D.3 Motion in electromagnetic fields flashcards: IB Physics SL and HL

All 63 D.3 Motion in electromagnetic fields cards

1. DefinitionSL & HL

   Define magnetic flux density B.

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   Magnetic flux density B is defined by the force experienced by a current-carrying conductor placed in the field: B is the force per unit current per unit length acting on a straight conductor placed perpendicular to the field, B = F/(IL). Equivalently, from F = qvB sinθ, it is the force per unit charge per unit velocity for a charge moving perpendicular to the field. Exam tip: the mark requires the words "perpendicular" (or "at 90° to the field") — a definition without the perpendicular condition is incomplete because F depends on sinθ. B is a vector; its direction is the direction a free north pole would be pushed, i.e. the direction of the field lines. Unit: tesla (T) = N A⁻¹ m⁻¹ = kg s⁻² A⁻¹.

2. DefinitionSL & HL

   Define the tesla.

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   One tesla is the magnetic flux density that produces a force of one newton on a conductor of length one metre carrying a current of one ampere placed perpendicular to the field: 1 T = 1 N A⁻¹ m⁻¹. Equivalently 1 T = 1 N s C⁻¹ m⁻¹ = 1 kg s⁻² A⁻¹, and 1 T = 1 Wb m⁻². Exam tip: state the perpendicular condition and give all three numbers (1 N, 1 m, 1 A) — dropping any of them loses the mark. The tesla is a large unit: the Earth's field is about 5 × 10⁻⁵ T, a school bar magnet about 0.01 T, an MRI magnet 1–3 T, so an answer of "B = 400 T" almost always signals an arithmetic slip.

3. DefinitionSL & HL

   State the rule for finding the direction of the magnetic force on a positive charge moving in a magnetic field.

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   Use Fleming's left-hand rule with the conventional current direction taken as the direction of motion of positive charge: first finger = Field B, second finger = Current (velocity of positive charge), thuMb = Motion (force). Equivalently F = qv × B, so F is perpendicular to BOTH v and B. Exam tip: for a NEGATIVE charge (an electron) the conventional current points opposite to the velocity, so the force reverses — reverse the second finger or reverse the final answer, but do not do both. State directions using the diagram's conventions: a dot (⊙) is a field or velocity out of the page, a cross (⊗) is into the page. Force is a vector, unit newton (N).

4. DefinitionSL & HL

   State and explain the work done by a magnetic force on a moving charged particle.

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   The work done is always zero. The magnetic force F = qv × B is at every instant perpendicular to the velocity, so the component of force along the displacement is zero and W = Fd cosθ = Fd cos90° = 0. Therefore the kinetic energy and the speed of the particle are constant; only the DIRECTION of the velocity changes, which is why the motion in a uniform field is uniform circular motion. Exam tip: the answer must link "force perpendicular to velocity" to "no component along displacement, so no work, so constant speed". A magnetic field can never speed a charge up — in a cyclotron the speeding-up is done by the ELECTRIC field in the gap between the dees.

5. DefinitionSL & HL

   Define the motor effect.

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   The motor effect is the production of a force on a current-carrying conductor placed in an external magnetic field, arising because the moving charges in the conductor each experience a magnetic force which is transmitted to the lattice of the wire. The magnitude is F = BIL sinθ, where θ is the angle between the current and the field, and the direction is given by Fleming's left-hand rule. Exam tip: the force is a maximum when the conductor is perpendicular to the field (θ = 90°) and ZERO when the conductor lies along the field (θ = 0). Students often forget that the wire must carry a current AND sit in an EXTERNAL field — a wire's own field exerts no net force on itself.

6. DefinitionSL & HL

   Define the charge-to-mass ratio (specific charge) of a particle.

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   The charge-to-mass ratio is q/m, the electric charge of a particle divided by its mass, with SI unit C kg⁻¹. For an electron q/m = e/m_e = 1.60 × 10⁻¹⁹ / 9.11 × 10⁻³¹ = 1.76 × 10¹¹ C kg⁻¹; for a proton it is 9.58 × 10⁷ C kg⁻¹, about 1836 times smaller. Exam tip: experiments with fields (Thomson's tube, the fine-beam tube, the mass spectrometer) measure q/m, not q and m separately, because the dynamics depend only on the ratio — a = qE/m and r = mv/(qB) both contain q and m only as q/m. Millikan's oil-drop experiment was needed to fix e independently and hence obtain m_e.

7. DefinitionSL & HL

   Distinguish between the deflection of a charged particle in a uniform electric field and in a uniform magnetic field.

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   Electric field: the force qE is parallel (or antiparallel) to E, is independent of the particle's speed, acts even on a stationary charge, does work on the particle, and produces a PARABOLIC path with changing speed and kinetic energy. Magnetic field: the force qvB sinθ is perpendicular to both v and B, is proportional to speed, vanishes for a stationary charge or for motion along B, does NO work, and produces a CIRCULAR (or helical) path at constant speed. Exam tip: a two-column comparison earns the "compare" marks. Both deflections reverse if the sign of the charge reverses, so neither on its own distinguishes a positive from a negative particle without knowing the field directions.

8. DefinitionSL & HL

   Define crossed fields.

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   Crossed fields are a uniform electric field and a uniform magnetic field set up in the same region of space with E perpendicular to B, and with the particle beam directed perpendicular to both. In this arrangement the electric force qE and the magnetic force qvB are antiparallel, so they can be made to balance. Exam tip: the condition for the beam to pass straight through is qE = qvB → v = E/B, which is the basis of the velocity selector and of Thomson's e/m measurement. Watch the geometry: if E is downwards (in the plane of the page) and B is into the page, then a positive charge moving to the right feels qE downwards and qvB upwards. Unit check: (V m⁻¹)/(T) = m s⁻¹.

9. DefinitionSL & HL

   Define a uniform electric field and state where one is produced in this sub-topic.

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   A uniform electric field is one whose field strength E has the same magnitude and direction at every point, represented by parallel, equally spaced field lines. It is produced in the region between two large, flat, parallel, oppositely charged plates separated by a small distance d, where E = V/d directed from the positive to the negative plate. Exam tip: uniformity fails near the edges of the plates ("edge effects"/fringing), where the lines bulge outwards — say this if asked to comment on a diagram. E is a vector with unit N C⁻¹ = V m⁻¹, and the force on a charge q is F = qE, in the direction of E for a positive charge and opposite to E for a negative charge.

10. DefinitionSL & HL

    Define the electronvolt and explain its use in electromagnetic-field problems.

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    One electronvolt is the kinetic energy gained by a particle carrying one elementary charge when it is accelerated through a potential difference of one volt: 1 eV = 1.60 × 10⁻¹⁹ J. From the work–energy relation qV = ΔE_k, a particle of charge ne accelerated through V volts gains nV electronvolts of kinetic energy. Exam tip: the eV is a unit of ENERGY, not of potential or of voltage — writing "a potential of 500 eV" loses the mark. ALWAYS convert to joules before substituting into ½mv², or you will get v wrong by a factor of about 2.5 × 10⁹. Useful multiples: 1 keV = 10³ eV, 1 MeV = 10⁶ eV; an electron through 1 kV has E_k = 1.60 × 10⁻¹⁶ J.

11. EquationSL & HLData booklet: Yes

    State and explain the equation for the magnetic force on a moving charge.

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    F = qvB sinθ (data booklet). F = magnitude of the magnetic force (N); q = magnitude of the charge (C); v = speed of the charge (m s⁻¹); B = magnetic flux density (T); θ = angle between the velocity v and the field B. Valid for a point charge moving in a magnetic field; the vector form is F = qv × B, with F perpendicular to the plane containing v and B. Common misuse: taking θ as the angle to the plates or to the normal — it is the angle between v and B, so F is MAXIMUM at θ = 90° and ZERO at θ = 0° or 180°. Sanity check: an electron at 1.0 × 10⁷ m s⁻¹ across B = 0.20 T feels F = 1.60 × 10⁻¹⁹ × 1.0 × 10⁷ × 0.20 = 3.2 × 10⁻¹³ N.

12. EquationSL & HLData booklet: Yes

    State and explain the equation for the force on a current-carrying conductor in a magnetic field.

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    F = BIL sinθ (data booklet). F = force (N); B = magnetic flux density (T); I = current (A); L = length of conductor IN the field (m); θ = angle between the conductor (current direction) and B. It follows from F = qvB sinθ summed over all the charge carriers, since I = nAvq. Conditions: uniform field, straight conductor, entire length L within the field. Common misuse: using the whole length of the wire when only part of it lies between the magnet poles — use only the length inside the field region. Sanity check: 3.0 A in a 5.0 cm length perpendicular to B = 0.40 T gives F = 0.40 × 3.0 × 0.050 = 6.0 × 10⁻² N, a force that a top-pan balance reads as about 6 g.

13. EquationSL & HLData booklet: No – derive

    Derive and state the equation for the radius of the circular path of a charged particle in a uniform magnetic field.

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    The magnetic force supplies the centripetal force: qvB = mv²/r → r = mv/(qB) = p/(qB). r = radius (m); m = mass (kg); v = speed (m s⁻¹); q = charge magnitude (C); B = flux density (T). Valid only when v is perpendicular to B and the speed is non-relativistic; if v makes an angle θ with B, replace v by v sinθ. Data-booklet status: No – derive from qvB = mv²/r (the booklet gives F = qvB sinθ). Common misuse: forgetting that r ∝ v, so doubling the accelerating pd multiplies v by √2 and r by √2, not by 2. Sanity check: an electron at 2.0 × 10⁷ m s⁻¹ in B = 1.0 × 10⁻³ T has r = (9.11 × 10⁻³¹ × 2.0 × 10⁷)/(1.60 × 10⁻¹⁹ × 1.0 × 10⁻³) = 0.11 m.

14. EquationSL & HLData booklet: No – derive

    State the period and frequency of a charged particle circulating in a uniform magnetic field.

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    From r = mv/(qB) and T = 2πr/v: T = 2πm/(qB), and the cyclotron frequency f = 1/T = qB/(2πm). T = period (s); f = frequency (Hz); m = mass (kg); q = charge (C); B = flux density (T). Valid for non-relativistic speeds with v perpendicular to B. Key feature: T and f are INDEPENDENT of the speed and of the radius — the particle takes the same time for a small fast-formed circle as for a large one, which is exactly why a fixed-frequency supply works in a cyclotron. Data-booklet status: No – derive. Common misuse: assuming a faster particle circulates more often. Sanity check: a proton in B = 0.50 T has f = (1.60 × 10⁻¹⁹ × 0.50)/(2π × 1.67 × 10⁻²⁷) = 7.6 × 10⁶ Hz.

15. EquationSL & HLData booklet: Yes

    State and explain the equation for the force per unit length between two long parallel current-carrying wires.

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    F/L = μ₀I₁I₂/(2πr) (data booklet). F/L = force per unit length on each wire (N m⁻¹); μ₀ = 4π × 10⁻⁷ T m A⁻¹ (permeability of free space); I₁, I₂ = the two currents (A); r = separation of the wires (m). Valid for long, straight, thin, parallel wires in vacuum with r much less than their length. The forces on the two wires are equal and opposite (Newton's third law) even when I₁ ≠ I₂; parallel currents attract, antiparallel currents repel. Common misuse: squaring or halving r — the dependence is 1/r, not 1/r². Sanity check: I₁ = I₂ = 1.0 A at r = 1.0 m gives F/L = (4π × 10⁻⁷ × 1)/(2π × 1) = 2 × 10⁻⁷ N m⁻¹, the old definition of the ampere.

16. EquationSL & HLData booklet: Yes

    State the equations relating electric field strength, force and potential difference for a uniform field.

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    E = F/q, so F = qE; and for parallel plates E = V/d (both in the data booklet). E = electric field strength (N C⁻¹ = V m⁻¹); F = force on the charge (N); q = charge (C); V = potential difference between the plates (V); d = plate separation (m). Conditions: E = V/d holds only for a UNIFORM field, i.e. between parallel plates away from the edges. Rearranged forms needed: V = Ed, d = V/E, F = qV/d. Common misuse: using E = V/d for the radial field of a point charge — there you need E = kq/r². Sanity check: 250 V across plates 5.0 mm apart gives E = 250/0.0050 = 5.0 × 10⁴ V m⁻¹, and the force on an electron is 1.60 × 10⁻¹⁹ × 5.0 × 10⁴ = 8.0 × 10⁻¹⁵ N.

17. EquationSL & HLData booklet: No – derive

    State the acceleration of a charged particle in a uniform electric field.

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    Newton's second law with F = qE gives a = qE/m = qV/(md) for parallel plates. a = acceleration (m s⁻²); q = charge (C); E = field strength (N C⁻¹ or V m⁻¹); m = mass (kg); V = plate pd (V); d = plate separation (m). The acceleration is constant while the particle is between the plates, so the suvat equations apply along the field direction. Data-booklet status: No – derive from F = qE and F = ma. Common misuse: adding g — for an electron a = qE/m is typically ~10¹⁵ m s⁻², so gravity is utterly negligible; only mention g for charged oil drops or dust. Sanity check: an electron in 5.0 × 10⁴ V m⁻¹ has a = (1.60 × 10⁻¹⁹ × 5.0 × 10⁴)/(9.11 × 10⁻³¹) = 8.8 × 10¹⁵ m s⁻².

18. EquationSL & HLData booklet: No – derive

    State the condition for a charged particle to pass undeflected through crossed electric and magnetic fields.

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    Balancing the two forces: qE = qvB → v = E/B. v = selected speed (m s⁻¹); E = electric field strength (V m⁻¹); B = magnetic flux density (T). Conditions: E, B and v mutually perpendicular, and the two forces must be arranged to oppose each other. The charge q cancels, so the selected speed is independent of the charge, the sign of the charge and the mass. Data-booklet status: No – derive from F = qE and F = qvB. Common misuse: writing v = B/E — always check units, (V m⁻¹)/(T) = (N C⁻¹)/(N s C⁻¹ m⁻¹) = m s⁻¹. Sanity check: E = 3.0 × 10⁴ V m⁻¹ with B = 0.020 T selects v = 3.0 × 10⁴/0.020 = 1.5 × 10⁶ m s⁻¹.

19. EquationSL & HLData booklet: No – derive

    State the equation for the speed of a charged particle accelerated from rest through a potential difference V.

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    Work done by the field = gain in kinetic energy: qV = ½mv² → v = √(2qV/m). q = charge (C); V = accelerating pd (V); m = mass (kg); v = final speed (m s⁻¹). Valid from rest, in vacuum, non-relativistic (v less than about 0.1c), with no energy losses. Useful rearrangements: momentum p = mv = √(2mqV), and combined with r = mv/(qB) it gives r = √(2mV/q)/B. Data-booklet status: No – derive (the booklet gives W = qΔV and E_k = ½mv²). Common misuse: forgetting the factor 2 or leaving V in eV. Sanity check: an electron through 2.0 kV reaches v = √(2 × 1.60 × 10⁻¹⁹ × 2000/9.11 × 10⁻³¹) = 2.7 × 10⁷ m s⁻¹.

20. EquationSL & HLData booklet: No – derive

    State the equations describing the parabolic path of a charge projected perpendicular to a uniform electric field.

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    Take x along the initial velocity v₀ and y along the field. Then x = v₀t and y = ½at² = ½(qE/m)t², so eliminating t: y = (qE/2mv₀²)x² — a parabola. The vertical deflection on leaving plates of length L is y = qEL²/(2mv₀²) = qVL²/(2mdv₀²), and the exit angle satisfies tanθ = v_y/v₀ = qEL/(mv₀²). Symbols: q (C), E (V m⁻¹), m (kg), v₀ (m s⁻¹), L = plate length (m), d = plate separation (m). Data-booklet status: No – derive using suvat. Common misuse: using the full plate length as the horizontal distance when the particle exits the side, or forgetting the ½. Note y ∝ 1/v₀², so a faster beam is deflected far less.

21. EquationSL & HLData booklet: No – derive

    State the equation used to determine the mass-to-charge ratio in a mass spectrometer.

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    For an ion accelerated through V and then bent on a radius r in field B: qV = ½mv² and r = mv/(qB) give m/q = B²r²/(2V), or equivalently r = (1/B)√(2mV/q). Symbols: m = ion mass (kg); q = ion charge (C); B = flux density in the deflecting region (T); r = radius of the semicircular path (m); V = accelerating pd (V). If a velocity selector is used instead, m/q = B′rB/E with v = E/B. Data-booklet status: No – derive. Common misuse: using the diameter (the detector distance from the slit is 2r) as r, which makes m four times too big. Sanity check: r ∝ √m, so ²⁰Ne and ²²Ne differ in radius by only √(22/20) = 1.05, about 5%.

22. EquationSL & HLData booklet: No – derive

    State the equation for the pitch of the helical path of a charged particle in a uniform magnetic field.

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    Resolve the velocity: v∥ = v cosθ along B, v⊥ = v sinθ across B. The radius is r = mv sinθ/(qB) and the period is T = 2πm/(qB), so the pitch (advance per revolution) is p = v cosθ × T = 2πmv cosθ/(qB). Symbols: p (m); m (kg); v (m s⁻¹); θ = angle between v and B; q (C); B (T). Valid for a uniform field and non-relativistic speeds. Data-booklet status: No – derive. Common misuse: putting v (not v sinθ) into the radius, or v sinθ (not v cosθ) into the pitch. Limits: θ = 90° gives p = 0 (a circle); θ = 0 gives r = 0 (a straight line) — quoting these two checks is a quick way to confirm you have the sines and cosines the right way round.

23. EquationSL & HLData booklet: Yes

    State the equation for the magnetic field of a long straight current-carrying wire and where it is needed here.

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    B = μ₀I/(2πr). B = flux density (T); μ₀ = 4π × 10⁻⁷ T m A⁻¹; I = current (A); r = perpendicular distance from the wire's axis (m). Valid outside a long, straight, thin wire in vacuum; the field lines are concentric circles whose sense is given by the right-hand grip rule (thumb along conventional current, fingers curl in the direction of B). Combining it with F = BIL sinθ for a second wire reproduces the booklet result F/L = μ₀I₁I₂/(2πr). Common misuse: quoting a 1/r² fall-off. Sanity check: I = 10 A at r = 5.0 cm gives B = (4π × 10⁻⁷ × 10)/(2π × 0.050) = 4.0 × 10⁻⁵ T, comparable with the Earth's field — which is why a compass near a mains cable deflects.

24. Graph/diagramSL & HL

    Describe the trajectory diagram of an electron entering a uniform electric field between parallel plates.

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    Axes: horizontal x = distance along the plates (m), vertical y = deflection towards a plate (m). Shape: a PARABOLA of the form y = (qE/2mv₀²)x², starting tangential to the initial velocity and curving towards the positive plate for an electron; beyond the plates the path is a straight line along the exit tangent. The gradient dy/dx = v_y/v_x at any point gives tanθ for the instantaneous direction; the curvature is set by a = qE/m. Changing a parameter: doubling V (hence E) doubles y at every x; doubling the entry speed v₀ quarters y since y ∝ 1/v₀²; reversing the plate polarity mirrors the curve. Linearisation: a plot of y against x² is a straight line of gradient qE/(2mv₀²), from which q/m can be found.

25. Graph/diagramSL & HL

    Describe the graph of the radius of a charged particle's circular path against its speed in a uniform magnetic field.

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    Axes: x = speed v (m s⁻¹), y = radius r (m). Shape: a straight line through the origin, since r = mv/(qB). Gradient = m/(qB) (units m/(m s⁻¹) = s), so the gradient gives the mass-to-charge ratio m/q = gradient × B; the zero intercept confirms that a stationary charge is undeflected. Changing a parameter: increasing B rotates the line towards the v-axis (smaller gradient, tighter circles, r ∝ 1/B); using a heavier ion of the same charge steepens the line. Alternative linearisation used in practical work: with particles accelerated through a pd V, r² = 2mV/(qB²), so a plot of r² against V is linear with gradient 2m/(qB²) — much easier to obtain than measuring v directly.

26. Graph/diagramSL & HL

    Describe the graph of the period of circular motion against speed for a charged particle in a uniform magnetic field.

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    Axes: x = speed v (m s⁻¹), y = period T (s). Shape: a HORIZONTAL straight line at T = 2πm/(qB) — the period is independent of speed and of radius, because r ∝ v so the extra path length exactly compensates the extra speed. Intercept on the T-axis = 2πm/(qB), giving m/q = T B/(2π); gradient = 0. Changing a parameter: doubling B halves the height of the line; using an ion of twice the mass doubles it. A companion graph of T against 1/B is a straight line through the origin with gradient 2πm/q, and this is the standard linearisation for a fine-beam-tube or cyclotron experiment. Exam trap: students draw a rising line by wrongly assuming faster means longer.

27. Graph/diagramSL & HL

    Describe the graph obtained when the force on a current-carrying conductor is measured as the current is varied.

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    Axes: x = current I (A), y = force F (N) — measured, for example, as the change in reading of a top-pan balance multiplied by g. Shape: a straight line through the origin, since F = BIL sinθ with B, L and θ constant. Gradient = BL sinθ; with the wire perpendicular to the field, B = gradient/L, which is the standard method of measuring the flux density between magnadur magnets. A non-zero intercept indicates a systematic error such as failing to zero the balance. Changing a parameter: stronger magnets or a longer wire in the field steepens the line; tilting the wire towards the field direction reduces the gradient by sinθ. Reversing the current gives a line of the same magnitude in the opposite sense (the balance reading falls instead of rises).

28. Graph/diagramSL & HL

    Describe the graph of force against angle for a conductor in a magnetic field.

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    Axes: x = θ, the angle between the conductor and the field (° or rad), y = force F (N). Shape: a sine curve, F = BIL sinθ, zero at θ = 0° and 180°, maximum F = BIL at θ = 90°. This is NOT a straight line, so to analyse it you linearise: plot F against sinθ, which gives a straight line through the origin of gradient BIL. From that gradient, B = gradient/(IL). Changing a parameter: increasing I or L increases the amplitude but not the positions of the zeros. Exam tip: the same shape applies to F = qvB sinθ for a charge, and the zero at θ = 0 is the graphical statement that a charge moving ALONG a magnetic field line experiences no force at all.

29. Graph/diagramSL & HL

    Describe the graph used to investigate the force between two parallel current-carrying wires.

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    Raw graph: axes x = separation r (m), y = force per unit length F/L (N m⁻¹); the shape is a decreasing curve (a hyperbola) tending to zero as r → ∞, since F/L = μ₀I₁I₂/(2πr). Because a curve cannot be used to extract a value reliably, linearise: plot F/L against 1/r (m⁻¹), giving a straight line through the origin of gradient μ₀I₁I₂/(2π); with the currents known this yields μ₀. Changing a parameter: doubling either current doubles the gradient; reversing one current changes attraction to repulsion, so the measured force changes sign while the magnitude graph is unchanged. Alternatively plot F/L against the product I₁I₂ at fixed r — again a straight line through the origin, gradient μ₀/(2πr).

30. Graph/diagramSL & HL

    Describe the field and force diagrams for a current-carrying wire and a moving charge, including the page conventions.

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    Conventions: a dot ⊙ represents a vector (B or v) out of the page; a cross ⊗ represents one into the page — think of the point and the flights of an arrow. Around a long straight wire the field lines are concentric circles, closer together near the wire (B ∝ 1/r), with the sense given by the right-hand grip rule. For a charge moving perpendicular to a uniform field the diagram shows equally spaced crosses (or dots) with a circular path drawn on them; the velocity is tangential and the magnetic force points to the centre. Changing a parameter: a stronger field is drawn with more closely spaced crosses and gives a smaller circle. Exam tip: label the direction of the CONVENTIONAL current and mark the charge's sign, or the direction marks are lost.

31. Graph/diagramSL & HL

    Describe the graph of the radius against the square root of the accelerating potential difference in a mass spectrometer or fine-beam tube.

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    Axes: x = √V (V^½), y = radius r (m). Shape: a straight line through the origin, because combining qV = ½mv² with r = mv/(qB) gives r = (1/B)√(2m/q) × √V. Gradient = (1/B)√(2m/q), so the charge-to-mass ratio is q/m = 2/(B × gradient)², a standard IB practical determination of e/m for the electron. Changing a parameter: increasing B reduces the gradient (r ∝ 1/B); a heavier ion of the same charge increases it. Equivalent linearisation: plot r² against V, gradient 2m/(qB²). Exam tip: quote the gradient with its unit (m V^−½), and take the gradient from a large triangle drawn on the best-fit line — never from a single data point.

32. Concept/explainSL & HL

    Explain why a charged particle entering a uniform electric field perpendicular to the field lines follows a parabolic path.

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    • In the uniform field the force F = qE is constant in magnitude and direction, so the acceleration a = qE/m is constant and always parallel to the field
    • The component of velocity perpendicular to the field is unaffected, so along that direction the particle moves with constant velocity, x = vt
    • Along the field direction the particle starts from rest relative to that axis and undergoes uniform acceleration, y = ½at²
    • Eliminating t gives y = (qE/2mv²)x², the equation of a parabola
    • The situation is mathematically identical to horizontal projectile motion under gravity, with qE/m replacing g. Exam tip: incomplete answers say the path is circular or that the particle slows down; the perpendicular velocity component never changes and the speed actually increases.
33. Concept/explainSL & HLData booklet: No – derive

    Explain why the period of circular motion of a charged particle in a uniform magnetic field is independent of its speed, and why this matters.

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    • From qvB = mv²/r the radius is r = mv/(qB), so a faster particle simply moves on a proportionally larger circle
    • The period is T = 2πr/v; substituting r gives T = 2πm/(qB), in which v has cancelled
    • Hence T (and the cyclotron frequency f = qB/2πm) depends only on the charge-to-mass ratio q/m and the flux density B
    • Physically, the extra distance travelled per revolution exactly compensates for the higher speed
    • This isochronism allows a cyclotron to use a fixed-frequency alternating pd across the dees to accelerate particles that speed up on every crossing. Exam tip: the independence fails at relativistic speeds because m increases, which is why synchrocyclotrons/synchrotrons vary the frequency or field.
34. Concept/explainSL & HLData booklet: No – derive

    Describe and explain the motion of a charged particle whose velocity enters a uniform magnetic field at an angle θ to the field, where 0 < θ < 90°.

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    • Resolve the velocity into a component v cosθ parallel to B and v sinθ perpendicular to B
    • The parallel component produces no magnetic force (sin0 = 0), so the particle drifts along the field direction at constant speed v cosθ
    • The perpendicular component gives a force qvB sinθ that is always perpendicular to it, producing circular motion of radius r = mv sinθ/(qB)
    • Superposing a uniform drift on circular motion gives a helix (spiral) whose axis lies along the field line
    • The pitch, the distance advanced per revolution, is p = v cosθ × 2πm/(qB)
    • Charged particles from the solar wind spiral in this way along Earth's field lines towards the poles, producing aurorae, and are trapped in the Van Allen belts. Exam tip: many answers say the path is simply circular; the parallel component is never removed.
35. Concept/explainSL & HLData booklet: Yes

    Explain how the force on a current-carrying conductor in a magnetic field, F = BIL sinθ, arises from the force on individual charge carriers.

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    • A current I in a wire consists of n charge carriers per unit volume, each of charge q, drifting with speed v, so that I = nAvq
    • Each carrier in the field experiences a magnetic force qvB sinθ
    • The number of carriers in a length L of wire is nAL, so the total force is F = nAL × qvB sinθ = (nAvq)BL sinθ = BIL sinθ
    • The carriers cannot leave the wire, so the force is transmitted to the lattice and the whole conductor is pushed
    • θ is the angle between the conductor (current) and the field, so the force is maximum when the wire is perpendicular to B and zero when it lies along B
    • This motor effect is the basis of the electric motor and the moving-coil loudspeaker. Exam tip: quoting I = nAvq as the link earns the connecting mark.
36. Concept/explainSL & HLData booklet: Yes

    Explain why two long straight parallel wires carrying currents in the same direction attract each other, and how this leads to the definition of the ampere.

    Show answer

    • Wire 1 produces a magnetic field at wire 2 of magnitude B₁ = μ₀I₁/(2πr), directed by the right-hand grip rule and perpendicular to wire 2
    • Wire 2, carrying current I₂ in that field, experiences a force per unit length F/L = B₁I₂ = μ₀I₁I₂/(2πr)
    • Applying the left-hand rule at wire 2 shows the force points towards wire 1 when the currents are parallel, so they attract; antiparallel currents repel
    • By Newton's third law the force on wire 1 is equal and opposite, so the pair attract mutually
    • Historically the ampere was defined as the constant current which, in two infinitely long thin parallel conductors 1 m apart in vacuum, produces a force of 2 × 10⁻⁷ N per metre. Exam tip: since 2019 the ampere is defined from the fixed value of e, but the force result is still examined.
37. Concept/explainSL & HLData booklet: No – derive

    Explain the operation of a velocity selector using crossed electric and magnetic fields.

    Show answer

    • Charged particles enter a region where a uniform electric field E and a uniform magnetic field B are perpendicular to each other and to the beam velocity v
    • The electric force qE acts along E; the magnetic force qvB acts in the opposite direction if the fields are correctly oriented
    • A particle travels undeflected only if the two forces balance: qE = qvB, so v = E/B
    • Both q and the sign of the charge cancel, so the selected speed is the same for all ions, whatever their charge or mass
    • Particles faster than E/B have the larger magnetic force and are deflected one way; slower ones are deflected the other way, and are removed by a collimating slit. Exam tip: the selector selects speed, not mass or charge — separation by mass happens later, in the second magnetic field.
38. Concept/explainSL & HLData booklet: No – derive

    Explain how a mass spectrometer with a velocity selector followed by a uniform magnetic field separates isotopes of an element.

    Show answer

    • Ions are produced and accelerated, then passed through crossed fields so that only those with v = E/B₁ emerge through the slit
    • They then enter a uniform field B₂ perpendicular to their velocity and follow a semicircular path of radius r = mv/(qB₂)
    • Since v, q and B₂ are the same for all the singly ionised isotopes, r ∝ m
    • Isotopes differ only in neutron number, so the heavier isotope has the larger radius and lands further from the entry slit; the detector position gives m directly
    • The relative intensity of each trace gives the relative abundance, from which the relative atomic mass is calculated
    • Applications include carbon dating, forensic and pharmaceutical analysis, and identifying molecules in planetary atmospheres. Exam tip: state explicitly that q is the same for singly charged ions before writing r ∝ m.
39. Concept/explainSL & HLData booklet: No – derive

    Outline the operation of a cyclotron and explain the factors that limit the maximum energy it can deliver.

    Show answer

    • Two hollow D-shaped electrodes (dees) sit in a uniform magnetic field perpendicular to their plane, inside a vacuum chamber
    • Inside a dee there is no electric field, so the particle moves in a semicircle of radius r = mv/(qB)
    • Across the gap an alternating pd accelerates the particle; because T = 2πm/(qB) is independent of speed, a fixed supply frequency stays in step with the particle
    • Each crossing increases the kinetic energy, so the radius grows and the path is a spiral; the particle is extracted at the outer radius R with v = qBR/m
    • Maximum energy is limited by the radius R of the magnet poles and by B, and ultimately by relativistic mass increase, which destroys the synchronism. Exam tip: no work is done inside the dees — all energy is gained in the gaps.
40. Concept/explainSL & HLData booklet: No – derive

    Explain why deflection experiments in magnetic fields determine the charge-to-mass ratio of a particle rather than its charge or mass separately.

    Show answer

    • The magnetic force gives qvB = mv²/r, so r = mv/(qB) and the measurable radius depends on q and m only through the ratio q/m
    • If the particle has been accelerated from rest through a pd V, then qV = ½mv², so v = √(2Vq/m), again containing only q/m
    • Substituting gives r² = 2Vm/(qB²), so q/m = 2V/(B²r²) — every measurable quantity (V, B, r) fixes the ratio and nothing else
    • To obtain q and m separately an independent measurement is needed, such as Millikan's oil-drop determination of e
    • Thomson's 1897 measurement of a very large e/m for cathode rays, independent of the gas or electrode material, was the evidence for a universal sub-atomic particle. Exam tip: quote e/m ≈ 1.76 × 10¹¹ C kg⁻¹ for the electron as a sanity check.
41. Concept/explainSL & HLData booklet: Yes

    Explain why a rectangular current-carrying coil in a uniform magnetic field experiences a turning effect but no net force.

    Show answer

    • Each side of the coil carries the same current I in the same uniform field B, and F = BIL sinθ applies to each side
    • The two sides parallel to the field (θ = 0) experience no force
    • The two sides perpendicular to the field carry current in opposite directions, so they experience equal and opposite forces of magnitude BIL
    • Since the two forces are equal, opposite and not collinear, the resultant force is zero but they form a couple, producing a torque that rotates the coil
    • The torque is maximum when the coil plane contains B and zero when the coil plane is perpendicular to B (the plane of the coil normal to B)
    • A split-ring commutator reverses the current every half turn so that rotation continues in one sense — the dc motor. Exam tip: say "equal and opposite forces, not in the same line" to secure the couple mark.
42. Worked problemSL & HLData booklet: No – derive

    An electron is accelerated from rest through a potential difference of 2.0 kV in an evacuated tube. Calculate its final speed.

    Show answer

    Principle: work done by the electric field equals gain in kinetic energy, qV = ½mv² (no other forces act in vacuum). Substitution: (1.60 × 10⁻¹⁹ C)(2.0 × 10³ V) = ½(9.11 × 10⁻³¹ kg)v². Left side = 3.20 × 10⁻¹⁶ J. So v² = 2(3.20 × 10⁻¹⁶)/(9.11 × 10⁻³¹) = 7.02 × 10¹⁴ m² s⁻². Therefore v = √(7.02 × 10¹⁴) = 2.65 × 10⁷ m s⁻¹ (3 s.f.). Check/Trap: v is about 9% of c so the non-relativistic treatment is acceptable at IB level; do not forget the factor ½, and note the answer is independent of the electron's starting position between the plates because only the pd traversed matters.

43. Worked problemSL & HLData booklet: Yes

    An electron travelling horizontally at 2.0 × 10⁷ m s⁻¹ enters midway between two parallel plates of length 6.0 cm separated by 2.0 cm with a pd of 200 V across them. Determine the vertical deflection as it leaves the plates.

    Show answer

    Principle: uniform field between plates, projectile-style analysis. Field: E = V/d = 200/0.020 = 1.0 × 10⁴ V m⁻¹. Acceleration: a = qE/m = (1.60 × 10⁻¹⁹)(1.0 × 10⁴)/(9.11 × 10⁻³¹) = 1.76 × 10¹⁵ m s⁻². Time in field: t = L/v = 0.060/(2.0 × 10⁷) = 3.0 × 10⁻⁹ s (horizontal speed constant). Deflection: y = ½at² = ½(1.76 × 10¹⁵)(3.0 × 10⁻⁹)² = 7.9 × 10⁻³ m = 7.9 mm (2 s.f.), towards the positive plate. Check/Trap: the half-gap is 10 mm, so the electron just escapes; students often use the full 2.0 cm gap or forget that the horizontal velocity is unchanged. Gravity is utterly negligible here.

44. Worked problemSL & HLData booklet: Yes

    A proton moves at 5.0 × 10⁶ m s⁻¹ at 30° to a uniform magnetic field of flux density 0.25 T. Calculate the magnetic force on it.

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    Principle: F = qvB sinθ, where θ is the angle between v and B (data booklet). Substitution: F = (1.60 × 10⁻¹⁹ C)(5.0 × 10⁶ m s⁻¹)(0.25 T)(sin30°). Intermediate: qv = 8.0 × 10⁻¹³ C m s⁻¹; × 0.25 T = 2.0 × 10⁻¹³ N; × 0.500 = 1.0 × 10⁻¹³ N. Final: F = 1.0 × 10⁻¹³ N (2 s.f.), directed perpendicular to both v and B, given by the left-hand rule with the second finger along v because the proton is positive. Check/Trap: θ is measured from the field, not from the perpendicular to it — using cos30° is the standard error. Note this force does no work: the proton's speed remains 5.0 × 10⁶ m s⁻¹ and its path is a helix.

45. Worked problemSL & HLData booklet: No – derive

    A proton enters a uniform magnetic field of 0.25 T at right angles with a speed of 5.0 × 10⁶ m s⁻¹. Determine the radius of its circular path. (m_p = 1.67 × 10⁻²⁷ kg)

    Show answer

    Principle: the magnetic force provides the centripetal force, qvB = mv²/r, so r = mv/(qB) (derive — not printed). Substitution: r = (1.67 × 10⁻²⁷ kg)(5.0 × 10⁶ m s⁻¹)/[(1.60 × 10⁻¹⁹ C)(0.25 T)]. Numerator: 8.35 × 10⁻²¹ kg m s⁻¹. Denominator: 4.0 × 10⁻²⁰ C T. Final: r = 0.209 m ≈ 0.21 m (2 s.f.). Check/Trap: quote the derivation line qvB = mv²/r for the method mark. A common error is to cancel v incorrectly and obtain r ∝ v²; another is to use the electron mass out of habit. Sanity check: units kg m s⁻¹/(C T) = kg m s⁻¹/(kg s⁻¹) = m, as required.

46. Worked problemSL & HLData booklet: No – derive

    For the proton of the previous problem (B = 0.25 T), calculate the period of its circular motion and hence the frequency of the alternating supply needed in a cyclotron using this field.

    Show answer

    Principle: T = 2πr/v with r = mv/(qB) gives T = 2πm/(qB), independent of speed and radius. Substitution: T = 2π(1.67 × 10⁻²⁷ kg)/[(1.60 × 10⁻¹⁹ C)(0.25 T)] = 1.049 × 10⁻²⁶/(4.0 × 10⁻²⁰) = 2.62 × 10⁻⁷ s (3 s.f.). Frequency: f = 1/T = 3.8 × 10⁶ Hz = 3.8 MHz. Check/Trap: because v cancels, the same 3.8 MHz supply keeps the proton in step as it spirals outwards — the whole principle of the cyclotron. Students often try to compute T from the radius they found earlier and lose accuracy; work symbolically first. Doubling the field halves the period; using a deuteron (twice the mass) doubles it.

47. Worked problemSL & HLData booklet: Yes

    A straight wire of length 25 cm carries a current of 3.0 A and lies at 40° to a uniform magnetic field of flux density 0.15 T. Calculate the magnitude of the force on the wire.

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    Principle: F = BIL sinθ (data booklet), with θ the angle between the conductor and the field. Substitution: F = (0.15 T)(3.0 A)(0.25 m)(sin40°). Intermediate: BIL = 0.15 × 3.0 × 0.25 = 0.1125 N; sin40° = 0.643. Final: F = 0.1125 × 0.643 = 0.072 N (2 s.f.), directed perpendicular to the plane containing the wire and the field, found by Fleming's left-hand rule. Check/Trap: convert 25 cm to 0.25 m; using θ = 50° (the angle to the perpendicular) is the classic slip. If the wire were aligned with the field the force would be zero, and the maximum possible force here is 0.11 N.

48. Worked problemSL & HLData booklet: Yes

    Two long parallel wires 4.0 cm apart each carry a current of 6.0 A in the same direction. Determine the magnitude and nature of the force on a 1.5 m length of one wire.

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    Principle: F/L = μ₀I₁I₂/(2πr) (data booklet), with μ₀ = 4π × 10⁻⁷ T m A⁻¹. Substitution: F/L = (4π × 10⁻⁷)(6.0)(6.0)/(2π × 0.040). Simplify 4π/2π = 2, so F/L = 2 × 10⁻⁷ × 36/0.040 = 7.2 × 10⁻⁶/0.040 = 1.8 × 10⁻⁴ N m⁻¹. For 1.5 m: F = (1.8 × 10⁻⁴)(1.5) = 2.7 × 10⁻⁴ N (2 s.f.). Nature: currents are parallel, so the force is attractive. Check/Trap: the separation must be in metres and the answer per metre must be multiplied by the length asked for. By Newton's third law the other wire feels the same 2.7 × 10⁻⁴ N pulling the opposite way, not double.

49. Worked problemSL & HLData booklet: No – derive

    In a velocity selector the parallel plates are 2.0 cm apart with 800 V across them, and the magnetic flux density is 0.050 T perpendicular to both the plate field and the beam. Calculate the speed of the ions transmitted.

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    Principle: an ion passes undeflected when the electric and magnetic forces balance, qE = qvB, so v = E/B. Field between the plates: E = V/d = 800/0.020 = 4.0 × 10⁴ V m⁻¹ (data booklet). Substitution: v = (4.0 × 10⁴ V m⁻¹)/(0.050 T). Final: v = 8.0 × 10⁵ m s⁻¹ (2 s.f.). Check/Trap: the charge q cancels, so the selected speed is the same for singly and doubly charged ions and for positive and negative ions — a favourite multiple-choice point. Units check: (V m⁻¹)/T = (N C⁻¹)/(N A⁻¹ m⁻¹) = m s⁻¹. Ions faster than 8.0 × 10⁵ m s⁻¹ are deflected towards the side the magnetic force acts on.

50. Worked problemSL & HLData booklet: No – derive

    Singly charged ions of speed 8.0 × 10⁵ m s⁻¹ enter a uniform field of 0.30 T at right angles and strike a detector after a semicircle of radius 0.60 m. Determine the mass of the ion in unified atomic mass units.

    Show answer

    Principle: qvB = mv²/r, so m = qBr/v. Substitution: m = (1.60 × 10⁻¹⁹ C)(0.30 T)(0.60 m)/(8.0 × 10⁵ m s⁻¹). Numerator: 2.88 × 10⁻²⁰. Mass: m = 3.6 × 10⁻²⁶ kg. Convert: 3.6 × 10⁻²⁶/(1.66 × 10⁻²⁷ kg u⁻¹) = 21.7 u ≈ 22 u (2 s.f.), consistent with the isotope neon-22. Check/Trap: "singly charged" means q = e = 1.60 × 10⁻¹⁹ C; for a doubly charged ion the mass would be twice as large. Note the detector measures the diameter 2r in many questions — read whether the quoted distance from the slit is r or 2r, a frequent 2-mark loss.

51. Worked problemSL & HLData booklet: No – derive

    In the spectrometer above, ions of neon-20 are also present. Calculate the separation of the two impact points on the detector.

    Show answer

    Principle: after the velocity selector all ions share v = 8.0 × 10⁵ m s⁻¹ and q = e, so r = mv/(qB) gives r ∝ m. Ratio: r₂₀ = r₂₂ × (20/22) = 0.60 × 0.9091 = 0.545 m. The ions travel semicircles, so each lands a distance 2r from the entry slit. Separation of impact points: Δ = 2(r₂₂ − r₂₀) = 2(0.600 − 0.545) = 2(0.055) = 0.109 m ≈ 0.11 m (2 s.f.). Check/Trap: the factor of 2 is essential — the detector spacing is the difference of diameters, not of radii. Using exact isotope masses (19.99 u and 21.99 u) changes the answer by well under 1%, so the whole-number approximation is safe. The heavier isotope has the larger radius.

52. Worked problemSL & HLData booklet: No – derive

    In a fine-beam tube, electrons accelerated through 250 V move in a circle of radius 4.5 cm in a magnetic field of 1.2 mT. Determine the specific charge e/m of the electron.

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    Principle: combine energy and circular-motion conditions. Acceleration: eV = ½mv² → v = √(2eV/m). Circular path: r = mv/(eB) → v = eBr/m. Equate and square: (eBr/m)² = 2eV/m, giving e/m = 2V/(B²r²). Substitution: e/m = 2(250)/[(1.2 × 10⁻³ T)²(0.045 m)²] = 500/[(1.44 × 10⁻⁶)(2.025 × 10⁻³)] = 500/(2.92 × 10⁻⁹). Final: e/m = 1.7 × 10¹¹ C kg⁻¹ (2 s.f.), against the accepted 1.76 × 10¹¹ C kg⁻¹. Check/Trap: convert mT to T and cm to m before squaring — an error here changes the answer by 10⁶. Because both B and r are squared, their fractional uncertainties are doubled in the result.

53. Worked problemSL & HLData booklet: No – derive

    A cyclotron of dee radius 0.50 m operates with a magnetic flux density of 0.80 T. Calculate the maximum kinetic energy, in MeV, given to a proton.

    Show answer

    Principle: the proton leaves at the outer radius where r = R, so from r = mv/(qB), v_max = qBR/m. Substitution: v = (1.60 × 10⁻¹⁹)(0.80)(0.50)/(1.67 × 10⁻²⁷) = 6.4 × 10⁻²⁰/1.67 × 10⁻²⁷ = 3.83 × 10⁷ m s⁻¹. Energy: E_k = ½mv² = ½(1.67 × 10⁻²⁷)(3.83 × 10⁷)² = ½(1.67 × 10⁻²⁷)(1.47 × 10¹⁵) = 1.23 × 10⁻¹² J. Convert: 1.23 × 10⁻¹²/(1.60 × 10⁻¹⁹) = 7.7 × 10⁶ eV = 7.7 MeV (2 s.f.). Check/Trap: the accelerating pd across the dees does not appear — it only sets how many turns are needed, not the final energy. Since E_k ∝ B²R², doubling B quadruples the output energy.

54. Worked problemSL & HLData booklet: No – derive

    A proton and an alpha particle move with the same speed perpendicular to the same uniform magnetic field. Determine the ratio of the radius of the alpha particle's path to that of the proton's.

    Show answer

    Principle: r = mv/(qB); with v and B identical, r ∝ m/q. For the proton m = 1u, q = e. For the alpha particle m = 4u, q = 2e. Ratio: r_α/r_p = (4u/2e)/(1u/1e) = 2/1 = 2. Final: the alpha particle's radius is twice the proton's. Check/Trap: this is a classic Paper 1 item — students often answer 4 (mass only) or ½ (charge only). If instead the two particles had been accelerated through the same pd rather than given the same speed, then v = √(2qV/m) and r = √(2Vm/q)/B, giving r_α/r_p = √(4/2) = √2 ≈ 1.4 — read the question wording carefully. Both curve in the same sense since both are positive.

55. Worked problemSL & HLData booklet: No – derive

    An electron with speed 3.0 × 10⁶ m s⁻¹ enters a uniform magnetic field of 0.020 T at 60° to the field direction. Calculate the radius and the pitch of its helical path.

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    Principle: resolve v into components. Perpendicular: v⊥ = v sin60° = 3.0 × 10⁶ × 0.866 = 2.60 × 10⁶ m s⁻¹. Parallel: v∥ = v cos60° = 1.50 × 10⁶ m s⁻¹. Radius: r = mv⊥/(eB) = (9.11 × 10⁻³¹)(2.60 × 10⁶)/[(1.60 × 10⁻¹⁹)(0.020)] = 2.37 × 10⁻²⁴/3.2 × 10⁻²¹ = 7.4 × 10⁻⁴ m. Period: T = 2πm/(eB) = 2π(9.11 × 10⁻³¹)/(3.2 × 10⁻²¹) = 1.79 × 10⁻⁹ s. Pitch: p = v∥T = (1.50 × 10⁶)(1.79 × 10⁻⁹) = 2.7 × 10⁻³ m (2 s.f.). Check/Trap: only v⊥ enters the radius, only v∥ enters the pitch, and T uses the full mass and field regardless of angle. The speed stays 3.0 × 10⁶ m s⁻¹ throughout.

56. Exam technique/trapSL & HL

    Identify the most common direction error in D.3 questions on electron beams, and state the correct procedure.

    Show answer

    The trap: applying Fleming's left-hand rule with the second finger along the electron's velocity. Why students fall for it: the rule is memorised as "second finger = current", and the velocity looks like the obvious current direction. Correct approach: conventional current is the direction positive charge would move, so for an electron it is opposite to v. Either point the second finger against the velocity, or apply the rule to v and then reverse the resulting force. Equivalently use F = qv × B and let the negative q flip the vector. Always check the final sketch: the force must point towards the centre of the circle you have drawn. If a question says "deduce the direction", a reasoned statement plus the named rule is required, not just an arrow.

57. Exam technique/trapSL & HL

    Explain the trap in questions that ask what happens to the kinetic energy or speed of a charged particle as it moves through a uniform magnetic field.

    Show answer

    The trap: writing that the particle accelerates and therefore gains kinetic energy, or that the magnetic field does work on it. Why students fall for it: the particle is genuinely accelerating (its velocity changes direction) and "force × distance" feels non-zero. Correct approach: the magnetic force is always perpendicular to the displacement, so W = Fd cos90° = 0; kinetic energy and speed are constant and only the direction changes. In a combined field question, attribute all energy changes to the electric field alone (ΔE_k = qΔV). For the command term "explain", the perpendicularity of F and v must be stated explicitly — simply asserting "speed is constant" scores no explanation mark.

58. Exam technique/trapSL & HLData booklet: No – derive

    Point out the errors students make when substituting into r = mv/(qB) for ions and nuclei.

    Show answer

    The trap: using q = 1.60 × 10⁻¹⁹ C for every particle and taking masses in u without conversion. Why students fall for it: e is the only charge in the data booklet, and u looks like a usable unit. Correct approach: for a doubly charged ion or an alpha particle q = 2e = 3.20 × 10⁻¹⁹ C; for a nucleus of mass number A, m = A × 1.66 × 10⁻²⁷ kg. Then r = mv/(qB) is dimensionally consistent in metres. Also watch whether the question quotes the radius, the diameter, or the distance between two isotope traces on a detector (a difference of diameters). Finally, r = mv/qB is not printed — show the line qvB = mv²/r to earn the derivation mark.

59. Exam technique/trapSL & HL

    Give guidance on answering "sketch the path" questions for a charged particle entering a bounded field region.

    Show answer

    The trap: drawing a curve that continues to bend after the particle has left the field, or drawing a spiral of decreasing radius. Why students fall for it: they picture friction or continual attraction. Correct approach: inside the field region the path is a circular arc of constant radius r = mv/(qB), because the speed is constant; on leaving the field the particle travels in a straight line tangential to the arc. Show the curvature in the direction given by the left-hand rule, and make the arc join the straight sections smoothly. For an electric field between plates, draw a parabola inside and a straight line beyond, again tangential. Label the direction of deflection and, if asked, annotate that the radius increases if v increases or B decreases.

60. Exam technique/trapSL & HLData booklet: Yes

    Explain the angle traps in F = qvB sinθ and F = BIL sinθ.

    Show answer

    The trap: using the angle between the velocity (or wire) and the normal to the field, or assuming sinθ = 1 whenever a diagram "looks" perpendicular. Why students fall for it: in mechanics θ is often measured from a surface, and diagrams are drawn in perspective. Correct approach: in both booklet equations θ is the angle between the field B and the direction of motion (or the current), so the force is maximum at θ = 90° and exactly zero at θ = 0° or 180°. A particle moving parallel to B experiences no magnetic force at all — this is the key to helical motion questions. When a wire is described as "at 30° to the vertical" and the field is horizontal, redraw and identify the correct included angle before substituting. State the angle you have used in your working.

61. Exam technique/trapSL & HLData booklet: No – derive

    Outline a practical determination of the charge-to-mass ratio of the electron using a fine-beam tube and Helmholtz coils, including variables and analysis.

    Show answer

    Apparatus: evacuated tube containing low-pressure gas, electron gun with variable accelerating pd V, Helmholtz coils giving a uniform field B ∝ I_coil, metre rule or internal scale to measure the circular beam diameter. Independent variable: coil current (hence B); dependent: radius r; controlled: accelerating pd V, coil geometry, gas pressure. Theory: e/m = 2V/(B²r²), so r = √(2Vm/e)/B, i.e. r ∝ 1/B. Analysis: plot r against 1/B; the gradient equals √(2Vm/e), from which e/m is found — a linearisation is expected rather than a single-point calculation. Limitations: measuring the beam diameter through glass introduces parallax, and the Earth's field (≈ 5 × 10⁻⁵ T) is a systematic error. Improvements: align the tube so Earth's field is along the axis, use a mirror scale, take readings for increasing and decreasing current.

62. Exam technique/trapSL & HLData booklet: Yes

    Describe how to measure the magnetic flux density between the poles of a magnet using a current-carrying wire and an electronic balance, and identify the main sources of error.

    Show answer

    Method: clamp a horizontal wire of measured length L between the poles of a magnet that rests on a top-pan balance, with the wire perpendicular to the field. Zero the balance, pass a current I, and record the change in reading Δm; by Newton's third law the downward force on the magnet equals BIL, so F = Δm g. Independent variable: I (varied with a rheostat); dependent: balance reading; controlled: L, position and orientation of the wire, magnet separation. Analysis: plot F against I; the graph should be a straight line through the origin with gradient BL, so B = gradient/L. Errors: heating changes the current (use short bursts), the wire may not be exactly perpendicular, the effective length in the fringe field is uncertain, and balance drift is systematic. Improvement: reverse the current and average magnitudes.

63. Exam technique/trapSL & HL

    Explain how uncertainties should be handled in D.3 calculations and graphs, using r = mv/(qB) and e/m = 2V/(B²r²) as examples.

    Show answer

    Rule: add fractional (percentage) uncertainties for quantities multiplied or divided, and multiply a fractional uncertainty by the power when a quantity is raised to that power. In r = mv/(qB), Δr/r = Δv/v + ΔB/B. In e/m = 2V/(B²r²), the fractional uncertainty is ΔV/V + 2ΔB/B + 2Δr/r, so a 3% error in radius becomes 6% in the result — measure r most carefully. Absolute uncertainty is then found from the fractional value and quoted to one significant figure, with the result rounded to the same decimal place. On graphs, plot error bars, draw best, maximum and minimum gradient lines through the bars, and take the uncertainty in the gradient as (max − min)/2. Distinguish random scatter (reduce by repeats) from systematic offsets such as Earth's field, which shift every point the same way and are revealed by a non-zero intercept.

Practise this topic with exam-style questions: D.3 Motion in electromagnetic fields questions (SL) · D.3 Motion in electromagnetic fields questions (HL) · all flashcards

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