Topic D. Fields HL TEST 1

Topic D Fields: Test 1 (HL)

D.1 Gravitational fields • D.2 Electric and magnetic fields

D.3 Motion in electromagnetic fields • D.4 Induction

Original IB-style topic assessment. Section A uses Paper 1A-style multiple-choice questions. Section B uses Paper 2-style structured questions and includes one investigation in the spirit of Paper 1B data analysis. This is a classroom test, not an official IB examination paper.

Instructions and data

Answer all questions. Each MCQ has one correct option and is worth 1 mark. No marks are deducted for incorrect answers.
A calculator and the IB Physics data booklet may be used.
Show working and include units in extended responses. Explain directions and signs when requested.
Give answers to 2–3 significant figures unless otherwise stated. Retain extra digits during calculations.
Use non-relativistic mechanics. Neglect fringing, resistance of ideal leads and self-induction unless stated otherwise.
All answers are in the boxed section at the bottom. Complete the test before consulting them.

$G=6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}},\qquad k=8.99\times10^9\,\mathrm{N\,m^2\,C^{-2}}$

$e=1.60\times10^{-19}\,\mathrm{C},\qquad m_p=1.67\times10^{-27}\,\mathrm{kg}$

$m_e=9.11\times10^{-31}\,\mathrm{kg},\qquad \mu_0=4\pi\times10^{-7}\,\mathrm{T\,m\,A^{-1}}$

Section A — Multiple-choice questions [25 marks]

Question 1 — D.1 [1 mark]

A spherical planet has radius $R$ and surface gravitational field strength $g_0$ . What is the field strength at a distance $3R$ from its centre?

$g_0/3$
$g_0/9$
$3g_0$
$9g_0$

Question 2 — D.1 [1 mark]

Gravitational potential is zero at infinity. A small mass moves from distance $r$ to $2r$ from the centre of an isolated spherical planet, with both positions outside the planet. What happens to its gravitational potential?

It becomes twice as negative.
It remains unchanged.
Its magnitude halves and it becomes positive.
Its magnitude halves and it becomes less negative.

Question 3 — D.1 [1 mark]

A mass $m$ is moved slowly from a planet’s surface, radius $R$ , to distance $2R$ from its centre. The planet has mass $M$ . What is the work done by the external agent? Neglect all other forces.

$GMm/(2R)$
$-GMm/(2R)$
$GMm/R$
$0$

Question 4 — D.1 [1 mark]

Two satellites have masses $m$ and $2m$ . Both are in circular orbits of the same radius around the same planet. How do their orbital periods compare?

The heavier satellite has twice the period.
The heavier satellite has half the period.
They are equal.
The heavier satellite has a period larger by a factor of square root of two.

Question 5 — D.1 [1 mark]

The circular orbital speed at distance $r$ from a planet’s centre is $v_c$ . What is the escape speed from that same position?

$\sqrt{2}v_c$
$v_c/\sqrt{2}$
$2v_c$
$v_c$

Question 6 — D.1 [1 mark]

A satellite of mass $m$ is in a circular orbit of radius $r$ around a planet of mass $M$ . Taking potential energy as zero at infinity, what is the satellite’s total mechanical energy?

$+GMm/(2r)$
$-GMm/r$
$0$
$-GMm/(2r)$

Question 7 — D.2 [1 mark]

An initially neutral object gains five electrons. What is its final charge?

$+8.0\times10^{-19}\,\mathrm{C}$
$-3.2\times10^{-20}\,\mathrm{C}$
$-8.0\times10^{-19}\,\mathrm{C}$
$+3.2\times10^{-20}\,\mathrm{C}$

Question 8 — D.2 [1 mark]

Point charges $+2.0\,\mu\mathrm{C}$ and $-3.0\,\mu\mathrm{C}$ are separated by $0.30\,\mathrm{m}$ in vacuum. What is the force between them?

$0.60\,\mathrm{N}$ , repulsive.
$0.60\,\mathrm{N}$ , attractive.
$0.18\,\mathrm{N}$ , attractive.
$0.18\,\mathrm{N}$ , repulsive.

Question 9 — D.2 [1 mark]

Two identical positive point charges are fixed at equal distances on opposite sides of point P. Potential is zero at infinity. Which statement about P is correct?

The electric field and potential are both zero.
The electric field is non-zero and the potential is zero.
The electric field is zero and the potential is negative.
The electric field is zero and the potential is positive.

Question 10 — D.2 [1 mark]

The electric potential at distance $r$ from an isolated negative point charge is $-V_0$ , where $V_0$ is positive. What is the potential at distance $2r$ ?

$-V_0/2$
$-2V_0$
$+V_0/2$
$0$

Question 11 — D.2 [1 mark]

Parallel plates have a potential difference of $240\,\mathrm{V}$ and separation $6.0\,\mathrm{mm}$ . What is the uniform electric field magnitude between them?

$1.4\,\mathrm{V\,m^{-1}}$
$40\,\mathrm{V\,m^{-1}}$
$4.0\times10^4\,\mathrm{V\,m^{-1}}$
$1.4\times10^3\,\mathrm{V\,m^{-1}}$

Question 12 — D.2 [1 mark]

The current in a long straight wire doubles. The observation point is also moved to twice its original distance from the wire. How does the magnetic field magnitude change?

It doubles.
It remains unchanged.
It halves.
It quadruples.

Question 13 — D.2 [1 mark]

Two long parallel wires carry currents in opposite directions. What is the magnetic interaction between them?

They repel.
They attract.
There is no force on either wire.
Only the wire carrying the larger current experiences a force.

Question 14 — D.3 [1 mark]

A proton initially at rest is accelerated through a potential difference of magnitude $1.50\times10^3\,\mathrm{V}$ . What is its speed? Neglect relativistic effects.

$2.68\times10^5\,\mathrm{m\,s^{-1}}$
$1.44\times10^{11}\,\mathrm{m\,s^{-1}}$
$7.58\times10^5\,\mathrm{m\,s^{-1}}$
$5.36\times10^5\,\mathrm{m\,s^{-1}}$

Question 15 — D.3 [1 mark]

A negative particle moves right through a magnetic field directed into the page. In which direction does the magnetic force act?

Right.
Left.
Down the page.
Up the page.

Question 16 — D.3 [1 mark]

An electron moves at $3.0\times10^6\,\mathrm{m\,s^{-1}}$ perpendicular to a uniform $0.20\,\mathrm{T}$ field. What is its circular path radius?

$1.7\times10^{-5}\,\mathrm{m}$
$8.5\times10^{-5}\,\mathrm{m}$
$8.5\times10^{-4}\,\mathrm{m}$
$1.7\times10^{-4}\,\mathrm{m}$

Question 17 — D.3 [1 mark]

A charged particle enters a uniform magnetic field with non-zero velocity components parallel and perpendicular to the field. Which statement describes its subsequent motion?

Both velocity components steadily decrease.
The perpendicular component remains constant in direction while the parallel component increases.
The particle moves in a straight line because the magnetic force does no work.
The parallel component remains constant while the perpendicular component produces circular motion.

Question 18 — D.3 [1 mark]

A velocity selector has crossed fields of magnitudes $E=6.0\times10^4\,\mathrm{V\,m^{-1}}$ and $B=0.20\,\mathrm{T}$ . The electric and magnetic forces oppose each other. Which speed gives an undeflected path?

$3.0\times10^5\,\mathrm{m\,s^{-1}}$
$1.2\times10^4\,\mathrm{m\,s^{-1}}$
$1.2\times10^5\,\mathrm{m\,s^{-1}}$
$3.0\times10^4\,\mathrm{m\,s^{-1}}$

Question 19 — D.3 [1 mark]

A non-relativistic particle moves in a circle perpendicular to a uniform magnetic field. Its speed doubles while its mass, charge and the field remain fixed. What happens to its orbital period?

It doubles.
It halves.
It remains unchanged.
It quadruples.

Question 20 — D.4 [1 mark]

A flat loop has area $0.015\,\mathrm{m^2}$ in a $0.40\,\mathrm{T}$ field. Its area normal makes 30° with the field. What is the flux magnitude through it?

$3.0\times10^{-3}\,\mathrm{Wb}$
$5.2\times10^{-3}\,\mathrm{Wb}$
$6.0\times10^{-3}\,\mathrm{Wb}$
$1.0\times10^{-2}\,\mathrm{Wb}$

Question 21 — D.4 [1 mark]

Flux through each turn of a 120-turn coil changes by $2.5\times10^{-4}\,\mathrm{Wb}$ in $0.050\,\mathrm{s}$ . What is the average induced emf magnitude?

$0.60\,\mathrm{V}$
$5.0\times10^{-3}\,\mathrm{V}$
$6.0\times10^{-3}\,\mathrm{V}$
$6.0\,\mathrm{V}$

Question 22 — D.4 [1 mark]

A magnet’s south pole moves away from a closed coil along its axis. Viewed from the magnet side, what is the induced conventional current direction?

Clockwise.
There is no induced current.
Anticlockwise.
The direction depends only on the coil’s resistance.

Question 23 — D.4 [1 mark]

A rod of active length $0.40\,\mathrm{m}$ moves at $2.5\,\mathrm{m\,s^{-1}}$ through a $0.50\,\mathrm{T}$ field. Rod, velocity and field are mutually perpendicular. What is the emf magnitude?

$0.20\,\mathrm{V}$
$1.25\,\mathrm{V}$
$5.0\,\mathrm{V}$
$0.50\,\mathrm{V}$

Question 24 — D.4 [1 mark]

The area of a rectangular generator coil doubles. Its number of turns, uniform field and rotation frequency remain fixed, and it rotates about an axis perpendicular to the field. What happens to peak emf and output frequency?

Both double.
Peak emf doubles and frequency is unchanged.
Both remain unchanged.
Peak emf is unchanged and frequency doubles.

Question 25 — D.4 [1 mark]

Current through a coil decreases. Which statement describes its self-induced emf?

It tends to sustain the original current.
It always acts against the original current.
It is zero because the coil is stationary.
It instantly reverses the original current.

Section B — Extended responses [52 marks]

Extended response 1 — D.1: A satellite and gravitational energy [10 marks]

A spherical planet has mass $M=6.00\times10^{24}\,\mathrm{kg}$ and radius $R=6.40\times10^6\,\mathrm{m}$ . A satellite of mass $m=600\,\mathrm{kg}$ travels in a circular orbit at height $h=1.60\times10^6\,\mathrm{m}$ above the surface. Neglect other bodies and atmospheric resistance. Take gravitational potential energy as zero at infinity.

(a) Calculate the orbital radius and gravitational field strength at the satellite. [2]

(b) Use Newton’s second law to derive the circular orbital speed expression and calculate the speed. [3]

(c) Calculate the orbital period. [2]

(d) Determine the total mechanical energy of the satellite and explain the significance of its sign. [2]

(e) State the minimum escape speed from the orbital position, neglecting all other bodies. [1]

Extended response 2 — D.2: Superposition, potential and work [10 marks]

Two fixed point charges lie on the x-axis in vacuum: $Q_1=+4.0\,\mathrm{nC}$ at $x=0$ and $Q_2=-1.0\,\mathrm{nC}$ at $x=0.30\,\mathrm{m}$ . Point P is at $x=0.10\,\mathrm{m}$ . Take electric potential as zero at infinity and positive direction to the right.

(a) Determine the resultant electric field at P, including its direction. [3]

(b) Calculate the electric potential at P. [2]

(c) A small test charge $q=-2.0\,\mathrm{nC}$ is placed at P. Calculate the electric force on it, including direction. [2]

(d) The test charge is moved slowly from P to infinity. Calculate the work done by an external agent and explain its sign. The two source charges remain fixed. [3]

Extended response 3 — D.3: A velocity selector and magnetic analyser [10 marks]

A beam of positive ions enters a velocity selector travelling right. The electric field has magnitude $E=4.0\times10^4\,\mathrm{V\,m^{-1}}$ and points down the page. The selector’s magnetic field $B_s=0.20\,\mathrm{T}$ points into the page. Undeflected ions then enter a separate uniform magnetic field $B_a=0.40\,\mathrm{T}$ out of the page, perpendicular to their velocity. Each ion has charge $q=+e$ and mass $m=3.34\times10^{-27}\,\mathrm{kg}$ . Neglect relativistic effects.

(a) Show that the selected speed is $v=E/B_s$ . [2]

(b) Calculate the selected speed. [1]

(c) Derive the radius expression for the analyser, calculate the radius and state whether the ions initially curve up or down the page. [4]

(d) Explain why the ions’ kinetic energy remains constant in the analyser. [2]

(e) Another ion has twice the mass and twice the positive charge. It passes through the same selector. State how its analyser radius compares with that of the original ion. [1]

Extended response 4 — D.4: Changing fields and generator output [10 marks]

A stationary rectangular coil has $N=80$ turns, each of area $A=0.010\,\mathrm{m^2}$ . Its plane lies in the page. A uniform field into the page increases linearly from $0.10\,\mathrm{T}$ to $0.50\,\mathrm{T}$ in $0.20\,\mathrm{s}$ . Total circuit resistance is $4.0\,\Omega$ . Neglect self-induction. Later, the same coil is rotated about an axis in its plane and perpendicular to the steady $0.50\,\mathrm{T}$ field at $10\,\mathrm{Hz}$ .

(a) Calculate the induced emf magnitude during the field increase. [2]

(b) Determine and explain the conventional current direction during the increase, viewed looking at the page. [2]

(c) Calculate the thermal energy dissipated during the 0.20 s increase. [2]

(d) For the rotating coil, calculate angular speed and peak emf. Use $\varepsilon_0=NBA\omega$ . [3]

(e) State the instantaneous emf when the rotating coil’s area normal is parallel to the field. [1]

Extended response 5 — Lab-style investigation: Motional emf [12 marks]

A vertical conducting rod slides right along horizontal rails through a uniform magnetic field directed into the page. The rod, velocity and field are mutually perpendicular. Active rod length is $L=0.300\,\mathrm{m}$ . A high-input-resistance voltmeter measures the emf with negligible circuit current. A motion sensor records the speed at the same time as each voltage measurement. The student obtains the following data:

Speed / m s⁻¹	Measured emf magnitude / V
0.40	0.060
0.80	0.119
1.20	0.181
1.60	0.239
2.00	0.300

(a) Identify the independent and dependent variables and state two quantities to keep constant. [2]

(b) Explain why a high-input-resistance voltmeter is used. [2]

(c) Describe the graph predicted by $|\varepsilon|=BLv$ and estimate the gradient using the first and last readings. [2]

(d) Use the gradient to determine $B$ . [1]

(e) A best-fit gradient is $S=(0.150\pm0.006)\,\mathrm{V\,s\,m^{-1}}$ and $L=(0.300\pm0.003)\,\mathrm{m}$ . Estimate the percentage and absolute uncertainty in $B$ . [2]

(f) State which end of the rod is positive. [1]

(g) In a later trial the voltmeter reads $+0.010\,\mathrm{V}$ when the rod is at rest. Explain the effect of this constant offset on the graph and how the readings should be corrected. [2]

Answers and worked solutions — Test 1

Section A: award 1 mark for each correct answer. Section B: suggested marking guidance accompanies the worked solutions. Accept equivalent correct methods and sensible rounding.

Section A — MCQ answers

Question	Answer	Explanation
1	B	$g=GM/r^2$ , so tripling the distance from the centre reduces the field to $g_0/9$ .
2	D	$V_g=-GM/r$ . At $2r$ , the potential is $-GM/(2r)$ , which is less negative.
3	A	$W_{\mathrm{ext}}=\Delta E_p=-GMm/(2R)-(-GMm/R)=GMm/(2R)$ . The external agent does positive work.
4	C	$T=2\pi\sqrt{r^3/(GM)}$ is independent of satellite mass.
5	A	$v_c=\sqrt{GM/r}$ and $v_{\mathrm{esc}}=\sqrt{2GM/r}$ , so $v_{\mathrm{esc}}=\sqrt{2}v_c$ .
6	D	$E_k=GMm/(2r)$ and $E_p=-GMm/r$ , giving $E_{\mathrm{total}}=-GMm/(2r)$ .
7	C	$Q=-5e=-5(1.60\times10^{-19})=-8.0\times10^{-19}\,\mathrm{C}$ .
8	B	$F=k\|q_1q_2\|/r^2=0.599\,\mathrm{N}\approx0.60\,\mathrm{N}$ . Unlike charges attract.
9	D	The vector fields cancel, but the scalar potentials add: $V_e=2kQ/r>0$ .
10	A	$V_e=kQ/r$ , so doubling distance halves the magnitude while retaining the negative sign.
11	C	$E=\|\Delta V_e\|/d=240/0.0060=4.0\times10^4\,\mathrm{V\,m^{-1}}$ .
12	B	$B=\mu_0 I/(2\pi r)$ . Doubling both $I$ and $r$ leaves $I/r$ unchanged.
13	A	Opposite currents produce repulsive forces. The forces on the wires have equal magnitudes and opposite directions.
14	D	$e\|\Delta V_e\|=\tfrac12m_pv^2$ , so $v=\sqrt{2e\|\Delta V_e\|/m_p}=5.36\times10^5\,\mathrm{m\,s^{-1}}$ .
15	C	For a positive charge, the force would be upwards. A negative charge experiences the opposite force, downwards.
16	B	$r=m_ev/(eB)=8.54\times10^{-5}\,\mathrm{m}$ . Use the charge magnitude.
17	D	The magnetic force acts only on the perpendicular component and changes its direction. The unchanged parallel component combines with circular motion to produce a helix.
18	A	$\|q\|E=\|q\|vB$ , giving $v=E/B=3.0\times10^5\,\mathrm{m\,s^{-1}}$ .
19	C	$T=2\pi m/(\|q\|B)$ . Radius and speed both double, so their ratio is unchanged.
20	B	$\Phi=BA\cos30^\circ=5.20\times10^{-3}\,\mathrm{Wb}$ .
21	A	$\|\varepsilon_{\mathrm{av}}\|=N\|\Delta\Phi\|/\Delta t=120(2.5\times10^{-4})/0.050=0.60\,\mathrm{V}$ .
22	C	The coil’s near face becomes north to attract the receding south pole and oppose the flux decrease. A north face corresponds to anticlockwise current.
23	D	$\|\varepsilon\|=BvL=(0.50)(2.5)(0.40)=0.50\,\mathrm{V}$ .
24	B	$\varepsilon_0=NBA\omega$ doubles with area. The number of rotations per second, and hence output frequency, remains unchanged.
25	A	Self-induced emf opposes the change in current and associated flux linkage. During a decrease it tends to maintain current in its original direction.

Section B — Worked solutions

Extended response 1 — A satellite and gravitational energy [10 marks]

(a) Distances in the field equations are measured from the planet’s centre.

$r=R+h=8.00\times10^6\,\mathrm{m}$

$g=\frac{GM}{r^2}=\frac{(6.67\times10^{-11})(6.00\times10^{24})}{(8.00\times10^6)^2}=6.25\,\mathrm{N\,kg^{-1}}$

(b) Gravity supplies the centripetal force:

$\frac{GMm}{r^2}=\frac{mv^2}{r}\quad\Rightarrow\quad v=\sqrt{\frac{GM}{r}}=7.07\times10^3\,\mathrm{m\,s^{-1}}$

$\text{(c)}\quad T=\frac{2\pi r}{v}=7.11\times10^3\,\mathrm{s}\approx1.97\,\mathrm{h}$

$\text{(d)}\quad E_{\mathrm{total}}=\frac12mv^2-\frac{GMm}{r}=-\frac{GMm}{2r}=-1.50\times10^{10}\,\mathrm{J}$

The negative total energy means the satellite is gravitationally bound. Additional energy is needed for it to reach infinity with zero final speed.

$\text{(e)}\quad v_{\mathrm{esc}}=\sqrt{\frac{2GM}{r}}=1.00\times10^4\,\mathrm{m\,s^{-1}}$

Suggested marking: (a) Orbital radius and field calculation, 1 each. (b) Force equality, rearrangement and speed, 1 each. (c) Period method and answer, 1 each. (d) Total energy and bound-state interpretation, 1 each. (e) Escape speed, 1.

Extended response 2 — Superposition, potential and work [10 marks]

(a) At P, the field of the positive charge points right, away from $Q_1$ . The field of the negative charge also points right, towards $Q_2$ . Add their magnitudes:

$E_1=\frac{k|Q_1|}{(0.10)^2}=3.596\times10^3\,\mathrm{N\,C^{-1}}$

$E_2=\frac{k|Q_2|}{(0.20)^2}=224.75\,\mathrm{N\,C^{-1}}$

$E=E_1+E_2=3.82\times10^3\,\mathrm{N\,C^{-1}}\quad\text{to the right}$

(b) Potentials are scalars, so retain the signs of the charges:

$V_e=k\left(\frac{Q_1}{0.10}+\frac{Q_2}{0.20}\right)=314.65\,\mathrm{V}\approx315\,\mathrm{V}$

$\text{(c)}\quad F=qE=(-2.0\times10^{-9})(3.82075\times10^3)=-7.64\times10^{-6}\,\mathrm{N}$

The force magnitude is $7.6\times10^{-6}\,\mathrm{N}$ , directed left because the test charge is negative.

$\text{(d)}\quad W_{\mathrm{ext}}=\Delta E_p=q(V_{e,f}-V_{e,i})$

$W_{\mathrm{ext}}=(-2.0\times10^{-9})(0-314.65)=+6.29\times10^{-7}\,\mathrm{J}$

Positive external work is required: the test charge’s potential energy increases from a negative value to zero. Slow motion means the kinetic energy change is negligible.

Suggested marking: (a) Correct field directions, individual contributions and resultant, 1 each. (b) Signed potential sum and answer, 1 each. (c) Force magnitude and direction, 1 each. (d) Energy-change method, numerical work and sign explanation, 1 each.

Extended response 3 — A velocity selector and magnetic analyser [10 marks]

(a) The electric force on a positive ion points down. The magnetic force in the selector points up. An undeflected ion has equal force magnitudes:

$qE=qvB_s\quad\Rightarrow\quad v=\frac{E}{B_s}$

$\text{(b)}\quad v=\frac{4.0\times10^4}{0.20}=2.0\times10^5\,\mathrm{m\,s^{-1}}$

(c) In the analyser, magnetic force supplies centripetal force:

$|q|vB_a=\frac{mv^2}{r}\quad\Rightarrow\quad r=\frac{mv}{|q|B_a}$

$r=\frac{(3.34\times10^{-27})(2.0\times10^5)}{(1.60\times10^{-19})(0.40)}=1.04\times10^{-2}\,\mathrm{m}$

The positive ions initially curve down the page, following clockwise circular motion in the analyser field out of the page.

(d) The magnetic force is perpendicular to the instantaneous velocity and does no work. Therefore kinetic energy and speed remain constant, although velocity changes direction.

(e) The radius is unchanged. Selected speed is independent of charge and mass, and doubling both mass and charge leaves $m/|q|$ unchanged.

Suggested marking: (a) Force balance and selected-speed expression, 1 each. (b) Speed, 1. (c) Centripetal-force equality, radius expression, numerical radius and curvature direction, 1 each. (d) Perpendicular force and no work/constant energy, 1 each. (e) Same radius, 1.

Extended response 4 — Changing fields and generator output [10 marks]

$\text{(a)}\quad |\varepsilon|=NA\frac{\Delta B}{\Delta t}=80(0.010)\frac{0.50-0.10}{0.20}=1.6\,\mathrm{V}$

(b) Anticlockwise. Flux into the page is increasing, so the induced current produces a field out of the page to oppose that increase.

$\text{(c)}\quad I=\frac{|\varepsilon|}{R}=\frac{1.6}{4.0}=0.40\,\mathrm{A}$

$E_{\mathrm{thermal}}=I^2R\Delta t=(0.40)^2(4.0)(0.20)=0.128\,\mathrm{J}\approx0.13\,\mathrm{J}$

$\text{(d)}\quad \omega=2\pi f=20\pi\,\mathrm{rad\,s^{-1}}\approx62.8\,\mathrm{rad\,s^{-1}}$

$\varepsilon_0=NBA\omega=80(0.50)(0.010)(20\pi)=25.1\,\mathrm{V}$

(e) Zero. Flux is at an extremum when the area normal is parallel to the field, so its instantaneous rate of change is zero.

Suggested marking: (a) Faraday method and emf, 1 each. (b) Current direction and Lenz explanation, 1 each. (c) Current/power method and energy, 1 each. (d) Angular speed, peak-emf substitution and answer, 1 each. (e) Zero emf, 1.

Extended response 5 — Investigating motional emf [12 marks]

(a) Independent variable: rod speed. Dependent variable: induced emf magnitude. Keep active rod length, magnetic field strength and perpendicular orientation fixed; state any two suitable controls.

(b) Its high resistance makes current negligible, so the measured terminal voltage closely approximates the induced emf. It also minimises current-dependent voltage drops and magnetic braking of the rod.

(c) Plot emf magnitude vertically against speed horizontally. The model predicts a straight line through the origin. Small deviations in the data are consistent with measurement scatter.

$S=\frac{0.300-0.060}{2.00-0.40}=0.150\,\mathrm{V\,s\,m^{-1}}$

$\text{(d)}\quad S=BL\quad\Rightarrow\quad B=\frac{S}{L}=\frac{0.150}{0.300}=0.500\,\mathrm{T}$

$\text{(e)}\quad \frac{\Delta B}{B}\approx\frac{\Delta S}{S}+\frac{\Delta L}{L}=\frac{0.006}{0.150}+\frac{0.003}{0.300}=0.050$

The estimated percentage uncertainty is $5.0\%$ .

$\Delta B=(0.050)(0.500)=0.025\,\mathrm{T}$

An appropriate rounded result is $B=(0.50\pm0.03)\,\mathrm{T}$ . The unrounded absolute uncertainty is $0.025\,\mathrm{T}$ .

(f) The upper end is positive. Electrons redistribute towards the lower end.

(g) A constant positive offset shifts the graph upwards, producing a positive vertical intercept. It does not change the gradient if it is constant at all speeds. Subtract $0.010\,\mathrm{V}$ from each signed reading, or zero the voltmeter before taking readings.

Suggested marking: (a) Both variables, 1; two suitable controls, 1. (b) Negligible current, 1; voltage close to emf or reduced loading/braking, 1. (c) Linear graph through origin, 1; gradient with units, 1. (d) Field strength, 1. (e) Percentage uncertainty and absolute uncertainty, 1 each. (f) Upper end positive, 1. (g) Constant vertical shift and correction, 1 each.

IB Physics HL Book 2025; Papers 1 & 2 Practice 5 Sections65 Lessons54 Weeks

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