one experiment the field is electric; in another it is magnetic.
Which row describes the path of the ion while it is in each field?
Uniform electric fieldUniform magnetic field
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Step 1In the electric field the force qE is constant in size and direction (like gravity on a projectile): a parabola. In the magnetic field the force qvB is always perpendicular to the velocity: a circular arc.
✓ 1
Answer D
Answer: D · 1 stage of work, one mark
Every option, and why
AThese are the wrong way round.
BThe magnetic force changes direction as the velocity changes, so it cannot give a parabola.
CThe electric force acts on a charge whatever its direction of motion; it is the magnetic force that is zero for motion along the field.
DCorrect: a constant force gives a parabola; a force always perpendicular to the velocity gives a circular arc.
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; the motion of a charged particle in a uniform magnetic field Command term: Identify
23D-1A-42
Parallel current-carrying wires·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
Two long parallel wires X and Y carry currents I and 3I respectively.
What is (force per unit length on Y)/(force per unit length on X)?
Show mark scheme
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Notes
Step 1F/L = μ0I1I2/2πr contains the product of both currents, so it is the same for both wires.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2The forces are a Newton's third law pair: equal in magnitude, so the ratio is 1.
✓ 1
Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis counts the factor 3 twice, once for the field and once for the current.
BThis uses Y's current 3I but assumes the same field at both wires; the field at Y comes from X's smaller current.
CCorrect: both forces equal μ0(I)(3I)/2πr per unit length — a Newton's third law pair.
DThis takes each force as proportional only to the current in the other wire, forgetting the wire's own current.
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/2πr where r is the separation between the two wires Command term: Deduce
24D-1A-43
Direction of the magnetic force·D.3 Motion in electromagnetic fields
Paper 1AMedium1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
An electron moves from left to right across the page. A uniform magnetic field is directed into the page.
What is the direction of the magnetic force on the electron as it enters the field?
Diagram NOT accurately drawnShow mark scheme
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Notes
Step 1The conventional current of a moving electron is opposite to its velocity: from right to left.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Fleming's left-hand rule: first finger into the page, second finger to the left, so the thumb points towards the bottom of the page.
✓ 1
Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: with the conventional current from right to left and the field into the page, the force is towards the bottom of the page.
BThis is the direction for a positive charge; the force on an electron is opposite.
CThe force is perpendicular to the field, so it cannot point along it.
DThe force is perpendicular to the velocity, so it cannot act along the direction of travel.
Syllabus understandingD.3 — the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θCommand term: Deduce
25D-1A-44
Circular motion in a magnetic field·D.3 Motion in electromagnetic fields
Paper 1AHard1 mark
Multiple choice · 1 mark2 steps to full marksDeduce
The diagram shows a positive ion moving in a circle in a uniform magnetic field directed into the page.
Which row describes the kinetic energy and the momentum of the ion as it moves round the circle?
Diagram NOT accurately drawn
Kinetic energyMomentum
Show mark scheme
Marking point
Mark
Notes
Step 1The magnetic force is always perpendicular to the velocity, so it does no work: the speed and the kinetic energy stay constant.
—
All 2 steps must be completed — there is no mark for a part-answer.
Step 2Momentum is a vector; its magnitude mv is constant but its direction changes continuously.
✓ 1
Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AMomentum is a vector. The force changes the direction of the velocity, so the momentum changes.
BAt constant speed the magnitude of the momentum, mv, is constant too; only its direction changes.
CThe magnetic force is perpendicular to the velocity, so it does no work and cannot increase the kinetic energy.
DCorrect: no work is done, so the kinetic energy is constant; the momentum has constant magnitude but changing direction.
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the kinetic energy of a charged particle stays constant in a magnetic field (guidance) Command term: Deduce
26D-1B-14
The velocity selector·D.3 Motion in electromagnetic fields
Paper 1BHard12 marks
Data-based question12 steps to full marksDetermine
In a deflection tube, electrons accelerated from rest through a potential difference Va travel from left to right across the page, midway between two horizontal parallel plates (50.0 ± 0.5) mm apart. The upper plate is positive. A pair of coils produces a uniform magnetic field of flux density (1.20 ± 0.02) mT, perpendicular to the page, in the region between the plates.
For each value of Va (uncertainty ±0.05 kV), the student adjusts the potential difference Vp between the plates until the beam passes through undeflected. Judging this gives an uncertainty in Vp of ±20 V. The student tests whether Vp = cVan, where c and n are constants, by plotting lg (Vp / V) against lg (Va / V). Four points have been plotted.
Va / kV
Vp / V
lg (Va / V)
lg (Vp / V)
1.50
1390
3.176
3.143
2.00
1580
3.301
3.199
2.50
1790
3.00
1940
3.477
3.288
3.50
2110
4.00
2240
3.602
3.350
Graph drawn to scale
(a)
State and explain the direction of the magnetic field.
(2)
(b)
Show that the speed of the electrons when Va = 2.00 kV is about 2.6 × 107 m s−1.
(1)
(c)
Complete the table.
(2)
(d)
Plot the two missing points on the graph and draw the line of best fit for the data.
(1)
(e)
Determine the gradient of the line and outline what it shows about the speed of the electrons.
(2)
(f)
Determine, using the data for Va = 2.00 kV, the charge-to-mass ratio e/m of the electron and its absolute uncertainty.
(3)
(g)
Compare your answer to (f) with the accepted value of 1.76 × 1011 C kg−1.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The electric force on the (negative) electrons is upwards, towards the positive plate, so the magnetic force must be downwards
✓ 1
Into the page
✓ 1
Award MP2 only with a correct reason (left-hand rule with conventional current to the left, or F = qv × B with q negative).
Part (b)
v = E/B = Vp/(dB) = 1580/(0.0500 × 1.20 × 10−3) = 2.63 × 107 m s−1
✓ 1
Must see eE = evB (or v = E/B) with full substitution OR an answer to at least 3 s.f.
Part (c)
lg (Va / V) = 3.398 AND 3.544
✓ 1
Accept 3 or 4 d.p.
lg (Vp / V) = 3.253 AND 3.324
✓ 1
Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (d)
Both points plotted correctly AND a single straight line of best fit
✓ 1
Allow ECF from (c).
Part (e)
gradient = 0.49
✓ 1
Accept 0.47–0.53. No unit.
v = Vp/(dB) ∝ Vp, so v ∝ Va0.5: v² ∝ Va, consistent with ½mv² = eVa
MP3 is for matching the precision of value and uncertainty. Accept (1.73 ± 0.18) × 1011 C kg−1.
Part (g)
The accepted value lies within the range 1.55–1.91 × 1011 C kg−1, so the result agrees with it
✓ 1
Allow ECF from (f).
Answers: (b) 2.63 × 107 m s−1 · (c) 3.398, 3.253; 3.544, 3.324 · (e) 0.49 · (f) (1.7 ± 0.2) × 1011 C kg−1(the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in perpendicularly orientated uniform electric and magnetic fields Command term: Determine
27D-1B-15
Circular motion in a magnetic field·D.3 Motion in electromagnetic fields
Paper 1BMedium11 marks
Data-based question11 steps to full marksDetermine
A student uses a computer simulation of a mass spectrometer. Singly charged positive ions, initially at rest, are accelerated through a potential difference of 2.00 kV and then enter a uniform magnetic field of flux density 0.250 T at right angles to the field. Each ion travels through a semicircle and strikes a detector. The student measures the diameter D of each semicircle with the simulation's on-screen ruler, to ±1 mm.
The masses m of the ions, in unified atomic mass units (u), are taken from a published table of isotope masses. The student plots lg (D / mm) against lg (m / u); five points have been plotted.
Ion
m / u
D / mm
lg (m / u)
lg (D / mm)
lithium-7
7.016
136
0.846
2.134
carbon-12
12.000
179
1.079
2.253
oxygen-16
15.995
207
neon-20
19.992
230
1.301
2.362
argon-40
39.962
326
1.602
2.513
iron-56
55.935
385
copper-63
62.930
409
1.799
2.612
Graph drawn to scale
(a)
Complete the table.
(2)
(b)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(c)
Determine the gradient of the line.
(2)
(d)
Explain, by considering the motion of an ion, why theory predicts this value of the gradient.
(2)
(e)
An unknown singly charged ion gives D = (268 ± 1) mm under the same conditions. Determine its mass, in u, with its absolute uncertainty, and deduce whether the ion is aluminium-27 (26.982 u) or silicon-28 (27.977 u).
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
lg (m / u) = 1.204 AND 1.748
✓ 1
Accept 3 or 4 d.p.
lg (D / mm) = 2.316 AND 2.585
✓ 1
Accept 3 or 4 d.p.; do not accept 2 d.p.
Part (b)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (a).
Single straight line of best fit through (or very close to) all seven points
✓ 1
Part (c)
Gradient calculated from a large triangle (at least half the line)
✓ 1
gradient = 0.50
✓ 1
Accept 0.48–0.52. No unit.
Part (d)
qvB = mv²/r gives r = mv/(qB) AND qV = ½mv² gives v = √(2qV/m)
✓ 1
so r = √(2mV/q)/B ∝ m½ (same q, V and B): D ∝ m0.5, so the gradient is 0.5
✓ 1
OWTTE
Part (e)
m = qB²r²/(2V) = 1.60 × 10−19 × 0.250² × 0.134²/(2 × 2000) = 4.49 × 10−26 kg = 27.0 u
✓ 1
Accept the ratio method m = 12.000 × (268/179)² = 26.9 u, or a reading from the graph (lg D = 2.428).
Δm/m = 2 × 1/268 = 0.75 %, so m = (27.0 ± 0.2) u
✓ 1
The fractional uncertainty in D must be doubled. Allow ECF.
The range 26.8–27.2 u includes 26.982 u but not 27.977 u, so the ion is aluminium-27
✓ 1
Award MP3 only if the decision follows from a stated range.
Answers: (a) 1.204, 2.316; 1.748, 2.585 · (c) 0.50 · (e) m = (27.0 ± 0.2) u — aluminium-27 (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ Command term: Determine
28D-1B-17
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 1BEasy9 marks
Data-based question9 steps to full marksDetermine
A straight horizontal wire of length (40.0 ± 0.5) mm lies between the wide, flat pole pieces of an electromagnet, where the magnetic field is uniform and horizontal. The wire is fixed to a holder on a force sensor and can be turned in the horizontal plane so that it makes an angle θ with the field; the leads to the wire are twisted together outside the field.
The current in the wire is kept at (2.00 ± 0.01) A. The sensor measures the vertical force F on the wire with an uncertainty of ±0.2 mN.
θ / °
sin θ
F / mN
0
0.000
0.1
15
0.259
4.2
30
7.9
45
0.707
11.4
60
0.866
13.8
75
15.6
90
1.000
15.9
Graph drawn to scale
(a)
Complete the table.
(1)
(b)
Plot the two missing points on the graph and draw the line of best fit for the data.
(2)
(c)
Determine the gradient of the line.
(2)
(d)
Calculate the magnetic flux density B between the pole pieces.
(1)
(e)
The uncertainty in the gradient is ±2 %. Calculate the percentage uncertainty in B.
(1)
(f)
The manufacturer states that B = 0.205 T for this setting. Deduce whether the student's value agrees with this.
(1)
(g)
The wire becomes warm during the experiment. Suggest how the student could make sure that this does not affect the results.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.500 AND 0.966
✓ 1
Both needed. Accept 0.5 for sin 30°.
Part (b)
Both points plotted correctly to within half a small square
✓ 1
Allow ECF from (a).
Single straight line through all the error bars, passing through or very close to the origin
✓ 1
Part (c)
Gradient calculated from a large triangle
✓ 1
gradient = 15.9 mN
✓ 1
Accept 15.4–16.4 mN. Unit required.
Part (d)
B = gradient/(IL) = 1.59 × 10−2/(2.00 × 0.0400) = 0.199 T
✓ 1
Allow ECF from (c). Award [0] if L is not in metres.
Part (e)
2 % + 0.5 % + 1.25 % = 3.8 %
✓ 1
Accept 3.8 % or 4 %.
Part (f)
ΔB = 3.8 % × 0.199 = ±0.007 T; the range 0.192–0.206 T includes 0.205 T, so the values agree
✓ 1
Allow ECF from (d) and (e).
Part (g)
Check the ammeter before each reading and adjust a variable resistor (rheostat) so that the current stays at 2.00 A (the resistance of the wire increases as it warms) OR switch off between readings so the wire stays cool
✓ 1
OWTTE
Answers: (a) 0.500, 0.966 · (c) 15.9 mN · (d) B = 0.199 T · (e) 3.8 % (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ Command term: Determine
29D-2-30
Charge in a uniform electric field·D.3 Motion in electromagnetic fields
Paper 2Medium13 marks
Short answer & extended response13 steps to full marksDetermine
In an ion-energy analyser, two parallel wire-mesh grids G1 and G2 are 12.0 mm apart in a vacuum. Protons, each with a kinetic energy of 450 eV, pass through G1 moving at right angles to the grids, towards G2 (Figure 1). A potential difference V is applied between the grids so that G2 is positive relative to G1. The field between the grids is uniform.
e = 1.60 × 10−19 C; mass of a proton = 1.67 × 10−27 kg.
Figure 1Figure 2 — axes for (c)(ii)
(a)
State the direction of the electric field between the grids.
(1)
(b)
V = 300 V.
(i)
Calculate the magnitude of the deceleration of a proton between the grids.
(2)
(ii)
Determine whether the protons reach G2 and, if they do, state their kinetic energy there in eV.
(2)
(c)
V is increased to 600 V.
(i)
Determine the distance from G1 at which a proton momentarily stops.
(2)
(ii)
Sketch, on Figure 2, a graph to show how the velocity v of a proton varies with time t from the moment it passes through G1, with speed v0, until it returns to G1. Take the direction from G1 to G2 as positive.
(3)
(iii)
State the kinetic energy, in eV, of the proton when it returns to G1.
(1)
(d)
A beam contains protons with a range of kinetic energies.
Suggest how the analyser could be used to find the number of protons with kinetic energies between 400 eV and 500 eV.
(2)
Show mark scheme
Marking point
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Notes
Part (a)
from G2 towards G1 «opposite to the direction of motion of the protons»
✓ 1
Part (b)(i)
E = 300 / 0.0120 = 2.5 × 104 V m−1
✓ 1
a = eE/m = 2.4 × 1012 m s−2
✓ 1
Award [2] for CNA.
Part (b)(ii)
work done against the field across the whole gap = 300 eV, which is less than 450 eV, so the protons reach G2
✓ 1
ALTERNATIVE: stopping distance = (450/300) × 12.0 mm = 18 mm > 12.0 mm.
kinetic energy at G2 = 450 − 300 = 150 eV
✓ 1
Part (c)(i)
450 eV = e × E × s with E = 600/0.0120 = 5.0 × 104 V m−1 OR s/12.0 mm = 450/600
✓ 1
s = 9.0 mm
✓ 1
Award [2] for CNA.
Part (c)(ii)
straight line starting at v0 with a constant negative gradient
✓ 1
line continues through v = 0 into negative values
✓ 1
line ends at −v0 at twice the time at which v = 0
✓ 1
Do not award MP3 if the graph is curved or the final speed differs from v0.
Part (c)(iii)
450 eV «the field does equal positive work on the return journey»
✓ 1
Part (d)
a detector beyond G2 counts the protons that pass through; only protons with kinetic energy «in eV» greater than V reach it
✓ 1
OWTTE
measure the count «rate» at V = 400 V and at V = 500 V; the difference is the number with energies between 400 eV and 500 eV
✓ 1
Answers: (b)(i) 2.4 × 1012 m s−2 · (b)(ii) yes; 150 eV · (c)(i) 9.0 mm · (c)(iii) 450 eV (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the uniform electric field strength between parallel plates as given by E = V/d (work done in electric fields in joules and electronvolts) Command term: Determine
30D-2-31
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 2Medium10 marks
Short answer & extended response10 steps to full marksDescribe
In a moving-coil loudspeaker, a coil of 60 turns, each of radius 12 mm, sits in the narrow gap of a magnet. The magnetic field in the gap is radial, with flux density 0.80 T, so the field is perpendicular to the wire at every point. The coil is fixed to a cone; together they have a mass of 8.0 g. The suspension of the cone behaves as a spring of spring constant 1.6 × 103 N m−1.
The diagram shows a cross-section through the axis of the coil. In the coil, the current is out of the page at the top and into the page at the bottom; the field points away from the axis.
Diagram NOT accurately drawn
(a)
Annotate the diagram with arrows to show the direction of the magnetic force on the coil at the top and at the bottom of the cross-section.
(2)
(b)
Show that the total length of wire in the coil is about 4.5 m.
(1)
(c)
There is a steady current of 0.50 A in the coil.
(i)
Calculate the magnitude of the magnetic force on the coil.
(1)
(ii)
Calculate the displacement of the cone from its equilibrium position.
(1)
(d)
Show that the natural frequency of oscillation of the cone is about 71 Hz.
(1)
(e)
The coil now carries a signal current of constant amplitude whose frequency is slowly increased from 20 Hz to 200 Hz.
(i)
Describe and explain how the amplitude of vibration of the cone changes.
(3)
(ii)
Suggest why loudspeaker designers make the motion of the cone heavily damped.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
arrow on the top wire pointing to the left, along the axis «towards the cone»
✓ 1
Use of F = IL × B or Fleming's left-hand rule.
arrow on the bottom wire also pointing to the left
the amplitude increases to a maximum and then decreases
✓ 1
the maximum occurs when the signal frequency is «close to» the natural frequency of about 71 Hz
✓ 1
resonance: when the driving frequency equals the natural frequency, energy is transferred to the cone most effectively
✓ 1
OWTTE
Part (e)(ii)
so that sounds with frequencies near 71 Hz are not much louder than other frequencies OR so that the cone stops vibrating as soon as the signal stops
✓ 1
OWTTE
Answers: (b) 4.52 m · (c)(i) 1.8 N · (c)(ii) 1.1 × 10−3 m · (d) 71.2 Hz (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ. Links: A.2 — Hooke's law; C.1 — T = 2π√(m/k); C.4 — resonance, natural frequency and the effect of damping Command term: Describe
31D-2-32
Circular motion in a magnetic field·D.3 Motion in electromagnetic fields
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDetermine
Electrons are accelerated from rest through a potential difference of 2.0 kV. They then enter, at right angles to its boundary, a region of uniform magnetic field of flux density 1.5 mT directed into the page. The field region has two parallel boundaries 50 mm apart and extends far above and below the beam.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT accurately drawn
(a)
Calculate the speed of the electrons as they enter the field.
(2)
(b)
Show that the radius of the path of the electrons in the field is about 0.10 m.
(1)
(c)
Draw, on the diagram, an arrow to show the direction of the magnetic force on an electron as it enters the field, and the path of the electron in the field and after it leaves the field.
(3)
(d)
Determine the angle through which the electrons are deflected by the field.
(3)
(e)
Calculate the time that an electron spends in the field.
(2)
(f)
The accelerating potential difference is reduced to 400 V.
Determine whether the electrons now leave the field through the far boundary.
(3)
Show mark scheme
Marking point
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Notes
Part (a)
eV = ½mv², so v = √(2 × 1.60 × 10−19 × 2.0 × 103 / 9.11 × 10−31)
Must see full substitution OR answer to 3 s.f. Allow ECF from (a).
Part (c)
arrow at the entry point pointing down the page «towards the bottom of the page»
✓ 1
Direction from F = qv × B with q negative.
path in the field is a circular arc curving downwards, tangential to the initial direction
✓ 1
path after leaving the field is a straight line, tangential to the arc at the exit point
✓ 1
Do not award MP3 if the path continues to curve outside the field.
Part (d)
the centre of the circular path lies on the entry boundary, a distance r from the entry point
✓ 1
sin θ = w/r = 0.050 / 0.10
✓ 1
Allow ECF from (b).
θ = 30°
✓ 1
Accept 29°–31°. Allow ECF from (b).
Part (e)
arc length = rθ = 0.10 × 0.52 rad = 0.052 m
✓ 1
θ must be in radians. Allow ECF from (b) and (d).
t = 0.052 / 2.65 × 107 = 2.0 × 10−9 s
✓ 1
Accept 1.9–2.0 × 10−9 s. Award [2] for CNA.
Part (f)
v ∝ √V, so r ∝ √V «for the same B»
✓ 1
r = 0.10 × √(400/2000) = 0.045 m
✓ 1
Accept 0.045 m. Award MP2 for a correct radius by direct calculation.
this is less than 50 mm, so the electrons do not reach the far boundary; they turn through a semicircle and leave back through the entry boundary
✓ 1
MP3 requires a comparison with 50 mm.
Answers: (a) 2.65 × 107 m s−1 · (b) 0.101 m · (d) 30° · (e) 2.0 × 10−9 s · (f) r = 0.045 m; no (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform magnetic field; the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θCommand term: Determine
32D-2-33
Parallel current-carrying wires·D.3 Motion in electromagnetic fields
Paper 2Medium12 marks
Short answer & extended response12 steps to full marksDetermine
Three long, straight, parallel cables X, Y and Z lie in a row on a horizontal cable tray. The diagram shows a cross-section. X carries a current of 200 A and Y carries 150 A, both out of the page; Z carries 300 A into the page. The distance between X and Y is 0.040 m and between Y and Z is 0.060 m.
μ0 = 4π × 10−7 T m A−1.
Diagram NOT accurately drawn
(a)
State and explain the direction of the magnetic force on Y due to X.
(2)
(b)
(i)
Draw, on the diagram, the magnetic field line due to the current in Z that passes through Y. Show its direction.
(1)
(ii)
Hence deduce the direction of the magnetic force on Y due to Z.
(1)
(c)
Determine the magnitude and direction of the resultant magnetic force per unit length on Y.
(3)
(d)
Calculate the magnitude of the resultant magnetic force per unit length on X and state its direction.
(3)
(e)
Determine whether there would be a resultant magnetic force on Y if the current in Z were reversed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
towards X «to the left»
✓ 1
the currents in X and Y are in the same direction, and parallel currents in the same direction attract
✓ 1
MP2 only scores if MP1 scores.
Part (b)(i)
circle centred on Z passing through Y, with a clockwise arrow «up the page at Y»
✓ 1
Part (b)(ii)
to the left, away from Z «F = IL × B with the current in Y out of the page and B up the page»
✓ 1
Allow ECF from (b)(i).
Part (c)
force per unit length due to X = 2 × 10−7 × 200 × 150 / 0.040 = 0.150 N m−1
✓ 1
Accept μ0/2π = 2 × 10−7 T m A−1.
force per unit length due to Z = 2 × 10−7 × 300 × 150 / 0.060 = 0.150 N m−1
✓ 1
resultant = 0.30 N m−1 towards X «to the left»
✓ 1
Both forces act in the same direction. Allow ECF from (a) and (b)(ii).
Part (d)
due to Y: 0.150 N m−1 towards Y «attraction»
✓ 1
due to Z: 2 × 10−7 × 200 × 300 / 0.100 = 0.120 N m−1 away from Z «repulsion»
✓ 1
Distance X to Z = 0.100 m.
resultant = 0.030 N m−1 towards Y «to the right»
✓ 1
Award [3] for CNA with direction.
Part (e)
with Z reversed, the force on Y due to Z is 0.150 N m−1 towards Z «attraction», opposite to the 0.150 N m−1 towards X
✓ 1
the two forces cancel, so there would be no resultant force on Y
✓ 1
Conclusion must follow from the comparison. Allow ECF from (c).
Answers: (c) 0.30 N m−1 towards X · (d) 0.030 N m−1 towards Y · (e) no resultant force (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the force per unit length between parallel wires as given by F/L = μ0I1I2/2πr where r is the separation between the two wires; the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θ (direction of the magnetic field from the current in a straight wire; parallel currents in the same direction attract) Command term: Determine
33D-2-34
Force on a current-carrying wire·D.3 Motion in electromagnetic fields
Paper 2Easy8 marks
Short answer & extended response8 steps to full marksCalculate
A rectangular coil PQRS of one turn lies in the plane of the page between the poles of a magnet. The uniform magnetic field of flux density 0.050 T is directed from left to right, in the plane of the page. Sides PQ and RS are 0.060 m long and perpendicular to the field. Sides QR and SP are 0.040 m long and parallel to the field. The current in the coil is 2.5 A in the direction P → Q → R → S.
Diagram NOT accurately drawn
(a)
Calculate the magnitude of the magnetic force on side PQ.
(2)
(b)
Annotate the diagram, using the symbols ⊙ (out of the page) and ⊗ (into the page), to show the direction of the magnetic force on side PQ and on side RS.
(2)
(c)
Explain why there is no magnetic force on side QR.
(1)
(d)
State the resultant magnetic force on the whole coil.
(1)
(e)
The coil is turned about a vertical axis until side QR makes an angle of 30° with the field.
Calculate the magnitude of the magnetic force on side QR in this position.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
F = BIL = 0.050 × 2.5 × 0.060
✓ 1
F = 7.5 × 10−3 N
✓ 1
Award [2] for CNA.
Part (b)
⊗ «into the page» on PQ
✓ 1
⊙ «out of the page» on RS
✓ 1
Award [1] max if both directions are reversed.
Part (c)
the current in QR is parallel to the field, so θ = 0 and F = BIL sin θ = 0
✓ 1
OWTTE
Part (d)
zero «the forces on PQ and RS are equal and opposite, and there is no force on QR or SP»
✓ 1
Part (e)
F = BIL sin θ = 0.050 × 2.5 × 0.040 × sin 30°
✓ 1
F = 2.5 × 10−3 N
✓ 1
Award [2] for CNA.
Answers: (a) 7.5 × 10−3 N · (d) zero · (e) 2.5 × 10−3 N (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a current-carrying conductor in a magnetic field as given by F = BIL sin θCommand term: Calculate
34D-2-35
Charge in a uniform electric field·D.3 Motion in electromagnetic fields
Paper 2Hard14 marks
Short answer & extended response14 steps to full marksDetermine
In an electron energy analyser, two large horizontal parallel plates are 10.0 mm apart in a vacuum. The potential difference between them is 150 V and the lower plate is positive. Electrons, each with a kinetic energy of 200 eV, enter the region between the plates through a small hole H in the lower plate, moving at 40° to the plate.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg.
Diagram NOT accurately drawn
(a)
State the direction of the electric force on an electron between the plates.
(1)
(b)
Calculate the magnitude of the acceleration of an electron between the plates.
(2)
(c)
Show that the speed of the electrons as they enter through H is about 8.4 × 106 m s−1.
(1)
(d)
Determine whether the electrons reach the upper plate.
(4)
(e)
Calculate the distance from H at which the electrons return to the lower plate.
(2)
(f)
Sketch, on the diagram, the path of an electron between the plates.
(2)
(g)
A second small hole is cut in the lower plate at the distance found in (e), and a detector is placed below it.
Suggest how this arrangement selects electrons of one kinetic energy.
(2)
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Notes
Part (a)
vertically downwards, towards the «positive» lower plate
component of velocity at right angles to the plates = 8.4 × 106 × sin 40° = 5.39 × 106 m s−1
✓ 1
the perpendicular component falls to zero at the greatest height: vy² = 2ah
✓ 1
greatest height = vy²/2a = 5.5 mm
✓ 1
ALTERNATIVE: energy for the perpendicular motion = 200 sin² 40° = 82.6 eV, so height = 82.6 / 1.5 × 104 m = 5.5 mm. Allow ECF.
5.5 mm is less than 10.0 mm, so the electrons do not reach the upper plate
✓ 1
Conclusion must follow from a calculation.
Part (e)
time between the plates = 2vy/a = 4.09 × 10−9 s
✓ 1
Allow ECF from (b) and (d).
distance = v cos 40° × t = 26 mm
✓ 1
Accept 26–27 mm. Award [2] for CNA.
Part (f)
a symmetrical curve «parabola» starting at H at about 40° and returning to the lower plate
✓ 1
highest point about halfway between the plates, clearly below the upper plate
✓ 1
Allow ECF from (d).
Part (g)
for a given angle and field, the landing distance is proportional to the kinetic energy «faster electrons land further from H»
✓ 1
Accept x = 2Ek sin 2θ/eE or a qualitative argument.
so only electrons with kinetic energy 200 eV reach the second hole and the detector; changing the potential difference selects a different energy
✓ 1
OWTTE
Answers: (b) 2.63 × 1015 m s−2 · (c) 8.38 × 106 m s−1 · (d) greatest height 5.5 mm; no · (e) 26 mm (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the uniform electric field strength between parallel plates as given by E = V/d (work done in electric fields in electronvolts). Link: A.1 — projectile motion launched above the horizontal Command term: Determine
35D-2-36
Force on a moving charge·D.3 Motion in electromagnetic fields
Paper 2Medium8 marks
Short answer & extended response8 steps to full marksDetermine
Close to the equator the Earth's magnetic field is horizontal and directed towards the north, with a flux density of 3.0 × 10−5 T. A cosmic-ray proton moves vertically downwards towards the surface at a speed of 2.0 × 107 m s−1. In the diagram, up the page is vertically upwards, the right of the page is north and east is out of the page.
e = 1.60 × 10−19 C; mass of a proton = 1.67 × 10−27 kg.
Diagram NOT accurately drawn
(a)
Calculate the magnitude of the magnetic force on the proton.
(1)
(b)
Determine the direction of this force.
(2)
(c)
Show that the radius of curvature of the path of the proton is about 7 km.
(1)
(d)
Calculate the angle, in radians, through which the velocity of the proton turns while it travels 1.0 km along its path. Assume that the field is uniform.
(1)
(e)
Near the Earth's magnetic poles the field is almost vertical.
Explain why protons arriving vertically there are hardly deflected by the field.
(2)
(f)
State the direction of the magnetic force on an electron moving vertically downwards at the equator.
(1)
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Notes
Part (a)
F = evB = 1.60 × 10−19 × 2.0 × 107 × 3.0 × 10−5 = 9.6 × 10−17 N
✓ 1
sin θ = 1 since v ⊥ B.
Part (b)
use of F = qv × B OR Fleming's left-hand rule with the current downwards and the field towards the north
✓ 1
towards the east «out of the page»
✓ 1
Award [1] max for “east” with no supporting reasoning.
Accept 0.14–0.15 rad. Allow ECF from (c). Do not accept an answer in degrees only.
Part (e)
the velocity is «almost» parallel to the field, so θ ≈ 0
✓ 1
F = qvB sin θ ≈ 0, so the path is hardly bent
✓ 1
Part (f)
towards the west «into the page»
✓ 1
Answers: (a) 9.6 × 10−17 N · (b) east · (c) 6.96 × 103 m · (d) 0.14 rad · (f) west (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the magnitude and direction of the force on a charge moving in a magnetic field as given by F = qvB sin θ; the motion of a charged particle in a uniform magnetic field Command term: Determine
36D-2-37
Motion in an electric field·D.3 Motion in electromagnetic fields
Paper 2Medium15 marks
Short answer & extended response15 steps to full marksDetermine
An electron is accelerated from rest through a potential difference of 2500 V. It then enters, along the centre line, the region between two horizontal parallel plates that are 40 mm long and 20 mm apart. The potential difference between the plates is 240 V and the upper plate is positive. Ignore edge effects.
e = 1.60 × 10−19 C; me = 9.11 × 10−31 kg; g = 9.81 m s−2.
Diagram NOT accurately drawn
(a)
Calculate the speed of the electron as it enters the region between the plates.
(2)
(b)
(i)
Calculate the electric field strength between the plates.
(1)
(ii)
Show that the acceleration of the electron between the plates is about 2.1 × 1015 m s−2.
(1)
(iii)
Determine the deflection of the electron, at right angles to the plates, as it leaves the region between them.
(3)
(c)
Sketch, on the diagram, the path of the electron between the plates and after it has left them.
(2)
(d)
Explain why the path of the electron between the plates is a parabola.
(2)
(e)
Determine whether the electron would still leave the region between the plates if the potential difference between the plates were increased to 2000 V.
(3)
(f)
Outline why the gravitational force on the electron can be ignored.
(1)
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Marking point
Mark
Notes
Part (a)
eV = ½mv², so v = √(2 × 1.60 × 10−19 × 2500 / 9.11 × 10−31)
✓ 1
v = 2.96 × 107 m s−1
✓ 1
Award [2] for CNA.
Part (b)(i)
E = 240 / 0.020 = 1.2 × 104 V m−1
✓ 1
Part (b)(ii)
a = eE / m = 1.60 × 10−19 × 1.2 × 104 / 9.11 × 10−31 = 2.11 × 1015 «m s−2»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(iii)
t = 0.040 / 2.96 × 107 = 1.35 × 10−9 s «horizontal velocity constant»
✓ 1
Allow ECF from (a).
y = ½at²
✓ 1
y = 1.9 × 10−3 m «1.9 mm»
✓ 1
Accept 1.9–2.0 mm. Award [3] for CNA.
Part (c)
between the plates the path curves upwards «towards the positive plate» with increasing gradient
✓ 1
after the plates the path is a straight line in the direction of motion at the exit «tangential»
✓ 1
Do not award MP2 if the path keeps curving or bends sharply at the exit.
Part (d)
there is no force parallel to the plates, so the component of velocity parallel to the plates is constant
✓ 1
the force at right angles to the plates is constant «uniform field», so the displacement at right angles is proportional to t² and hence to (distance along the plates)², as for a horizontally launched projectile
✓ 1
OWTTE
Part (e)
y ∝ potential difference between the plates «a ∝ E ∝ V and the time between the plates is unchanged»
✓ 1
ALTERNATIVE: full recalculation with E = 1.0 × 105 V m−1.
y = 1.92 × 2000/240 = 16 mm
✓ 1
Allow ECF from (b)(iii).
this is greater than 10 mm «half the separation», so the electron hits the positive plate and does not leave
✓ 1
Conclusion must follow from the comparison with 10 mm.
Part (f)
the weight «8.9 × 10−30 N» is negligible compared with the electric force «1.9 × 10−15 N»
✓ 1
Must see a comparison. Do not accept “the electron has negligible mass” alone.
Answers: (a) 2.96 × 107 m s−1 · (b)(i) 1.2 × 104 V m−1 · (b)(iii) 1.9 mm · (e) 16 mm; no (the remaining parts are explanations — see the table above)
Syllabus understandingD.3 — the motion of a charged particle in a uniform electric field; D.2 — the uniform electric field strength between parallel plates as given by E = V/d. Link: A.1 — projectile motion with a constant acceleration at right angles to the initial velocity Command term: Determine
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