D.3 Motion in electromagnetic fields – IB Physics exam questions with mark scheme

1Standard5 marks

A proton enters a region of uniform magnetic field of flux density 0.40 T at a speed of 2.5 × 10⁶ m s⁻¹. The velocity is perpendicular to the field, which is directed into the page.

(a) Calculate the magnetic force on the proton. [2]

(b) Calculate the radius of the proton's path. [2]

(a) F = qvB with correct substitution(1)
(a) F = 1.6 × 10⁻¹³ N(1)
(b) Magnetic force provides centripetal force: qvB = mv²/r so r = mv/(qB)(1)
(b) r = 0.065 m (6.5 × 10⁻² m)(1)
(c) Force is always perpendicular to velocity, so no work is done / speed unchanged(1)
Answer: (a) 1.6 × 10⁻¹³ N; (b) 0.065 m; (c) force ⟂ velocity so no work done
Examiner tip: Say the force is perpendicular to the motion so does no work – 'speed is constant' alone just restates the question.

2Hard5 marks

In a mass spectrometer, singly charged argon-40 ions (mass 6.64 × 10⁻²⁶ kg) are accelerated from rest through a potential difference of 2.0 kV and then enter a region of uniform magnetic field of flux density 0.25 T, perpendicular to their velocity.

(a) Calculate the speed of the ions on entering the field. [2]

(b) Calculate the radius of their circular path. [2]

(a) Gain in kinetic energy = work done by field: qV = ½mv²(1)
(a) v = 9.8 × 10⁴ m s⁻¹ (9.82 × 10⁴)(1)
(b) qvB = mv²/r so r = mv/(qB)(1)
(b) r = 0.16 m (allow 0.163 m)(1)
(c) r = 0.15 m (allow 0.155 m; allow 0.16 × √(5.98/6.64) = 0.152 m by ECF); r ∝ √m, so smaller than for Ar-40(1)
Answer: (a) 9.8 × 10⁴ m s⁻¹; (b) 0.16 m; (c) 0.15 m
Examiner tip: Use qV = ½mv² (not mv²); for the isotope in (c), r ∝ √m for the same charge, p.d. and B, so the lighter ion has the smaller radius.

3Core4 marks

Two parallel metal plates are 2.5 cm apart in a vacuum. A potential difference of 750 V is applied between them. An electron is released from rest at the surface of the negative plate.

(a) Calculate the magnitude of the electric field strength between the plates. [1]

(b) Calculate the magnitude of the acceleration of the electron and state its direction. [2]

(a) $E=\dfrac{V}{d}=\dfrac{750}{0.025}=3.0\times10^{4}\ \text{V m}^{-1}$ (accept N C⁻¹)(1)
(b) $a=\dfrac{eE}{m_e}=\dfrac{1.60\times10^{-19}\times3.0\times10^{4}}{9.11\times10^{-31}}=5.3\times10^{15}\ \text{m s}^{-2}$ (allow ECF from (a))(1)
(b) towards the positive plate / opposite to the direction of the electric field (do not accept 'towards the negative plate')(1)
(c) $eV=\tfrac12m_ev^2$ so $v=\sqrt{\dfrac{2\times1.60\times10^{-19}\times750}{9.11\times10^{-31}}}=1.6\times10^{7}\ \text{m s}^{-1}$ ; accept $v^2=2ad$ with ECF from (b)(1)
Answer: (a) $3.0\times10^4$ V m⁻¹; (b) $5.3\times10^{15}$ m s⁻² towards the positive plate; (c) $1.6\times10^{7}$ m s⁻¹
Examiner tip: Convert 2.5 cm to 0.025 m before using $E=V/d$, and remember the electron accelerates against the field direction because its charge is negative.

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4Standard6 marks

An electron travelling horizontally at $3.2\times10^{7}$ m s⁻¹ enters the region between two horizontal parallel plates, midway between them. The plates are 6.0 cm long and 3.0 cm apart, and the potential difference between them is 600 V, as shown. The region is a vacuum and gravitational effects are negligible.

(a) Show that the electric field strength between the plates is $2.0\times10^{4}$ V m⁻¹. [1]

(b) Explain why the path of the electron between the plates is parabolic. [2]

(a) $E=\dfrac{600}{0.030}=2.0\times10^{4}\ \text{V m}^{-1}$ (substitution must be seen for a 'show that')(1)
(b) the electric force is constant and vertical (upwards, towards the positive plate) so the vertical acceleration is constant / vertical velocity increases uniformly(1)
(b) there is no horizontal force so the horizontal velocity is constant; constant horizontal velocity with constant perpendicular acceleration gives a parabola (allow 'like a projectile')(1)
(c) time between plates $t=\dfrac{0.060}{3.2\times10^{7}}=1.875\times10^{-9}$ s(1)
(c) $a=\dfrac{eE}{m_e}=\dfrac{1.60\times10^{-19}\times2.0\times10^{4}}{9.11\times10^{-31}}=3.5\times10^{15}\ \text{m s}^{-2}$(1)
(c) $s=\tfrac12at^{2}=6.2\times10^{-3}$ m (6.2 mm) upwards / towards the positive plate (allow ECF; accept 6.1–6.2 mm)(1)
Answer: (b) constant vertical force with constant horizontal velocity; (c) $6.2\times10^{-3}$ m (towards the positive plate)
Examiner tip: Treat the two directions separately: the horizontal motion only fixes the time in the field; never put the 3.2 × 10⁷ m s⁻¹ into the vertical equation.

5Standard5 marks

A student uses a 'current balance' to measure the magnetic flux density between the poles of a magnet. The magnet sits on a top-pan balance and a rigid horizontal copper wire, clamped so it cannot move, passes between the poles at 90° to the horizontal magnetic field. The length of wire in the field is 5.0 cm.

When a current of 4.0 A flows in the wire, the balance reading increases by 1.84 g.

(a) Explain why the balance reading increases. [2]

(b) Calculate the magnetic flux density between the poles. [2]

(c) The wire is now rotated in the horizontal plane so that it makes an angle of 40° with the field. The current is unchanged. Calculate the new increase in the balance reading. [1]

(a) the magnetic field exerts a (vertical) force on the current-carrying wire, upwards (allow 'force $BIL$ on the wire')(1)
(a) by Newton's third law the wire exerts an equal and opposite (downward) force on the magnet, so the magnet pushes harder on the balance(1)
(b) $F=mg=1.84\times10^{-3}\times9.81=1.8\times10^{-2}$ N and $B=\dfrac{F}{IL}$ seen(1)
(b) $B=\dfrac{1.805\times10^{-2}}{4.0\times0.050}=0.090$ T (accept 0.09 T; allow 0.092 T with g = 10)(1)
(c) $F=BIL\sin40^\circ$ so increase $=1.84\sin40^\circ=1.2$ g (1.18 g; accept answer as 0.012 N)(1)
Answer: (b) 0.090 T; (c) 1.2 g
Examiner tip: The balance reads a mass, so convert the change in reading to a force with $F=mg$ (in kg) before using $F=BIL$; in (c) the angle is between the wire and the field, so use $\sin\theta$.

6Hard7 marks

The diagram shows a velocity selector. Positive ions enter from the left, moving horizontally. Between the plates there is a uniform electric field, due to the potential difference of 450 V across plates 1.5 cm apart, and a uniform magnetic field of flux density 0.060 T directed into the page. Ions travelling at one particular speed pass undeflected through the slit.

(a) On entering the fields, state the direction of (i) the electric force and (ii) the magnetic force on a positive ion. [1]

(b) Show that the speed of the undeflected ions is given by $v=\dfrac{E}{B}$, and calculate this speed. [3]

(d) State, with a reason, whether the same speed would be selected for doubly charged ions. [1]

(a) electric force downwards (towards the 0 V plate) and magnetic force upwards (both needed)(1)
(b) undeflected so forces balance: $qE=qvB$, giving $v=E/B$(1)
(b) $E=\dfrac{450}{0.015}=3.0\times10^{4}$ V m⁻¹(1)
(b) $v=\dfrac{3.0\times10^{4}}{0.060}=5.0\times10^{5}$ m s⁻¹ (allow ECF from E)(1)
(c) the magnetic force $qvB$ increases with speed but the electric force $qE$ does not, so the magnetic force is greater than the electric force(1)
(c) so faster ions are deflected upwards (towards the positive plate) and do not pass through the slit / hit the top plate or the barrier(1)
(d) yes, the same speed: $q$ cancels in $qE=qvB$ (both forces are proportional to charge) (accept 'selected speed is independent of charge and mass')(1)
Answer: (a) electric down, magnetic up; (b) $5.0\times10^5$ m s⁻¹; (c) deflected upwards, miss the slit; (d) same speed, $q$ cancels
Examiner tip: For 'explain' in (c), compare the two forces: say which one depends on speed and therefore which way the faster ions bend.

7StandardHL only4 marks

Two long, straight, parallel wires in air are 4.0 cm apart. Wire X carries a current of 12 A and wire Y carries a current of 5.0 A in the opposite direction.

(a) Calculate the magnitude of the force per unit length on wire Y. [2]

(b) State whether the wires attract or repel. [1]

(c) A student predicts that the force per unit length on wire X is larger because X carries the larger current. Comment on this prediction. [1]

(a) $\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}=\dfrac{4\pi\times10^{-7}\times12\times5.0}{2\pi\times0.040}$ (substitution, with r in m)(1)
(a) $3.0\times10^{-4}$ N m⁻¹(1)
(b) repel (currents in opposite directions)(1)
(c) prediction is wrong: the forces are equal in magnitude (and opposite) by Newton's third law / the formula is symmetrical in $I_1$ and $I_2$ (field of X × current of Y = field of Y × current of X)(1)
Answer: (a) $3.0\times10^{-4}$ N m⁻¹; (b) repel; (c) wrong: equal forces (Newton's third law)
Examiner tip: Use r in metres and both currents in the product; the force on each wire is the same even when the currents differ.

8Hard5 marks

In a fine-beam tube, electrons are accelerated from rest through a potential difference $V$ and then move at right angles to a uniform magnetic field of flux density 1.20 mT. The electrons travel in a circle of radius $r$, which is measured for several values of $V$.

$V$ / V	$r$ / cm
150	3.43
200	3.99
250	4.44
300	4.88
350	5.25

(a) Show that $r^{2}=\dfrac{2m}{eB^{2}}V$, where $m$ is the mass and $e$ the charge of the electron. [2]

(b) Use all the data to determine the specific charge $\dfrac{e}{m}$ of the electron. [3]

(a) $eV=\tfrac12mv^{2}$ and $evB=\dfrac{mv^{2}}{r}$ (so $r=\dfrac{mv}{eB}$)(1)
(a) eliminating $v$: $v=\dfrac{eBr}{m}$ substituted into $eV=\tfrac12mv^2$, rearranged to $r^{2}=\dfrac{2m}{eB^{2}}V$(1)
(b) $r^{2}$ values calculated (11.8, 15.9, 19.7, 23.8, 27.6 cm²) and plotted against $V$ / a best-fit line used (allow use of the mean of $2V/(r^2B^2)$ for all five rows)(1)
(b) gradient $\approx7.9\times10^{-6}$ m² V⁻¹ (accept $7.6$ to $8.2\times10^{-6}$; do not accept cm² unless converted)(1)
(b) $\dfrac{e}{m}=\dfrac{2}{\text{gradient}\times B^{2}}=1.8\times10^{11}$ C kg⁻¹ (accept 1.7 to $1.9\times10^{11}$; allow ECF from gradient)(1)
Answer: (b) $e/m\approx1.8\times10^{11}$ C kg⁻¹
Examiner tip: Linearise first: plot $r^2$ against $V$ and remember to convert cm² to m² (× 10⁻⁴) and mT to T before using the gradient.

9Hard6 marks

A narrow beam of protons, all moving with the same speed, enters a region of vacuum at right angles to a uniform field. In experiment 1 the field is a uniform electric field. In experiment 2 it is a uniform magnetic field.

Compare and contrast the motion of the protons in the two fields. In your answer refer to the shape of the path, the direction of the force, any change in speed and kinetic energy, and how the path would change if the protons entered with a greater speed. [6]

Electric field: force $qE$ is constant in magnitude and direction (parallel to the field lines)(1)
Electric field: constant acceleration perpendicular to the initial velocity, while velocity parallel to the plates is constant, so the path is a parabola(1)
Electric field: the force has a component along the velocity, so work is done on the protons; speed and kinetic energy increase(1)
Magnetic field: force $qvB$ is always perpendicular to the velocity (and to the field) so its direction continually changes(1)
Magnetic field: force of constant magnitude perpendicular to velocity is a centripetal force, so the path is a circular arc, radius $r=\dfrac{mv}{qB}$(1)
Magnetic field: no work is done (force ⊥ displacement), so speed and kinetic energy are constant(1)
Greater speed in E field: less time in the field so smaller deflection (deflection ∝ 1/v²); force unchanged(1)
Greater speed in B field: magnetic force larger but radius $\propto v$ is larger, so the path is less curved; period $2\pi m/(qB)$ unchanged(1)
Similarity: in both cases the force is proportional to the charge, and the protons are deflected (accelerate) in the plane perpendicular to the field (for B) / along the field (for E)(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, correct comparison of both fields covering path shape with reasons, force direction, energy (work done or not) and the effect of greater speed; uses correct physics terms and links ideas logically.
Level 2	3–4	Correctly describes the path in both fields and at least one of energy change or effect of speed, but some reasoning is missing or one field is treated only briefly.
Level 1	1–2	Some relevant points, e.g. identifies a parabola or a circle, with little explanation; may contain errors such as stating that the magnetic field changes the speed.
Level 0	0	No relevant physics.

Answer: E field: constant force, parabola, speed increases, faster protons deflect less. B field: force ⊥ velocity, circular arc, speed constant, faster protons have larger radius.
Examiner tip: The most common loss is saying the magnetic force speeds the protons up: it is always perpendicular to the velocity, so it does no work.

10Stretch6 marks

A proton (mass $1.67\times10^{-27}$ kg, charge $+e$) and an alpha particle (mass $6.64\times10^{-27}$ kg, charge $+2e$) each have a kinetic energy of 1.0 MeV. They enter, separately, a region of uniform magnetic field of flux density 0.50 T, moving perpendicular to the field, and each completes a semicircle before leaving the field.

(a) Show that the time taken for a particle of mass $m$ and charge $q$ to complete the semicircle is independent of its speed. [2]

(b) Calculate the time taken by the proton to complete the semicircle. [1]

(d) Calculate the time taken by the alpha particle to complete the semicircle. [1]

(a) $qvB=\dfrac{mv^{2}}{r}$ so $r=\dfrac{mv}{qB}$(1)
(a) time $=\dfrac{\pi r}{v}=\dfrac{\pi m}{qB}$, which contains no $v$ (allow $T=\dfrac{2\pi m}{qB}$ and time $=T/2$)(1)
(b) $t=\dfrac{\pi\times1.67\times10^{-27}}{1.60\times10^{-19}\times0.50}=6.6\times10^{-8}$ s(1)
(c) $mv=\sqrt{2mE_k}$ so $r=\dfrac{\sqrt{2mE_k}}{qB}$, i.e. $r\propto\dfrac{\sqrt{m}}{q}$ for the same $E_k$ and $B$(1)
(c) alpha: $\dfrac{\sqrt{4m_p}}{2e}=\dfrac{\sqrt{m_p}}{e}$ so radii (approximately) equal, OR both calculated: proton 0.289 m, alpha 0.288 m (≈ 0.29 m)(1)
(d) $t=\dfrac{\pi\times6.64\times10^{-27}}{3.20\times10^{-19}\times0.50}=1.3\times10^{-7}$ s (allow 'twice the proton's time' with value)(1)
Answer: (b) $6.6\times10^{-8}$ s; (c) both $r\approx0.29$ m; (d) $1.3\times10^{-7}$ s
Examiner tip: Use $q = 2e$ for the alpha particle and convert MeV to J (× 1.60 × 10⁻¹³) before finding the momentum.

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