ExaminerPrep › Question bank › IAL Physics › 4.5 Magnetic fields and forces
Edexcel International A Level Physics · Unit 4 · Further Mechanics, Fields and Particles
40 free, high-quality past-paper-style questions on Magnetic fields and forces for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which of the following is equivalent to the tesla, the unit of magnetic flux density?
| Answer | Mark | Notes |
|---|---|---|
| From F = BIl for a wire at right angles to the field, B = F/Il, so 1 T = 1 N ÷ (1 A × 1 m) = 1 N A−1 m−1. (It is also 1 Wb m−2, flux per unit area.) | ✓ 1 | Answer D |
Answer: D
Exam tip: Derive units from a defining equation rather than memorising them: rearrange F = BIl for B and substitute the units.
A flat circular coil of diameter 4.0 cm has 150 turns. It is placed with its plane at right angles to a uniform magnetic field of flux density 0.12 T.
What is the flux linkage of the coil?
| Answer | Mark | Notes |
|---|---|---|
| Radius = 0.020 m, so A = π × 0.0202 = 1.26 × 10−3 m2. Nφ = NBA = 150 × 0.12 × 1.26 × 10−3 = 2.3 × 10−2 Wb turns (0.0226). | ✓ 1 | Answer C |
Answer: C
Common error: Using the diameter in πr2. Halve it first — using the diameter makes the area four times too big.
A straight wire carries a current of 3.0 A. A 0.15 m length of the wire lies in a uniform magnetic field of flux density 0.040 T, at an angle of 30° to the field lines.
What is the size of the force on the wire?
| Answer | Mark | Notes |
|---|---|---|
| F = BIl sin θ = 0.040 × 3.0 × 0.15 × sin 30° = 0.0090 N. | ✓ 1 | Answer A |
Answer: A
Exam tip: In F = BIl sin θ, θ is the angle between the wire and the field. Only the component of the wire across the field feels a force.
A small rectangular search coil measures 1.2 cm × 2.5 cm and has 400 turns. It is held between the poles of an electromagnet with its plane at right angles to a uniform field of flux density 0.36 T.
Calculate the magnetic flux through one turn of the coil.
Calculate the flux linkage of the coil.
The coil is rotated through 90° so that its plane is parallel to the field. State the new flux linkage.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = 0.012 × 0.025 = 3.0 × 10−4 m2 | ✓ 1 | |
| φ = BA = 0.36 × 3.0 × 10−4 = 1.1 × 10−4 Wb (1.08 × 10−4 Wb) | ✓ 1 | |
| Part (b) | ||
| Nφ = 400 × 1.08 × 10−4 = 0.043 Wb turns (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| zero — no field lines pass through the coil | ✓ 1 | |
Common error: Squaring or multiplying lengths still in centimetres. 1 cm2 = 10−4 m2, so convert each side to metres first.
In a moving-coil loudspeaker, a coil of wire sits in a strong magnetic field that points radially outwards from the centre, so the field is everywhere at right angles to the wire. The total length of wire in the field is 3.2 m and the flux density is 0.90 T. The coil is attached to a light paper cone.
Explain why there is a force on the coil when there is a current in it.
Calculate the size of the force on the coil when the current is 0.25 A.
The loudspeaker is supplied with an alternating current. Explain why the cone vibrates.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a current-carrying conductor at an angle to / at right angles to a magnetic field experiences a force (the motor effect) | ✓ 1 | |
| Part (b) | ||
| F = BIl = 0.90 × 0.25 × 3.2 | ✓ 1 | |
| = 0.72 N | ✓ 1 | |
| Part (c) | ||
| the current keeps reversing, so the direction of the force reverses / the force changes direction at the frequency of the current, pushing the cone back and forth | ✓ 1 | |
Exam tip: When a field is described as radial or as at right angles to the wire everywhere, sin θ = 1 and F = BIl uses the whole length of wire in the field.
An electron travels at 4.0 × 106 m s−1 at right angles to a uniform magnetic field of flux density 5.0 mT.
Calculate the size of the magnetic force on the electron.
Explain why the speed of the electron does not change while it is in the field.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = Bqv = 5.0 × 10−3 × 1.60 × 10−19 × 4.0 × 106 | ✓ 1 | |
| = 3.2 × 10−15 N | ✓ 1 | |
| Part (b) | ||
| the magnetic force is always at right angles to the electron's velocity / direction of motion | ✓ 1 | |
| so no work is done on the electron and its kinetic energy (and speed) stays constant | ✓ 1 | |
Exam tip: Convert mT to T (× 10−3) before substituting. For the explanation, the key link is perpendicular force → no work done → constant kinetic energy.
The figure shows a flat rectangular wire loop drawn on a grid of 1 cm squares. Part of the loop is inside a region of uniform magnetic field of flux density 0.25 T, directed into the page. The plane of the loop is at right angles to the field.
What is the magnetic flux through the loop?
| Answer | Mark | Notes |
|---|---|---|
| Only the part of the loop inside the field counts. From the grid, that part is 3 cm × 4 cm = 12 cm2 = 1.2 × 10−3 m2. φ = BA = 0.25 × 1.2 × 10−3 = 3.0 × 10−4 Wb. | ✓ 1 | Answer A |
Answer: A
Exam tip: In φ = BA, A is the area that the field actually passes through. When a loop is only partly in the field, use the overlap, and convert cm2 to m2 by multiplying by 10−4.
A straight wire is placed at right angles to a uniform magnetic field, with 5.0 cm of the wire in the field. The graph shows how the force F on the wire varies with the current I in it.
What is the magnetic flux density?
| Answer | Mark | Notes |
|---|---|---|
| From the marked point, F = 30 mN = 0.030 N when I = 4.0 A. B = F/Il = 0.030 ÷ (4.0 × 0.050) = 0.15 T. (Equivalently, gradient = Bl = 7.5 × 10−3 N A−1, and B = gradient ÷ l.) | ✓ 1 | Answer B |
Answer: B
Exam tip: Read a point that lies on the line at a gridline intersection, and convert both mN to N and cm to m before you divide.
A flat coil of 120 turns, each of area 25 cm2, is turned slowly in a uniform magnetic field. The graph shows how the flux linkage Nφ of the coil varies with the angle θ between the field and the normal to the coil.
What is the magnetic flux density?
| Answer | Mark | Notes |
|---|---|---|
| The greatest flux linkage, when the field is along the normal (θ = 0), is NBA = 0.060 Wb turns. A = 25 × 10−4 m2 = 2.5 × 10−3 m2, so B = 0.060 ÷ (120 × 2.5 × 10−3) = 0.20 T. | ✓ 1 | Answer C |
Answer: C
Exam tip: The flux linkage is greatest when the coil faces the field (normal along B), so read the peak of the graph and use Nφ = NBA.
A flat coil of 80 turns, each a square of side 5.0 cm, is held in a uniform magnetic field of flux density 0.15 T. The figure shows the coil edge-on, with its normal at 40° to the field.
Calculate the magnetic flux through one turn of the coil in the position shown.
Calculate the flux linkage of the coil.
State the smallest angle through which the coil must be turned to give the greatest possible flux linkage, and calculate this flux linkage.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A = 0.0502 = 2.5 × 10−3 m2 | ✓ 1 | |
| φ = BA cos 40° = 0.15 × 2.5 × 10−3 × cos 40° = 2.9 × 10−4 Wb (2.87 × 10−4 Wb) | ✓ 1 | |
| Part (b) | ||
| Nφ = 80 × 2.87 × 10−4 = 0.023 Wb turns (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| turn through 40°, so that the normal is along the field / the plane of the coil is at right angles to the field | ✓ 1 | |
| NBA = 80 × 0.15 × 2.5 × 10−3 = 0.030 Wb turns | ✓ 1 | |
Common error: Using cos of the angle between the field and the plane of the coil. Flux uses the angle between the field and the normal: here the field makes 50° with the plane and 40° with the normal.
The figure shows an electron about to enter a region of uniform magnetic field at point X, moving at right angles to the field.
State the direction of the magnetic force on the electron at X.
Calculate the size of this force.
A proton enters the field at X with the same velocity. State two ways in which its path in the field differs from the path of the electron, giving a reason for each.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| towards the top of the page (conventional current to the left, field out of the page, Fleming's left-hand rule) | ✓ 1 | |
| Part (b) | ||
| F = Bqv = 0.080 × 1.60 × 10−19 × 3.5 × 107 | ✓ 1 | |
| = 4.5 × 10−13 N (4.48 × 10−13 N) | ✓ 1 | |
| Part (c) | ||
| it curves the opposite way, towards the bottom of the page, because its charge is positive / of opposite sign | ✓ 1 | |
| its path has a much larger radius, because r = mv/BQ and its mass is much greater (same v, B and size of charge) | ✓ 1 | |
Exam tip: For a negative charge, point the second finger of the left hand opposite to the velocity. Then check the answer: reversing the sign of the charge must reverse the force.
The figure shows a simple electric motor seen from one end. The rectangular coil has 60 turns and carries a current of 0.80 A. Sides P and Q of the coil are each 5.0 cm long (at right angles to the page) and lie in a uniform magnetic field of flux density 0.12 T.
State the direction of the magnetic force on side P and on side Q.
Calculate the size of the force on side P.
State the effect of the two forces on the coil.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P: towards the top of the page | ✓ 1 | |
| Q: towards the bottom of the page | ✓ 1 | |
| Part (b) | ||
| F = NBIl = 60 × 0.12 × 0.80 × 0.050 | ✓ 1 | |
| = 0.29 N (0.288 N) | ✓ 1 | |
| Part (c) | ||
| they form a couple / turning effect, so the coil turns clockwise (as seen in the figure) about the axle | ✓ 1 | |
Common error: Leaving out the number of turns. Each turn carries the current through the field, so the force on one side of the coil is N times BIl.
An electron moves from left to right across the page through a uniform magnetic field directed into the page.
In which direction is the magnetic force on the electron when it enters the field?
| Answer | Mark | Notes |
|---|---|---|
| For an electron the conventional current is opposite to its velocity: right to left. Fleming's left-hand rule with the field into the page and current to the left gives a force towards the bottom of the page. (A proton moving the same way would be pushed towards the top.) | ✓ 1 | Answer A |
Answer: A
Common error: Pointing the second finger along the electron's velocity. Fleming's rule uses conventional current, which is opposite to the motion of a negative charge.
An alpha particle (charge +3.2 × 10−19 C) moves at 1.5 × 106 m s−1 in a uniform magnetic field of flux density 0.20 T. Its velocity makes an angle of 30° with the field.
What is the size of the magnetic force on it?
| Answer | Mark | Notes |
|---|---|---|
| F = Bqv sin θ = 0.20 × 3.2 × 10−19 × 1.5 × 106 × sin 30° = 4.8 × 10−14 N. | ✓ 1 | Answer B |
Answer: B
Exam tip: Only the component of the velocity at right angles to the field, v sin θ, contributes to the magnetic force.
A proton moving at right angles to a uniform magnetic field follows a circular path of radius 12 cm.
The proton's speed is doubled and the magnetic flux density is halved. What is the new radius?
| Answer | Mark | Notes |
|---|---|---|
| r = mv/BQ, so r ∝ v/B. Doubling v doubles r; halving B doubles it again: 12 × 2 × 2 = 48 cm. | ✓ 1 | Answer D |
Answer: D
Exam tip: For a ratio question, write the equation as a proportionality first (r ∝ v/B here) and apply each change as a factor.
A magnet stands on an electronic balance. A stiff, clamped horizontal wire passes between its poles at right angles to the field, with 6.0 cm of the wire in the field. When the current in the wire is 3.0 A, the balance reading increases by 2.4 g.
What is the magnetic flux density between the poles?
| Answer | Mark | Notes |
|---|---|---|
| Force = change in reading × g = 2.4 × 10−3 × 9.81 = 0.0235 N. B = F/Il = 0.0235 ÷ (3.0 × 0.060) = 0.13 T. | ✓ 1 | Answer C |
Answer: C
Common error: Treating the change in balance reading as a force. Convert grams to kilograms and multiply by g to get newtons.
A flat coil of 200 turns and area 3.0 × 10−3 m2 is in a uniform magnetic field of flux density 0.050 T. The plane of the coil makes an angle of 30° with the field lines.
What is the flux linkage of the coil?
| Answer | Mark | Notes |
|---|---|---|
| The plane is at 30° to the field, so the normal to the coil is at 60° to the field. Nφ = NBA cos 60° = 200 × 0.050 × 3.0 × 10−3 × 0.50 = 1.5 × 10−2 Wb turns. | ✓ 1 | Answer B |
Answer: B
Exam tip: Check whether an angle is measured to the plane of the coil or to its normal. Flux uses the component of the area at right angles to the field: A cos(angle between field and normal).
Two long, straight, parallel wires P and Q are a few centimetres apart in a laboratory. Both carry currents in the same direction. A student also has a long solenoid connected to a d.c. supply.
Describe the pattern of the magnetic field inside and outside the solenoid.
State how the right-hand grip rule shows which end of the solenoid behaves as a north pole.
Explain why wires P and Q attract each other.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| inside: strong and uniform — parallel, equally spaced field lines along the axis | ✓ 1 | |
| outside: weaker; lines spread out from one end and loop round to the other, like the field of a bar magnet | ✓ 1 | |
| Part (b) | ||
| curl the fingers of the right hand in the direction of the (conventional) current in the turns: the thumb points to the north-pole end / the end the field lines leave | ✓ 1 | |
| Part (c) | ||
| the current in P produces a (circular) magnetic field at Q, and Q is a current-carrying wire in that field, so there is a force on Q | ✓ 1 | |
| applying the grip rule and then Fleming's left-hand rule shows the force on Q is towards P (and by Newton's third law the force on P is towards Q) | ✓ 1 | |
Exam tip: For a force between two currents, use two rules in turn: the right-hand grip rule for the field one wire makes at the other, then Fleming's left-hand rule for the force on the other wire.
A student uses a current balance to measure the flux density between the poles of a magnet. The magnet rests on an electronic balance and a clamped horizontal wire passes between its poles, at right angles to the field, with 4.5 cm of the wire in the field. The balance is zeroed with no current.
Results: I = 1.0 A, increase in reading Δm = 0.56 g; 2.0 A, 1.09 g; 3.0 A, 1.66 g; 4.0 A, 2.19 g; 5.0 A, 2.76 g.
Explain why the balance reading increases when there is a current in the wire, even though the wire does not touch the magnet.
The gradient of her best-fit line of Δm against I is 0.550 g A−1. Determine the magnetic flux density.
The student repeats each reading with the current reversed and averages the sizes of the two changes. Suggest why.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the field exerts an (upward) force on the current-carrying wire | ✓ 1 | |
| by Newton's third law the wire exerts an equal and opposite (downward) force on the magnet, which pushes harder on the balance | ✓ 1 | |
| Part (b) | ||
| force per unit current = 0.550 × 10−3 × 9.81 = 5.40 × 10−3 N A−1 | ✓ 1 | |
| B = (F/I) ÷ l = 5.40 × 10−3 ÷ 0.045 | ✓ 1 | |
| = 0.12 T (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| to cancel the effect of any change in the reading that does not reverse with the current, e.g. convection currents or expansion caused by the wire heating up / a drift in the balance zero | ✓ 1 | |
Model sketch
Exam tip: A current balance gives a change in mass. Convert to kilograms and multiply by g: only then is it a force you can use in F = BIl.
In an instrument used to separate isotopes, singly charged positive lithium ions all moving at 3.0 × 105 m s−1 enter a uniform magnetic field of flux density 0.20 T at right angles to it. They travel through a semicircle and strike a detector plate. Mass of a lithium-7 ion = 1.16 × 10−26 kg; mass of a lithium-6 ion = 9.98 × 10−27 kg.
Show that the radius of the path of an ion is given by r = mv/BQ.
Calculate the radius of the path of a lithium-7 ion.
Calculate the distance between the points where the two isotopes strike the detector plate.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the magnetic force provides the centripetal force: BQv = mv2/r | ✓ 1 | |
| cancelling v and rearranging gives r = mv/BQ | ✓ 1 | |
| Part (b) | ||
| r = (1.16 × 10−26 × 3.0 × 105) ÷ (0.20 × 1.60 × 10−19) | ✓ 1 | |
| = 0.11 m (0.109 m) | ✓ 1 | |
| Part (c) | ||
| radius for lithium-6 = (9.98 × 10−27 × 3.0 × 105) ÷ (3.2 × 10−20) = 0.0936 m | ✓ 1 | |
| separation = 2 × (0.109 − 0.0936) = 0.030 m (allow error carried forward) | ✓ 1 | |
Common error: Subtracting the radii instead of the diameters. After a semicircle each ion lands one diameter, 2r, from the entry point.
A horizontal metal rod of mass 25 g hangs from two light, flexible vertical leads. The rod lies east–west, and 0.15 m of it is in a uniform horizontal magnetic field of flux density 0.40 T directed towards the north.
Calculate the current in the rod needed to make the tension in the leads zero.
State the direction of the current in the rod.
The field is now turned, still horizontal, so that it makes an angle of 30° with the rod. Calculate the current now needed to make the tension zero.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| BIl = mg, so I = (0.025 × 9.81) ÷ (0.40 × 0.15) | ✓ 1 | |
| = 4.1 A (4.09 A) | ✓ 1 | |
| Part (b) | ||
| towards the east (field north, force upwards, by Fleming's left-hand rule) | ✓ 1 | |
| Part (c) | ||
| I = mg/(Bl sin 30°) | ✓ 1 | |
| = 4.09 ÷ 0.50 = 8.2 A (allow error carried forward) | ✓ 1 | |
Exam tip: In a balance-of-forces question, write the condition first (magnetic force = weight), then substitute. With the field at an angle, only the component of the rod at right angles to the field counts.
The figure shows four charged particles P, Q, R and S moving in the plane of the page through a uniform magnetic field directed into the page. The arrows show their velocities.
Which particles experience a magnetic force towards the top of the page?
| Answer | Mark | Notes |
|---|---|---|
| Fleming's left-hand rule uses conventional current, which is along the velocity of a positive charge and opposite to the velocity of a negative one. P (proton to the right): current to the right, field into the page, force up. Q (electron to the right): current to the left, force down. R (electron to the left): current to the right, force up. S (proton to the left): force down. So P and R. | ✓ 1 | Answer A |
Answer: A
Common error: Applying the rule to an electron's velocity. For a negative charge point the second finger the opposite way to its motion, or apply the rule as for a positive charge and then reverse the force.
Protons enter a uniform magnetic field of flux density 0.40 T through a slit S, moving at right angles to the field. The figure, drawn on a grid of 1 cm squares, shows the path of the protons, which strike a detector plate at D.
What is the speed of the protons? (mass of proton = 1.67 × 10−27 kg)
| Answer | Mark | Notes |
|---|---|---|
| SD = 18 cm is a diameter, so r = 9.0 cm = 0.090 m. BQv = mv2/r gives v = BQr/m = (0.40 × 1.60 × 10−19 × 0.090) ÷ 1.67 × 10−27 = 3.4 × 106 m s−1. | ✓ 1 | Answer C |
Answer: C
Common error: Taking the distance from S to D as the radius. After half a circle the particle is one diameter, 2r, from where it entered.
A right-angled triangular loop of wire ABC lies in the plane of a uniform magnetic field of flux density 0.20 T, as shown in the figure. The current in the loop is 3.0 A.
What is the size of the magnetic force on side AC?
| Answer | Mark | Notes |
|---|---|---|
| AC = √(302 + 402) = 50 cm and makes an angle θ with the field where sin θ = 40/50 = 0.80. F = BIl sin θ = 0.20 × 3.0 × 0.50 × 0.80 = 0.24 N. This equals BI × (the 40 cm of AC measured across the field). | ✓ 1 | Answer D |
Answer: D
Exam tip: For a sloping wire, l sin θ is the length of wire measured at right angles to the field. Here that is simply the 40 cm height of the triangle.
Positive ions pass between two parallel metal plates as shown in the figure. There is a uniform magnetic field between the plates, and the ions travel in a straight line without being deflected.
What is the magnetic flux density?
| Answer | Mark | Notes |
|---|---|---|
| Electric field strength E = V/d = 1500 ÷ 0.030 = 5.0 × 104 V m−1. No deflection means the forces balance: BQv = EQ, so B = E/v = 5.0 × 104 ÷ 2.5 × 105 = 0.20 T. (The electric force is down, towards the 0 V plate; Fleming's rule gives the magnetic force up.) | ✓ 1 | Answer C |
Answer: C
Exam tip: In crossed fields with no deflection, set the magnetic force equal to the electric force. The charge cancels, so v = E/B whatever the ion.
The figure shows two long, straight, parallel wires X and Y seen end-on. They carry equal currents out of the page. Point P is directly above M, the midpoint of XY.
What is the direction of the resultant magnetic field at P?
| Answer | Mark | Notes |
|---|---|---|
| By the right-hand grip rule each field line is an anticlockwise circle round its wire. At P the field of X is along the tangent to a circle centred on X: up and to the left. The field of Y at P is down and to the left. The two are equal in size, so the vertical parts cancel and the horizontal parts add: the resultant is towards the left of the page. (The fields cancel completely only at M.) | ✓ 1 | Answer B |
Answer: B
Exam tip: The field of a long straight wire is along the tangent to a circle round the wire, never along the line joining the wire to the point. Draw both tangents at the point, then add them as vectors.
The figure shows the magnetic field lines in a cross-section through a flat circular coil carrying a steady current. P, Q and R are three points in the plane of the page.
State the direction of the magnetic field at P, and state which face of the coil, left or right, acts as a north pole.
Put the points P, Q and R in order of magnetic flux density, largest first. Explain how the figure shows this.
A short straight wire is placed through P at right angles to the page. It carries a current into the page. State the direction of the force on this wire.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at P the field is to the right (along the axis) | ✓ 1 | |
| the right-hand face is the north pole, because the field lines leave the coil there | ✓ 1 | |
| Part (b) | ||
| Q, P, R | ✓ 1 | |
| the flux density is greater where the field lines are closer together: they are most crowded near the conductor (Q) and most widely spaced far from the coil (R) | ✓ 1 | |
| Part (c) | ||
| towards the bottom of the page (field to the right, current into the page, Fleming's left-hand rule) | ✓ 1 | |
Exam tip: On a field-line diagram the direction of the field at a point is along the line through it (the arrows), and the flux density is shown by how close the lines are.
The figure shows, from above, a metal rod resting across two horizontal metal rails connected to a battery. The rod is at right angles to the rails and there is a uniform vertical magnetic field. Friction may be ignored.
State the direction in which the rod starts to move.
Calculate the magnetic force on the rod.
Calculate the initial acceleration of the rod.
The rails are now tilted so that they slope upwards away from the battery, at an angle α to the horizontal. The field is still vertical and the current is unchanged. Calculate the value of α for which the rod stays at rest.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| to the right / away from the battery (current down the page in the rod, field into the page) | ✓ 1 | |
| Part (b) | ||
| F = BIl = 0.18 × 5.0 × 0.060 | ✓ 1 | |
| = 0.054 N | ✓ 1 | |
| Part (c) | ||
| a = F/m = 0.054 ÷ 0.030 = 1.8 m s−2 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| the magnetic force is still horizontal; along the slope F cos α = mg sin α, so tan α = F/mg = 0.054 ÷ (0.030 × 9.81) | ✓ 1 | |
| α = 10° (10.4°) | ✓ 1 | |
Exam tip: The magnetic force is at right angles to both the current and the field. A vertical field therefore gives a horizontal force, even when the rails are tilted: resolve it along the slope.
A beam of electrons moves at right angles to a uniform magnetic field. A student varies the speed v of the electrons and measures the radius r of their circular path. The graph shows the results and the best-fit line. (mass of electron = 9.11 × 10−31 kg)
Show that r = (m/eB)v, and explain why the graph is a straight line through the origin.
Use the graph to determine the magnetic flux density.
The flux density is doubled and the experiment is repeated. State how the new line differs from the one shown.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the magnetic force provides the centripetal force: Bev = mv2/r, so r = (m/eB)v | ✓ 1 | |
| m/eB is constant, so r ∝ v: a straight line through the origin with gradient m/eB | ✓ 1 | |
| Part (b) | ||
| gradient from the line, e.g. 0.050 m ÷ 5.0 × 106 m s−1 = 1.0 × 10−8 s (allow 0.98–1.02 × 10−8 s) | ✓ 1 | |
| B = m/(e × gradient) = 9.11 × 10−31 ÷ (1.60 × 10−19 × 1.0 × 10−8) | ✓ 1 | |
| = 5.7 × 10−4 T (allow 5.5–5.9 × 10−4 T) | ✓ 1 | |
| Part (c) | ||
| still a straight line through the origin, but with half the gradient (radius halved at each speed) | ✓ 1 | |
Common error: Reading the gradient with the axis units ignored. Convert cm to m and include the 106 from the speed axis before you use the gradient in B = m/(e × gradient).
The graph shows how the magnetic flux density B varies with distance x along the axis of a long solenoid carrying a steady current. x is measured from the centre of the solenoid, and its ends are at x = ±15 cm.
Describe what the graph shows about the field inside the solenoid, away from its ends.
Compare the flux density at an end of the solenoid with the flux density at its centre.
A small search coil of 30 turns, each of area 1.5 cm2, is placed at the centre of the solenoid with its plane at right angles to the axis. Calculate its flux linkage.
The search coil is moved along the axis to one end of the solenoid, keeping the same orientation. State its new flux linkage.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the field is (almost) uniform: B stays at about 4.0 mT from about x = −10 cm to +10 cm | ✓ 1 | |
| Part (b) | ||
| at an end B ≈ 2.0 mT, about half the value at the centre | ✓ 1 | |
| Part (c) | ||
| Nφ = NBA = 30 × 4.0 × 10−3 × 1.5 × 10−4 | ✓ 1 | |
| = 1.8 × 10−5 Wb turns | ✓ 1 | |
| Part (d) | ||
| about half: 9.0 × 10−6 Wb turns (allow 8.8–9.2 × 10−6; allow error carried forward) | ✓ 1 | |
Exam tip: Read values at gridline intersections: the centre value and the value at the marked end. Convert mT to T and cm2 to m2 before using Nφ = NBA.
A proton and an alpha particle have the same kinetic energy. Each moves at right angles to the same uniform magnetic field. An alpha particle has four times the mass and twice the charge of a proton.
What is the ratio radius of the alpha particle's path ÷ radius of the proton's path?
| Answer | Mark | Notes |
|---|---|---|
| r = p/BQ and p = √(2mEk). With equal Ek, r ∝ √m/Q. Ratio = √4 ÷ 2 = 1.0: the paths have the same radius. | ✓ 1 | Answer B |
Answer: B
Exam tip: Equal kinetic energy is not equal speed or equal momentum. Convert with p = √(2mEk) before using r = p/BQ.
A rectangular coil of 50 turns in a simple motor has two sides, each 4.0 cm long, at right angles to a uniform field of flux density 0.30 T. These sides are 3.0 cm apart. The current is 0.20 A and the plane of the coil is parallel to the field.
What is the moment of the couple acting on the coil?
| Answer | Mark | Notes |
|---|---|---|
| Force on each 4.0 cm side = NBIl = 50 × 0.30 × 0.20 × 0.040 = 0.12 N. The two forces are equal, opposite and 3.0 cm apart, so the moment of the couple = 0.12 × 0.030 = 3.6 × 10−3 N m. | ✓ 1 | Answer C |
Answer: C
Exam tip: The moment of a couple is one of the forces × the perpendicular distance between them. Remember the factor N: every turn of the coil carries the current through the field.
In a glass bulb containing low-pressure gas, a narrow beam of electrons travels at 8.0 × 106 m s−1. A pair of coils produces a uniform magnetic field of 1.2 mT at right angles to the beam, and the glowing gas shows that the beam follows a circular path.
Calculate the radius of the circular path.
Explain why the beam follows a circular path, and describe and explain what happens to the path if the speed of the electrons is doubled and, separately, if the direction of the magnetic field is reversed.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| r = mv/BQ = (9.11 × 10−31 × 8.0 × 106) ÷ (1.2 × 10−3 × 1.60 × 10−19) | ✓ 1 | |
| = 0.038 m (0.0380 m) | ✓ 1 | |
| Part (b) | ||
| Indicative content: | — | |
| the field exerts a force Bev on each moving electron, at right angles to both the field and the velocity | ✓ 1 | |
| the force is always perpendicular to the velocity, so no work is done and the speed stays constant | ✓ 1 | |
| a force of constant size perpendicular to the velocity provides a centripetal force, so the path is a circle | ✓ 1 | |
| Bev = mv2/r gives r = mv/Be, so r is proportional to v | ✓ 1 | |
| doubling the speed doubles the radius (to about 0.076 m) | ✓ 1 | |
| reversing the field reverses the direction of the force, so the beam curves the opposite way (same radius) | ✓ 1 | |
Exam tip: In a six-mark explanation, link each step: perpendicular force → no work → constant speed → centripetal force → circle → r = mv/BQ. Then use the equation to justify each change.
A photograph from a particle detector shows the track of a charged particle in a uniform magnetic field of flux density 1.5 T directed into the plane of the photograph. Where the particle enters, moving from left to right, its track curves towards the top of the photograph with a radius of curvature of 0.35 m. The particle carries a charge of magnitude 1.60 × 10−19 C. Further along, the track becomes a spiral of decreasing radius.
Calculate the momentum of the particle where it enters.
The particle is a proton. Calculate its speed.
Explain what the spiral shows about the particle's motion.
Use Fleming's left-hand rule to deduce the sign of the particle's charge.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = BQr = 1.5 × 1.60 × 10−19 × 0.35 | ✓ 1 | |
| = 8.4 × 10−20 kg m s−1 | ✓ 1 | |
| Part (b) | ||
| v = p/m = 8.4 × 10−20 ÷ 1.67 × 10−27 = 5.0 × 107 m s−1 | ✓ 1 | |
| Part (c) | ||
| r = p/BQ, with B and Q constant, so a decreasing radius means the momentum is decreasing | ✓ 1 | |
| the particle is losing (kinetic) energy, e.g. by ionising the material of the detector, which also shows it travels towards the centre of the spiral | ✓ 1 | |
| Part (d) | ||
| field into the page and force towards the top of the page: the rule gives a conventional current from left to right | ✓ 1 | |
| this is the direction the particle moves, so its charge is positive | ✓ 1 | |
Exam tip: Detector tracks are read with r = p/BQ: the radius gives momentum, a shrinking radius means energy loss, and the direction of curvature with the field direction gives the sign of the charge.
A student investigates how the force on a current-carrying wire depends on the angle θ between the wire and a uniform magnetic field. The current is kept at 4.0 ± 0.1 A and the length of wire in the field is 50 ± 1 mm. She measures the force F with a current balance.
Results: θ = 15°, F = 4.5 mN; 30°, 8.4 mN; 45°, 12.1 mN; 60°, 14.6 mN; 75°, 16.5 mN; 90°, 17.0 mN.
Explain why she should plot a graph of F against sin θ, and what the graph should look like if F = BIl sin θ is obeyed.
The gradient of her line is 0.0170 N. Calculate the magnetic flux density.
The percentage uncertainty in the gradient is 3%. Determine the absolute uncertainty in her value of B.
State the force on the wire when it is parallel to the field.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| F = (BIl) sin θ has the form y = mx with BIl constant | ✓ 1 | |
| so the graph should be a straight line through the origin with gradient BIl | ✓ 1 | |
| Part (b) | ||
| B = gradient ÷ Il = 0.0170 ÷ (4.0 × 0.050) | ✓ 1 | |
| = 0.085 T | ✓ 1 | |
| Part (c) | ||
| % uncertainty in I = 2.5% and in l = 2.0%, so in B = 3 + 2.5 + 2.0 = 7.5% | ✓ 1 | |
| ΔB = 0.075 × 0.085 = 0.006 T (0.0064 T), so B = 0.085 ± 0.006 T | ✓ 1 | |
| Part (d) | ||
| zero, since sin 0° = 0 | ✓ 1 | |
Model sketch
Exam tip: When a result is found by multiplying and dividing measured quantities, add their percentage uncertainties. Convert to an absolute uncertainty only at the end.
The figure shows a current balance. A light wire frame is pivoted at its centre. Its far end, a straight section 4.0 cm long, lies inside a solenoid at right angles to the uniform field along the solenoid's axis. With no current, the frame balances horizontally. When the current in the frame is 2.5 A, the frame balances horizontally again with a 0.40 g rider placed as shown.
What is the magnetic flux density inside the solenoid?
| Answer | Mark | Notes |
|---|---|---|
| Taking moments about the pivot: BIl × 0.150 = mg × 0.090. B = (0.40 × 10−3 × 9.81 × 0.090) ÷ (2.5 × 0.040 × 0.150) = 0.024 T (0.0235 T). | ✓ 1 | Answer B |
Answer: B
Exam tip: A balance with a pivot needs moments, not just forces: the magnetic force and the rider's weight act at different distances from the pivot.
An electron enters a strip of uniform magnetic field of flux density 2.0 mT, moving at right angles both to the field and to the edge of the strip. The figure shows its path: it leaves the strip, of width 5.0 cm, deflected through 30°.
What is the momentum of the electron?
| Answer | Mark | Notes |
|---|---|---|
| The electron enters at right angles to the edge, so the centre of its circle lies on the entry edge. When it has turned through 30° its horizontal distance from the entry point is r sin 30°, which equals the width: r = 0.050 ÷ sin 30° = 0.10 m. p = BQr = 2.0 × 10−3 × 1.60 × 10−19 × 0.10 = 3.2 × 10−23 kg m s−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: In a deflection problem, sketch the radii to the entry and exit points. The angle between them equals the angle of deflection, which turns the geometry into a right-angled triangle.
Protons moving at 4.0 × 106 m s−1 enter a region of uniform magnetic field through a small hole H in its boundary, as shown in the figure. The field is at right angles to the page and exists only above the boundary. The protons follow a circular arc and leave the field through the boundary. (mass of proton = 1.67 × 10−27 kg)
Calculate the radius of the path of the protons in the field.
Use Fleming's left-hand rule to state whether the protons leave the field to the left or to the right of H. Hence state the angle through which the protons turn while they are in the field.
Show that the protons leave the field about 0.29 m from H.
Calculate the time the protons spend in the field.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| r = mv/BQ = (1.67 × 10−27 × 4.0 × 106) ÷ (0.25 × 1.60 × 10−19) | ✓ 1 | |
| = 0.17 m (0.167 m) | ✓ 1 | |
| Part (b) | ||
| at H the force is at right angles to the velocity, up and to the left, so the protons curve anticlockwise and leave to the left of H | ✓ 1 | |
| by symmetry they leave at 60° to the boundary, having turned through 360° − 2 × 60° = 240° | ✓ 1 | |
| Part (c) | ||
| the radius at H is at right angles to the velocity, so it makes 30° with the boundary; the chord along the boundary is 2r cos 30° = 2r sin 60° | ✓ 1 | |
| = 2 × 0.167 × sin 60° = 0.29 m | ✓ 1 | |
| Part (d) | ||
| time for a full circle = 2πr/v = 2π × 0.167 ÷ 4.0 × 106 = 2.62 × 10−7 s | ✓ 1 | |
| t = (240/360) × 2.62 × 10−7 = 1.7 × 10−7 s (allow error carried forward) | ✓ 1 | |
Exam tip: For a particle crossing a straight boundary, draw the radius at the entry point at right angles to the velocity. The centre of the circle then fixes the chord, the side the particle leaves on, and the angle it turns through.
A horizontal metal rod of mass 12 g hangs from two light, flexible threads in a uniform vertical magnetic field of flux density 0.30 T. 8.0 cm of the rod is in the field. When there is a current in the rod, it moves sideways and comes to rest with the threads at 20° to the vertical. The figure shows the arrangement seen along the rod.
Name the forces acting on the rod in its new position and explain why the magnetic force on the rod is horizontal.
Calculate the current in the rod.
State the direction of the current in the rod.
The current is doubled. Calculate the new angle between the threads and the vertical.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| weight (down), tension (along the threads) and the magnetic force | ✓ 1 | |
| the magnetic force is at right angles to both the field (vertical) and the current (along the horizontal rod), so it is horizontal | ✓ 1 | |
| Part (b) | ||
| horizontal and vertical balance give tan 20° = BIl/mg, so I = (0.012 × 9.81 × tan 20°) ÷ (0.30 × 0.080) | ✓ 1 | |
| = 1.8 A (1.79 A) | ✓ 1 | |
| Part (c) | ||
| into the page (force to the right, field upwards, Fleming's left-hand rule) | ✓ 1 | |
| Part (d) | ||
| tan θ = 2 × tan 20° = 0.728 | ✓ 1 | |
| θ = 36° (not 40°: the angle does not double) | ✓ 1 | |
Common error: Assuming the angle is proportional to the current. It is tan θ that is proportional to the magnetic force, so doubling the current doubles tan θ, not θ.
A beam containing protons and alpha particles, all moving at the same speed, enters a uniform magnetic field of flux density 0.15 T at P, at right angles to the field. The figure, drawn on a grid of 1 cm squares, shows the two tracks produced. (mass of proton = 1.67 × 10−27 kg; mass of alpha particle = 6.64 × 10−27 kg; charge of alpha particle = +2e)
Deduce whether the magnetic field is directed into or out of the page.
Explain which track is made by the alpha particles.
Use track 1 to calculate the speed of the particles.
Show that track 2 is consistent with your answer to (c).
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| both particles are positive and moving to the right at P, and the force on them at P is towards the top of the page (towards the centre of curvature) | ✓ 1 | |
| Fleming's left-hand rule then gives a field into the page | ✓ 1 | |
| Part (b) | ||
| r = mv/BQ, and m/Q for an alpha particle is about twice that for a proton (four times the mass, twice the charge) | ✓ 1 | |
| so at the same speed the alpha particles have the larger radius: track 2 | ✓ 1 | |
| Part (c) | ||
| radius of track 1 = 6.0 cm (allow 5.9–6.1 cm) | ✓ 1 | |
| v = BQr/m = 0.15 × 1.60 × 10−19 × 0.060 ÷ 1.67 × 10−27 = 8.6 × 105 m s−1 | ✓ 1 | |
| Part (d) | ||
| r = (6.64 × 10−27 × 8.6 × 105) ÷ (3.20 × 10−19 × 0.15) = 0.119 m ≈ 12 cm, which matches the furthest distance of track 2 from the left edge | ✓ 1 | |
Exam tip: To get a radius from a drawn track, find two points on it that are a diameter apart, or the furthest point from the line through the centre, and use the grid. Then r = p/BQ links the radius to the particle.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy