IB Physics SL D.3 Motion in electromagnetic fields Questions

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Practise IB Physics D.3 by predicting charged-particle motion in uniform electric or magnetic fields, using crossed-field balance, and calculating forces on conductors and parallel wires.

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

Exam points & syllabus

Exam points

Use charge sign to set electric acceleration: an electron moves towards the positive plate, opposite to the electric-field direction.
Combine E=V/d, F=qE and F=ma using the full plate gap, correct particle charge and mass, and explicit steps in show-that calculations.
When a charge is held against gravity, compare weight with the changed electric force before predicting the direction of acceleration.
Separate plate-parallel motion from transverse acceleration; find collision time using the initial distance to the plate, then apply x=vₓt.
For acceleration from rest through a given voltage, equate |q|V to ½mv² and take the square root to obtain speed.
Distinguish the tangent velocity from the inward magnetic-force arrow on a circular track, and infer B direction using charge sign.
Explain circular magnetic motion by a force perpendicular to velocity and directed towards the centre; explain constant energy through zero work.
Equate |q|vB to mv²/r to derive r=mv/(|q|B), then substitute consistently and use the precision required by the question.
Use r=mv/(|q|B) to compare changes in mass, speed, charge and field; speed versus radius is linear through the origin for fixed m, q and B.
Use T=2πr/v for a complete revolution and combine it with the magnetic-force equation to derive f=|q|B/(2πm).
For particles with the same initial speed and charge magnitude, compare masses through radius and charge signs through curvature direction.
For a magnetic track spiralling inward in air, link decreasing energy or speed to decreasing radius without assigning the loss to magnetic work.
For undeflected crossed-field motion, balance opposing forces |q|E=|q|vB and use E=vB with the requested electric or magnetic field unit.
Choose B direction so the magnetic force opposes the electric force, checking the particle sign and the actual plate polarity.
Explain same-speed proton-to-alpha replacement in crossed fields: both forces scale with charge and remain balanced, independent of mass.
When plate data determine a crossed-field balance, use B=V/(vd) with the full separation, or use a verified prior electric-force result.
Check the angle in F=|q|vB sinθ before scaling force: motion parallel to B or a stationary charge gives zero magnetic force.
Near a current-carrying wire, find its local B first and then the particle force; reverse the positive-charge rule for negative carriers.
For a single carrier in a wire, use its drift velocity and charge in qvB rather than substituting the conductor’s total current.
Identify single-frame magnetic interactions of moving charges, and use flow velocity and carrier sign to predict charge collection in a liquid.
Use F=BIL sinθ with conventional current; determine force perpendicular to both I and B and assess each coil side’s actual angle.
Balance conductor magnetic force against weight; cancel length when mass per length is supplied, and use the stated field component.
Use ΔF=BLΔI for force-change data, and scale B and I together when comparing direct conductor forces.
Show the substitution or required precision in a current show-that task, and use the active perpendicular rod length for a sliding rod’s initial force.
For a circular coil with moving negative carriers, convert to conventional current and combine local radial forces before predicting its shape change.
For parallel wires, use relative current directions to identify mutual attraction or repulsion and explain the interaction through their magnetic fields.
Add every neighbouring wire’s force: equal opposite repulsions can cancel, while the nearer interaction can dominate an opposing farther one.
Calculate force per length with μ₀I₁I₂/(2πr), retaining N/m; scale the current product and separation and track reversals independently.

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 an electron between charged plates, direct force and acceleration opposite to the electric field and towards the positive plate.
Combine E=V/d, F=qE and F=ma to obtain a=|q|V/(md), using the full plate gap in the field calculation.
For an alpha particle, use charge 2e and mass about 4u explicitly; a correct charge-to-mass ratio alone may hide the wrong particle data.
For a charge held against gravity, compare electric force with weight after the change; reducing the upward electric force produces downward acceleration.
Separate plate-parallel motion from transverse acceleration: find collision time from the starting distance to a plate, then use x=vₓt.
In a show-that acceleration task, show the electric-force expression and division by mass rather than giving only the supplied rounded target.
For a particle accelerated from rest through a given voltage, equate |q|V to ½mv² and retain the square root when solving for speed.

[Maximum number: 2]

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.

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

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

Allow ECF from a).

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

[2]

Exam points

Use the charge sign, velocity and inward force to infer the magnetic-field direction; state into or out of the page explicitly when required.
Draw velocity tangent to the circular path in the correct sense, starting at the particle; label it clearly if other vectors are shown.
Explain a circular path by linking velocity perpendicular to B with a magnetic force that stays perpendicular to motion and acts towards the centre.
Explain constant speed or kinetic energy through perpendicular force and zero work; naming centripetal acceleration alone may be insufficient.
Derive the circular radius by equating |q|vB to mv²/r before rearranging to r=mv/(|q|B).
In radius calculations, show the substitution or required precision and round to the significant figures justified by the given data.
Use r=mv/(|q|B) to compare parameter changes and recognize that speed versus radius is linear through the origin for a fixed particle and field.
Calculate a full revolution time with T=2πr/v; derive f=|q|B/(2πm) by combining circular speed with the magnetic-force equation.
For same-speed particles with equal charge magnitudes, use radius to compare masses and the bending direction to identify charge sign.
For a track spiralling inward in air, link energy or speed loss to decreasing radius; do not attribute the energy loss to work by the magnetic force.

[Maximum number: 4]

(a)

A proton moves on a circular path in a region of uniform magnetic field of magnetic flux density B that is directed into the plane of the page.

[ 4 ]

(i)

On the diagram, draw an arrow to indicate the velocity of the proton at the position shown.

[ 1 ]

(ii)

Show that the frequency of revolution of the proton is given by $f=\frac{e B}{2 \pi m_{\mathrm{p}}}$ .

[ 3 ]

$\begin{aligned} e v B=\frac{m_{\mathrm{p}} v^{2}}{R} f=\frac{1}{T}=\frac{v}{2 \pi R} \end{aligned}$

Both q and e are acceptable for the
charge.

[3]

Algebra leading to required expression $\boldsymbol{<}=\frac{e B}{2 \pi m_{\mathrm{p}}} \boldsymbol{} \boldsymbol{}$

Exam points

For undeflected motion, equate the opposing electric and magnetic forces, |q|E=|q|vB, before using E=vB or B=E/v.
Set the magnetic-field direction so its force opposes the electric force; use the particle’s charge sign and state into or out of the page.
When plate voltage and separation are given, combine E=V/d with force balance to obtain B=V/(vd), using the full gap in metres.
Explain why replacing a proton by an alpha particle at the same velocity preserves undeflected motion: both forces scale with charge and still cancel.
Give the requested field unit: electric field may be N/C or V/m, while magnetic flux density is measured in tesla.

[Maximum number: 2]

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.

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

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

Allow ECF from a) and b) i)

[2]

Exam points

Use F=|q|vB for perpendicular motion and include sinθ for other angles; parallel motion gives zero force despite non-zero speed and field.
Determine magnetic-force direction from velocity and B, then reverse the positive-charge rule for electrons or negative ions.
For a charge near a current-carrying wire, find the wire’s local B direction first, then apply the moving-charge force rule with its charge sign.
For a force on one wire charge carrier, use its drift speed and elementary charge; do not substitute wire current for the individual carrier’s velocity.
Draw the magnetic-force arrow from the particle towards the centre of a circular path, rather than tangent to the path.
A stationary charge has no magnetic force because v=0 in the force law; this alone does not imply that its total force is zero.
In a neutral current-carrying wire’s laboratory description, distinguish the magnetic interaction with a moving charge from electrostatic attraction.
Explain the magnetic attraction of parallel-moving like charges through their moving-charge fields or the parallel-current analogy.
For charge carriers in flowing liquid, apply the force rule to their flow velocity and sign to identify the side where charge accumulates.

[Maximum number: 2]

In an experiment a student constructs the circuit shown in the diagram. The ammeter and the voltmeter are assumed to be ideal.

The diagram shows a cross-sectional view of the connecting wire in (c).

The wire which carries a current of 3.5 A into the page, is placed in a region of uniform magnetic field of flux density 0.25 T . The field is directed at right angles to the wire.

Determine the magnitude and direction of the magnetic force on one of the charge carriers in the wire.

$F=<q v B=1.6 \times 10^{-19} \times 2.0 \times 10^{-4} \times 0.25=>8.1 \times 10^{-24} \mathrm{~N}$

directed down OR south

[Maximum number: 3]

The long straight conductor is formed into a coil consisting of two separate turns, X and Y . The coil hangs with its axis vertical.

Assume that the turns of the coil each behave as a long straight conductor.

The current in the coil is 15 A and the circumference of one turn is 0.48 m . In order to restore X and Y to their original separation, a mass of $2.8 \times 10^{-4} \mathrm{~kg}$ is suspended from turn Y . Estimate the magnetic field strength at X due to Y .

A3. This question is about radioactivity.

Caesium-137 ( ${ }_{55}^{137} \mathrm{Cs}$ ) is a radioactive waste product with a half-life of 30 years that is formed during the fission of uranium. Caesium-137 decays by the emission of a beta-minus $\left(\beta^{-}\right)$ particle to form a nuclide of barium (Ba).

$F=\left(0.280 \times 10^{-3} \times 9.81=\right) 2.75 \times 10^{-3} \mathrm{~N}$ ;
$B=\frac{F}{I l}$ or correct substitution $2.75 \times 10^{-3}=B \times(15) \times 0.48$ ;
$B=\left(\frac{2.75 \times 10^{-3}}{15 \times 0.48}=\right) 3.8 \times 10^{-4} \mathrm{~T} ;$

[Maximum number: 3]

An electrically heated pad is designed to keep a pet warm.

The pad is heated using a resistor that is placed inside the pad. The dimensions of the resistor are shown on the diagram. The resistor has a resistance of $4.2 \Omega$ and a total length of 1.25 m .

When there is a current in the resistor, the temperature in the pad changes from a room temperature of $20^{\circ} \mathrm{C}$ to its operating temperature at $35^{\circ} \mathrm{C}$ .

(a)

When there is a current in the resistor, magnetic forces act between the resistor strips.

For the part of the resistor labelled RS,

[ 3 ]

(i)

outline the magnetic force acting on it due to the current in PQ .

[ 1 ]

(ii)

state and explain the net magnetic force acting on it due to the currents in PQ and TU .

[ 2 ]

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