IB Physics HL D.3 Motion in electromagnetic fields Questions

SLSL HLHL

Practise IB Physics D.3 by tracing charged-particle motion in uniform electric, magnetic and crossed fields, then calculating forces on current-carrying conductors and parallel wires.

121 questions in this syllabus scope, with question parts counted together.

Exam points & syllabus

Exam points

In a uniform plate field, combine E = V/d with F = qE and a = F/m; use the full gap and the actual charge-to-mass ratio.
An electron accelerates towards the positive plate, opposite to E, even when its initial velocity is parallel to the plates.
A nonparallel entry velocity in a uniform electric field produces a parabola: transverse acceleration is constant and perpendicular velocity is unchanged.
For midpoint plate collisions, use half the gap as the transverse displacement, then multiply flight time by unchanged plate-parallel speed.
Identical transit times from rest across the same gap in the same field imply equal charge-to-mass ratios, not equal masses and charges separately.
In the supplied drift-tube apparatus, acceleration occurs in the gaps; reverse polarity while the particle is inside a conducting tube with zero field.
In the supplied ring apparatus, electric fields in the cavity add particle energy while magnetic fields bend the beam into the circular path.
For v perpendicular to B, equate |q|vB to mv²/R and derive R = mv/(|q|B); do not merely quote a show-that target.
Draw velocity tangent to a circular path and magnetic force or acceleration towards the centre at the actual particle position.
Use charge sign to determine the curvature or B direction; reverse the positive-charge force rule for an electron.
Magnetic force alone does no work, so speed and kinetic energy remain constant while the momentum vector changes direction.
At fixed speed, field and charge magnitude, a larger circular radius indicates larger mass; determine charge sign independently from curvature.
A neutral particle travels straight through the magnetic region, while positive and negative particles curve in opposite directions.
Use T = 2πR/v for a full revolution and derive f = |q|B/(2πm) by substituting the magnetic circular radius.
The straight-line gradient of an R–v graph at fixed B is m/(|q|B).
After voltage acceleration, first use |q|V = mv²/2; for fixed V the acceleration distance does not determine magnetic curvature radius.
If air removes kinetic energy from a particle in the magnetic field, speed falls and the radius shrinks into a spiral; B itself does no work.
For a given local field, qvB = mv²/r can determine initial curvature even if the field later varies; do not infer a uniform field or full circular path.
For an undeflected beam in crossed uniform fields, make electric and magnetic forces opposite and set E = vB.
Calculate the plate field with the full gap before finding B = E/v, converting voltages and lengths to SI units.
For a rightward positive beam in a downward electric field, B into the page supplies the upward magnetic force required for balance.
Increasing electron speed in fixed selector fields increases magnetic force while electric force stays fixed; state the actual direction of deflection.
A new particle with the same selected velocity remains undeflected because both forces scale with its charge, including a simultaneous sign reversal.
Separate a spectrometer’s crossed-field velocity selector from its magnetic-only curved deflection region.
For perpendicular magnetic force use |q|vB; a stationary charge has zero magnetic force regardless of field strength.
Reverse the positive-charge force for negative carriers before identifying the surface where electrons accumulate.
For an external proton near a wire, reverse electron flow to find conventional current, determine B at the proton, then apply qv × B.
Distinguish a carrier’s force from the whole conductor’s force: electrons in a right-moving rod with B into the page feel magnetic force down.
In the ideal neutral-wire laboratory model, magnetic-only interaction is explained by current with no net wire charge; no field transformation is needed.
A moving muon creates a magnetic field at another moving muon in the given ground frame; this short part asks no quantitative field transformation.
For the moving rod, combine ε = BvL with electron force eBv: divide emf by full rod length, then multiply by e.
Use F = BIL for a perpendicular conductor and F/L = BI when force per unit length is requested.
Use the real pole and conventional-current directions to choose conductor force; locate the arrow on the requested wire or coil.
Explain wire force through moving charge carriers in B, explicitly identifying magnetic force as acting on those carriers.
Rotation about the magnetic-field axis can leave a wire perpendicular to B, so a 30° rotation need not change force magnitude.
Clockwise current in an into-page field gives outward forces on a planar loop; apply the rule separately to each side.
For a multiturn circular coil, use total exposed length NπD and convert both magnetic-field and current prefixes before calculating BIL.
For an entering coil, obtain induced current from emf and resistance, then include the turns in the force on the exposed side.
Explain constant-speed rod motion through balanced external and magnetic forces, with rod current arising from the induced emf.
For a new battery-driven rail experiment, determine the current afresh; it can drive the rod in the opposite direction to prior induced drag.
Use ΔF = BLΔI for a fixed perpendicular conductor when only the current and force changes are given.
In the supplied motor, section XY remains perpendicular to B: its force magnitude stays constant while the commutator reverses its sign each half-turn.
Parallel same-direction currents attract and opposite currents repel; their mutual forces are equal and opposite even when current magnitudes differ.
For parallel cables driven in phase by one AC source, both currents reverse together, preserving attraction while force reaches two maxima per cycle.
In the balance apparatus, antiparallel currents push the lower wire down and increase the reading; explain the intermediate force direction.
For the specified rotated-wire arrangement, weaker field on parts of the wire gives smaller force and a lower balance reading; keep this qualitative.
For the shown oblique wires, find the source-wire field at Q before using Q’s current direction to infer its upward force.
If the mutual field at a wire is given, use F/L = BI; for several parallel wires combine pairwise forces with their inverse-distance dependence.
Use the supplied long-straight-conductor approximation for coil turns: each turn’s field acts on the other, producing attraction and reduced separation.

Covered syllabus points

D.3.1—Charge in electric fields D.3.2—Charge in magnetic fields D.3.3—Crossed electric and magnetic fields D.3.4—Magnetic force on charge D.3.5—Force on current conductor D.3.6—Force between parallel wires

Exam points

For a uniform plate field, combine E = V/d, F = qE and a = F/m to obtain acceleration magnitude |q|V/(md).
An electron accelerates towards the positive plate, opposite to the electric field; plate-parallel initial velocity does not change that direction.
Use the full gap to calculate E, then the actual charge and mass; an alpha particle has charge +2e and mass about 4u.
Nonparallel entry into a uniform electric field gives a parabola: perpendicular velocity stays constant while acceleration along the field is constant.
Equal transit times from rest over the same distance in the same uniform field imply the same charge-to-mass ratio, not identical charges and masses.
From a plate midpoint, use half the gap for transverse travel time, while retaining the full gap when calculating the electric field.
Find plate-parallel travel from x = vxt after calculating the transverse collision time; the electric field leaves vx unchanged.
For a show-that acceleration, display a correct force expression and division by mass, keeping enough precision to justify the stated scale.
In the supplied drift-tube accelerator, protons gain speed in the gaps; reverse polarity while they are inside conducting tubes where the field is zero.

[Maximum number: 1]

This question is in two parts. Part 1 is about photoelectricity. Part 2 is about electrical and magnetic force fields.
Part 1 Photoelectricity

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

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

Exam points

For velocity perpendicular to a uniform magnetic field, equate |q|vB to mv²/R to derive R = mv/(|q|B).
Draw velocity tangent to the circular path and force or acceleration towards its centre; use the actual particle position.
Use the magnetic-force rule with charge sign to infer curvature or field direction; reverse the positive-charge rule for an electron.
A magnetic force perpendicular to velocity does no work: speed and kinetic energy stay constant, while momentum direction changes.
Explain a circular orbit through a force perpendicular to velocity and directed towards the centre, with constant magnitude in the uniform field.
At fixed m, v and B, increasing charge magnitude reduces orbit radius; at fixed charge and speed, a larger mass gives a larger radius.
Infer charge sign and mass separately from a trajectory: curvature direction gives the sign, while radius gives mass only when speed and |q| match.
A neutral particle is undeflected by the magnetic field; distinguish its straight path from oppositely curved positive and negative tracks.
Calculate revolution time from T = 2πR/v; for the given proton orbit R = 0.54 m and v = 2.16 × 10⁶ m/s give T ≈ 1.6 μs.
Derive f = |q|B/(2πm) by combining circular magnetic force with f = v/(2πR); show the algebra rather than quoting the target.
For protons at B = 2.5 T, f = eB/(2πmp) gives about 3.8 × 10⁷ Hz.
On an R–v graph for fixed particle and field, the straight-line gradient is m/(|q|B).
After acceleration through voltage V, use |q|V = mv²/2 before finding magnetic radius; the acceleration distance does not set the gained energy.
For a magnetic orbit in air, attribute energy loss to the air: decreasing speed reduces radius and produces an inward spiral.
In the shown spectrometer deflection region, identify the magnetic field as the source of the curved ion trajectories.

[Maximum number: 5]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

(a)

An electron is moving at constant velocity $v=9.0 \times 10^{5} \mathrm{~ms}^{-1}$ parallel to a rod T , at a perpendicular distance d. T is connected to a cell via a switch. When the switch is closed the electron's path curves with initial radius r. The magnetic field strength at the position of the electron at d is $4.0 \times 10^{-6} \mathrm{~T}$ .

[ 5 ]

(i)

Calculate the initial radius of the electron's path.

[ 3 ]

use of $\mathrm{qvB}=\mathrm{mv}^{2} / \mathrm{r}$

OR
$\begin{aligned} & r=\frac{m v}{q B} & 1.6 \times 10^{-19} \times 4.0 \times 10^{-6}=\frac{9.11 \times 10^{-31} \times 9 \times 10^{5}}{r} \end{aligned}$

OR
$\begin{aligned} & r=\frac{\left(9.11 x 10^{-31}\right)\left(9 x 10^{5}\right)}{\left[\left(1.6 \times 10^{-19}\right)\left(4 x 10^{-6}\right)\right]} & r=1.28 \text { OR } 1.3 \text { «m» } \end{aligned}$

(ii)

Explain the subsequent change in the radius of the electron's path.

[ 2 ]

(b)

Calculate the magnitude of the force per unit length on R.

[ 2 ]

$\begin{aligned} \text { Use of } F=\text { BIL OR F/L = BI } =1.0 \text { « } \mathrm{N} \mathrm{~m}^{-1} \text { » } \text { } \end{aligned}$

(c)

Identify on the diagram the direction of the force on R.

[ 1 ]

Exam points

For an undeflected beam in perpendicular uniform fields, make electric and magnetic forces opposite and set |q|E = |q|vB.
Use v = E/B or B = E/v after calculating a plate field E = V/d with the full gap and SI units.
For a rightward positive beam with electric field downwards, a magnetic field into the page provides the upward force needed for balance.
Determine a requested field unit explicitly: E = vB can give 897 N/C, about 900 N/C, for the supplied proton selector.
If electron speed rises while the selector fields stay fixed, magnetic force grows and can exceed electric force; name the resulting deflection direction.
At the same selected velocity, changing particle mass or charge does not break balance; electric and magnetic forces scale by the same charge factor.
Distinguish a mass spectrometer’s crossed-field velocity selector from its magnetic-only curved deflection region.
Use a complete force-balance or field-ratio route for B and retain enough precision before rounding; do not combine marks from two alternative routes.

[Maximum number: 3]

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

(a)

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

Calculate the magnitude of the magnetic field.

[ 2 ]

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

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

Allow ECF from a) and b) i)

□

□

□

(b)

The alpha particle in (c) is replaced by an electron. The electron enters the region between the plates with the same velocity as the alpha particle.

Draw, on the diagram, the path of the electron.

[ 1 ]

Exam points

For motion perpendicular to B, calculate magnetic-force magnitude as |q|vB, using the actual charge, speed and field in SI units.
A stationary charge has zero magnetic force even in a nonzero magnetic field.
Use the charge sign after finding v × B; a falling negative charge in a rightward field experiences force into the page.
Draw magnetic force towards the centre of the given circular orbit and use |q|vB = mv²/R when a radius is requested.
For moving electrons in a conductor, reverse the positive-charge force direction before labelling the surface where negative charge accumulates.
Find a wire’s field from conventional current, reversing supplied electron flow first, then apply the force rule to the separate moving charge.
For electrons in a rod moving right through an into-page field, magnetic force is down; distinguish it from the force on the whole rod.
Use the rod relation ε = BvL to obtain Bv, then calculate the magnetic force on one electron as eBv.
In the ideal neutral-wire laboratory model, explain magnetic-only interaction by the wire current and absence of net wire charge.
In the ground frame of the given moving-muon example, a moving charge creates a magnetic field at the other moving charge.

[Maximum number: 1]

An electron moves in circular motion in a uniform magnetic field.

The velocity of the electron at point P is $6.8 \times 10^{5} \mathrm{~ms}^{-1}$ in the direction shown. The magnitude of the magnetic field is 8.5 T .

Calculate, in N , the magnitude of the magnetic force acting on the electron.

$1.60\times10^{-19}\times6.8\times10^{5}\times8.5=9.2\times10^{-13}\ \mathrm N$

[Maximum number: 5]

A rod, R , lies perpendicular to a uniform magnetic field B of strength 0.50 T directed into the plane of the page. R is connected to a circuit and the electric current $I_{\mathrm{R}}$ is 2.0 A .

(a)

[ 5 ]

(i)

Calculate the initial radius of the electron's path.

[ 3 ]

(ii)

Explain the subsequent change in the radius of the electron's path.

[ 2 ]

(b)

Calculate the magnitude of the force per unit length on R.

[ 2 ]

$\begin{aligned} \text { Use of } F=\text { BIL OR F/L = BI } =1.0 \text { « } \mathrm{N} \mathrm{~m}^{-1} \text { » } \text { } \end{aligned}$

(c)

Identify on the diagram the direction of the force on R.

[ 1 ]

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) generator to an electrical load. The copper wires are protected by an insulator.

Two identical cables connected in parallel are suspended a constant distance apart. Explain how the magnetic forces acting between the cables vary during the course of one cycle of the alternating current (ac).

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