Summary to follow. 6 syllabus statements · 22 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Uniform electric field between parallel plates
Uniform electric field between parallel plates
A region in which the electric field strength E has the same magnitude and direction at every point. Between two parallel plates with a potential difference V and separation d (away from the edges), E = V/d, directed from the positive plate to the negative plate. E is measured in V m⁻¹, which is equivalent to N C⁻¹.
Force and acceleration of a charge in a uniform electric field
A charge q in a field of strength E experiences a force F = qE, in the direction of the field for a positive charge and opposite to it for a negative charge. The force is the same at every point of a uniform field and does not depend on the speed of the charge, so the charge has a constant acceleration a = qE/m.
Parabolic path in a uniform electric field
A charged particle that enters a uniform electric field at right angles to the field lines has a constant velocity component perpendicular to the field (no force acts in that direction) and a uniform acceleration qE/m parallel to the field. Its path is a parabola, exactly like a horizontally launched projectile in a uniform gravitational field. The work done by the field increases its kinetic energy.
Energy gained by a charge accelerated through a potential difference
When a charge q moves from rest through a potential difference V in a vacuum, the work done by the field, qV, becomes kinetic energy: qV = ½mv². The final speed v = √(2qV/m) depends on the potential difference, not on the separation of the plates.
Students often think A charged particle deflected by any field moves in a circular arc at constant speed. In fact A parabola. The force qE is constant in magnitude and direction, so the particle has a constant velocity component across the field and a uniform acceleration along it.
Students often think A force makes a body move at a steady velocity in the direction of the force, so the particle heads off in a straight line at an angle. In fact No. The force changes the velocity gradually: the velocity component along the field grows steadily while the component across the field is unchanged, so the path curves.
Circular motion in a uniform magnetic field
Circular motion in a uniform magnetic field
A charged particle moving perpendicular to a uniform magnetic field moves in a circle at constant speed, the magnetic force qvB providing the centripetal force: qvB = mv²/r, so r = mv/(qB). The time for one revolution, T = 2πr/v = 2πm/(qB), is independent of the speed: a faster particle moves in a proportionally larger circle.
Constant kinetic energy in a magnetic field
The magnetic force on a charged particle is always perpendicular to its velocity, so it does no work on the particle. The kinetic energy, and therefore the speed, of a charged particle stays constant in a magnetic field, even though its velocity changes direction continuously.
Determination of the charge-to-mass ratio from the path in a uniform magnetic field
If a charged particle of known speed v moves in a circle of radius r in a uniform field B, then q/m = v/(Br). If the particles are first accelerated from rest through a potential difference V, combining qV = ½mv² with r = mv/(qB) gives q/m = 2V/(B²r²). The unit of q/m is C kg⁻¹.
Students often think Because the magnetic force accelerates the particle, it must do work on it and increase its kinetic energy. In fact No. The magnetic force is always perpendicular to the velocity, so the work done is zero and the kinetic energy is unchanged.
Students often think Kinetic energy depends on velocity, so if the velocity changes (even only in direction) the kinetic energy changes too. In fact No. Kinetic energy ½mv² depends only on the speed, a scalar; a change of direction at constant speed leaves it unchanged.
Velocity selector (crossed electric and magnetic fields)
Velocity selector (crossed electric and magnetic fields)
A region in which a uniform electric field and a uniform magnetic field are perpendicular to each other and to the velocity of the particles, arranged so that the electric and magnetic forces are opposite. A particle passes through undeflected when qE = qvB, that is when v = E/B, whatever its charge or mass. A faster particle is deflected in the direction of the magnetic force and a slower one in the direction of the electric force.
Students often think A faster particle follows a larger circle, so the magnetic force on it must be smaller. In fact No. The magnetic force F = qvB increases with speed. A faster particle is bent less only because its greater momentum needs an even larger force to turn it through the same radius.
Students often think The magnetic force on a charge depends only on the charge and the field, like the electric force qE, so it is the same at any speed. In fact Yes. F = qvB sin θ is proportional to the speed; a stationary charge experiences no magnetic force at all.
Magnetic force on a moving charge, F = qvB sin θ
Magnetic force on a moving charge, F = qvB sin θ
A charge q moving with speed v in a magnetic field of strength B experiences a force of magnitude F = qvB sin θ, where θ is the angle between the velocity and the field. The force is zero for motion parallel to the field and greatest (qvB) for motion perpendicular to it. B is measured in tesla (T); 1 T = 1 N A⁻¹ m⁻¹.
Direction of the magnetic force on a moving charge
The magnetic force is perpendicular both to the velocity of the charge and to the magnetic field. For a given velocity and field, the force on a negative charge is opposite in direction to the force on a positive charge. Because it is always perpendicular to the velocity, the magnetic force changes the direction of motion but not the speed.
Students often think The magnetic force acts along the field lines, in the direction of the field, as the electric force on a positive charge does. In fact No. The magnetic force is perpendicular to the field (and to the velocity or current).
Students often think The direction of the magnetic force depends only on the direction of motion and of the field, not on the sign of the charge. In fact No. For the same velocity and field, the force on a negative charge is opposite to the force on a positive charge.
Force on a current-carrying conductor, F = BIL sin θ
Force on a current-carrying conductor, F = BIL sin θ
A straight conductor of length L carrying a current I in a uniform magnetic field of strength B experiences a force of magnitude F = BIL sin θ, where θ is the angle between the conductor (the current) and the field, and L is the length of conductor inside the field. The force is zero when the conductor is parallel to the field.
Direction of the force on a current-carrying conductor
The force on the conductor is perpendicular both to the current and to the magnetic field. The current direction used is the conventional current, the direction of flow of positive charge, which is opposite to the direction in which electrons move in a metal wire. Reversing either the current or the field reverses the force.
Students often think L is the total length of the wire, including any part outside the field. In fact No. L is the length of the conductor that is inside the magnetic field; only that part experiences a force.
Students often think The force on a current-carrying wire acts along the wire, in the direction in which the current flows. In fact No. The force on the wire is perpendicular to the current as well as to the field.
Magnetic field of a long straight current-carrying wire
Magnetic field of a long straight current-carrying wire
The field lines are concentric circles centred on the wire, in planes perpendicular to it; the field is strongest close to the wire. The sense of the circles is given by the right-hand grip rule. For a wire in the plane of the page, the field is into the page on one side of the wire and out of the page on the other side.
Force per unit length between parallel current-carrying wires
Each wire lies in the magnetic field of the other, so each experiences a force F = BIL. For long parallel wires separated by r, the force per unit length is F/L = μ₀I₁I₂/(2πr), measured in N m⁻¹. The force is attractive when the currents are in the same direction and repulsive when they are in opposite directions. By Newton's third law the forces on the two wires are equal in magnitude, even when the currents differ.
Permeability of free space, μ₀
The constant that relates the magnetic field produced by a current to that current in a vacuum; μ₀ = 4π × 10⁻⁷ T m A⁻¹. The permeability of air is very close to μ₀.
Students often think All field forces fall off with the square of the distance, so the force between wires is proportional to 1/r². In fact No. For long parallel wires the force per unit length is inversely proportional to the separation: F/L = μ₀I₁I₂/(2πr).
Students often think The quantity μ₀I₁I₂/(2πr) is the force itself, so no length is needed. In fact No. It is the force per unit length, in N m⁻¹; the force on a length L of wire is (F/L) × L.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 An electron moves horizontally into the region between two horizontal parallel plates, midway between them and at right angles to the uniform electric field between the plates. It is deflected towards the positive plate. Gravity is negligible. Which describes the motion of the electron while it is between the plates?
Answer and reasoning
A parabola: its horizontal velocity is constant as it accelerates towards the positive plate — The electric force qE has a fixed magnitude and direction, so the electron has a uniform acceleration towards the positive plate and no acceleration horizontally. Constant horizontal velocity combined with uniform perpendicular acceleration gives a parabola, as for a projectile.
A circular arc at constant speed, with the force at right angles to its velocity — A student who expects every field to bend a particle into a circle picks this. That happens in a magnetic field, where the force stays perpendicular to the velocity. The electric force has a fixed direction, so the path is a parabola and the electron speeds up.
A straight line at an angle, moving towards the positive plate at a steady velocity — A student who thinks a steady force produces a steady velocity picks this. The force produces an acceleration: the velocity component towards the plate keeps increasing, so the path keeps curving.
A curve along which its horizontal velocity decreases as it is pulled sideways — A student who thinks the sideways pull uses up some of the forward motion picks this. The force acts only along the field lines, so the horizontal component of velocity does not change.
2 A proton moves in a circle in a uniform magnetic field that is perpendicular to its velocity. No other forces act on it. Which statement about the kinetic energy of the proton is correct?
Answer and reasoning
It varies continuously around the circle, as the direction of the proton's velocity keeps changing. — A student who thinks kinetic energy changes whenever velocity changes picks this. Kinetic energy is ½mv², a scalar that depends on the speed only. A change in direction at constant speed leaves it unchanged.
It stays constant, because the magnetic force is perpendicular to its velocity. — The magnetic force is always at right angles to the velocity, so it does no work on the proton. The force changes the direction of motion but not the speed, so the kinetic energy of a charged particle stays constant in a magnetic field.
It increases steadily, because the magnetic force accelerates the proton and so does work on it. — A student who equates acceleration with gaining energy picks this. The proton does accelerate, but centripetally: the force is at 90° to the displacement at every instant, so the work done is Fs cos 90° = 0.
It decreases steadily, because the magnetic force acts against the proton's motion. — A student who pictures the field resisting motion like friction picks this. The magnetic force has no component along the velocity, so it cannot slow the proton; the speed stays constant.
3 Protons pass undeflected through a velocity selector in which two parallel plates 1.5 cm apart have a potential difference of 600 V between them and a uniform magnetic field of strength 0.050 T acts at right angles both to the electric field and to the velocity of the protons. The protons then enter a region containing only a uniform magnetic field of strength 0.050 T, at right angles to their velocity, travel through a semicircle and strike a detector. How far is the detector from the point where the protons entered this region? Gravity is negligible. Take e = 1.60 × 10⁻¹⁹ C and m_p = 1.67 × 10⁻²⁷ kg.
Answer and reasoning
3.3 × 10⁻¹ m — Undeflected protons have eE = evB, so v = E/B = (600 V ÷ 0.015 m) ÷ 0.050 T = 8.0 × 10⁵ m s⁻¹, whatever their charge or mass. In the second region r = m_pv/(eB) = 0.167 m, and a semicircle ends one diameter from the entry point: 2r = 3.3 × 10⁻¹ m.
1.7 × 10⁻¹ m — A student who takes the distance to the detector as the radius stops at r = m_pv/(eB) = 0.17 m. After a semicircle the proton is one diameter from where it entered, so the distance is 2r = 3.3 × 10⁻¹ m.
5.0 × 10⁻³ m — A student who uses the potential difference as the field strength calculates v = 600 ÷ 0.050 = 1.2 × 10⁴ m s⁻¹. The field strength is E = V/d = 4.0 × 10⁴ V m⁻¹, so v = E/B = 8.0 × 10⁵ m s⁻¹ and the distance is 2r = 3.3 × 10⁻¹ m.
7.5 × 10⁻⁵ m — A student who multiplies the potential difference by the separation takes E = 600 × 0.015 = 9 V m⁻¹ and v = 180 m s⁻¹. The field strength is V/d = 4.0 × 10⁴ V m⁻¹, giving v = 8.0 × 10⁵ m s⁻¹ and a distance of 2r = 3.3 × 10⁻¹ m.
Working Selector: E = V/d = 600 V / 0.015 m = 4.0 × 10⁴ V m⁻¹. Undeflected ⇒ resultant force zero ⇒ eE = evB ⇒ v = E/B = (4.0 × 10⁴ V m⁻¹)/(0.050 T) = 8.0 × 10⁵ m s⁻¹. Second region: evB = m_pv²/r ⇒ r = m_pv/(eB) = (1.67 × 10⁻²⁷ kg)(8.0 × 10⁵ m s⁻¹)/((1.60 × 10⁻¹⁹ C)(0.050 T)) = 0.167 m. After a semicircle the detector is one diameter from the entry point: 2r = 0.334 m ≈ 3.3 × 10⁻¹ m.
4 A proton moves with a velocity parallel to the field lines of a uniform magnetic field. No other fields act. Which describes its subsequent motion?
Answer and reasoning
It moves in a circle around the field lines, as any moving charge in a magnetic field does. — A student who has met only the perpendicular case picks this. The force is qvB sin θ, and for motion along the field θ = 0°, so the force is zero and there is no circular motion.
It slows down steadily and stops, because the magnetic field acts against its motion. — A student who pictures the field resisting motion like friction picks this. A magnetic force is always perpendicular to the velocity and, here, is zero anyway; the speed cannot change.
It carries on in a straight line at constant speed, as no magnetic force acts. — F = qvB sin θ and θ = 0° for a velocity parallel to the field, so the force is zero. With no resultant force, the proton continues at constant velocity.
It accelerates along the field lines, as the magnetic force acts along the field. — A student who thinks the magnetic force acts along the field picks this. The magnetic force is perpendicular to the field; for a charge moving along the field it is zero.
5 A straight wire lies in the plane of the page and carries a current towards the right of the page. A uniform magnetic field in the plane of the page is directed towards the top of the page. What is the direction of the force on the wire?
Answer and reasoning
Towards the top of the page — A student who thinks the magnetic force acts along the field lines picks this. The force on a current is perpendicular to the field, not along it.
Towards the right of the page — A student who pictures the current pushing the wire along picks this. The force on a current-carrying wire is perpendicular to the current.
Out of the plane of the page — The force is perpendicular to both the current (right) and the field (towards the top), so it is along the line into or out of the page. For a current to the right in a field towards the top of the page, it is out of the page.
Into the plane of the page — A student who takes the given current direction as the direction the electrons move reverses the current, and so reverses the force. The current given is conventional current; the force is out of the page.
6 A long straight wire lies in the plane of the page and carries a current towards the top of the page. Point P is 3.0 cm to the right of the wire and point Q is 3.0 cm to the left of it, both in the plane of the page. What is the direction of the magnetic field due to the current at P and at Q?
Answer and reasoning
At P it is to the right; at Q it is to the left — A student who pictures the field lines spreading radially outwards, like the electric field of a charged wire, picks this. Magnetic field lines around a current are circles, so the field is perpendicular to the line from the wire.
At both P and Q it points towards the top of the page — A student who thinks the field points along the current picks this. The field circles the wire in planes perpendicular to it, so it is never along the wire.
At both P and Q it is directed into the page — A student who thinks the field of a wire points the same way all around it picks this. A circular field line that goes into the page on one side of the wire comes out of the page on the other side.
At P it is into the page; at Q it is out of the page — The field lines are circles around the wire. By the right-hand grip rule, for a current towards the top of the page the circles pass into the page on the right of the wire and out of the page on the left.
7 Two horizontal parallel plates, 2.5 cm apart in a vacuum, have a potential difference of 150 V between them. What is the magnitude of the acceleration of an electron between the plates? Gravity is negligible. Take e = 1.60 × 10⁻¹⁹ C and m_e = 9.11 × 10⁻³¹ kg.
Answer and reasoning
1.1 × 10¹³ m s⁻² — A student who substitutes the separation in centimetres uses E = 150 ÷ 2.5 = 60, which is 100 times too small. Converting first, d = 0.025 m, gives E = 6.0 × 10³ V m⁻¹ and a = 1.1 × 10¹⁵ m s⁻².
2.6 × 10¹³ m s⁻² — A student who treats the potential difference as the field strength uses E = 150 V m⁻¹. The field strength is the potential difference per unit distance: E = V/d = 6.0 × 10³ V m⁻¹.
6.6 × 10¹¹ m s⁻² — A student who multiplies the potential difference by the separation uses E = 150 × 0.025 = 3.75 V m⁻¹. The field strength is V/d, not Vd: moving the plates closer together makes the field stronger.
1.1 × 10¹⁵ m s⁻² — E = V/d = 150 V ÷ 0.025 m = 6.0 × 10³ V m⁻¹. Then a = eE/m_e = (1.60 × 10⁻¹⁹ C × 6.0 × 10³ V m⁻¹) ÷ 9.11 × 10⁻³¹ kg = 1.1 × 10¹⁵ m s⁻².
Working E = V/d = 150 V / 0.025 m = 6.0 × 10³ V m⁻¹. F = eE and a = F/m_e = eE/m_e = (1.60 × 10⁻¹⁹ C)(6.0 × 10³ V m⁻¹) / (9.11 × 10⁻³¹ kg) = 1.05 × 10¹⁵ m s⁻² ≈ 1.1 × 10¹⁵ m s⁻².
8 Positive ions of a single type, initially at rest, are accelerated through a potential difference of 1500 V. They then enter a uniform magnetic field of strength 0.25 T at right angles to their velocity, travel through a semicircle and strike a detector 6.3 cm from the point where they entered the field. What is the charge-to-mass ratio q/m of the ions?
Answer and reasoning
2.4 × 10⁷ C kg⁻¹ — A student who writes qV = mv², dropping the ½, obtains q/m = V/(B²r²), half the correct value. With qV = ½mv², q/m = 2V/(B²r²) = 4.8 × 10⁷ C kg⁻¹.
1.2 × 10⁷ C kg⁻¹ — A student who uses 6.3 cm as the radius gets a value four times too small, because r is squared. The ions travel a semicircle, so 6.3 cm is the diameter and r = 3.15 cm.
4.8 × 10³ C kg⁻¹ — A student who substitutes r = 3.15 in centimetres makes r² 10⁴ times too large, so q/m comes out 10⁴ times too small. Convert first: r = 0.0315 m.
4.8 × 10⁷ C kg⁻¹ — The detector is one diameter from the entry point, so r = 3.15 cm = 0.0315 m. From qV = ½mv² and r = mv/(qB), q/m = 2V/(B²r²) = 2 × 1500 ÷ (0.25² × 0.0315²) = 4.8 × 10⁷ C kg⁻¹.
Working Semicircle ⇒ 6.3 cm is the diameter, so r = 3.15 cm = 0.0315 m. Acceleration: qV = ½mv² ⇒ v² = 2qV/m. In the field: qvB = mv²/r ⇒ r = mv/(qB) ⇒ r² = m²v²/(q²B²) = 2mV/(qB²). Hence q/m = 2V/(B²r²) = (2 × 1500 V)/((0.25 T)² × (0.0315 m)²) = 3000/(0.0625 × 9.92 × 10⁻⁴) = 4.8 × 10⁷ C kg⁻¹.
9 In a velocity selector, uniform electric and magnetic fields act at right angles to each other and to the velocity of the particles, so that the electric and magnetic forces on a proton are in opposite directions. Protons moving at speed v₀ pass straight through undeflected. Gravity is negligible. Which statement about this velocity selector is correct?
Answer and reasoning
A proton of any speed passes undeflected, as neither of the forces depends on its speed. — A student who treats the magnetic force like the electric force, independent of speed, picks this. The electric force qE is the same at any speed, but the magnetic force qvB is proportional to v, so the two forces balance only at one speed, v₀ = E/B.
Only particles with a proton's charge-to-mass ratio can pass undeflected at speed v₀. — A student who carries the charge-to-mass dependence of r = mv/(qB) over to the selector picks this. Undeflected particles have qE = qvB, so the charge cancels and the mass never enters: any charged particle moving at v₀ = E/B passes straight through.
A proton faster than v₀ is deflected in the direction of the magnetic force on it. — The electric force qE does not depend on speed, but the magnetic force qvB is proportional to it. At v₀ = E/B they balance; above v₀ the magnetic force is the larger, so a faster proton is deflected in the direction of the magnetic force.
A proton faster than v₀ is deflected in the direction of the electric force acting on it. — A student who reads the larger radius r = mv/(qB) of a faster particle as evidence of a weaker magnetic force picks this. The magnetic force qvB increases with speed, so above v₀ it exceeds the electric force and the proton is deflected in the direction of the magnetic force.
10 A proton moves at 3.0 × 10⁶ m s⁻¹ in a uniform magnetic field of strength 40 mT. The velocity of the proton makes an angle of 25° with the direction of the field. What is the magnitude of the magnetic force on the proton? Take e = 1.60 × 10⁻¹⁹ C.
Answer and reasoning
8.1 × 10⁻¹⁵ N — F = qvB sin θ with θ the angle between the velocity and the field: F = 1.60 × 10⁻¹⁹ C × 3.0 × 10⁶ m s⁻¹ × 0.040 T × sin 25° = 8.1 × 10⁻¹⁵ N.
1.7 × 10⁻¹⁴ N — A student who uses the component of velocity along the field uses cos 25°. The force depends on the component perpendicular to the field, v sin θ, with θ measured from the field: F = qvB sin 25° = 8.1 × 10⁻¹⁵ N.
1.9 × 10⁻¹⁴ N — A student who ignores the angle calculates qvB, which applies only when the velocity is perpendicular to the field. Here θ = 25°, so F = qvB sin 25° = 8.1 × 10⁻¹⁵ N.
8.1 × 10⁻¹² N — A student who substitutes B = 40 instead of 0.040 T gets a force 1000 times too large. Convert first: 40 mT = 0.040 T, giving F = 8.1 × 10⁻¹⁵ N.
Working B = 40 mT = 0.040 T; θ = 25° (angle between v and B). F = qvB sin θ = (1.60 × 10⁻¹⁹ C)(3.0 × 10⁶ m s⁻¹)(0.040 T)(sin 25°) = (1.92 × 10⁻¹⁴ N)(0.423) = 8.1 × 10⁻¹⁵ N.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
12 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Two parallel plates, 2.0 cm apart in a vacuum, have a potential difference of 500 V between them. An electron is released from rest at the negative plate and accelerates across the gap. What is its speed when it reaches the positive plate? Take e = 1.60 × 10⁻¹⁹ C and m_e = 9.11 × 10⁻³¹ kg.
Answer and reasoning
9.4 × 10⁶ m s⁻¹ — A student who writes the kinetic energy as mv² instead of ½mv² gets v = √(eV/m_e). With the factor ½, eV = ½m_ev² and v = √(2eV/m_e) = 1.3 × 10⁷ m s⁻¹.
1.9 × 10⁶ m s⁻¹ — A student who treats the potential difference as the field strength takes E = 500 V m⁻¹, so a = eE/m_e and v² = 2ad = 2 × (e × 500 ÷ m_e) × 0.020. The field is E = V/d = 2.5 × 10⁴ V m⁻¹; used correctly, the separation cancels and v = √(2eV/m_e) = 1.3 × 10⁷ m s⁻¹.
1.3 × 10⁷ m s⁻¹ — The work done by the field on the electron, eV, becomes kinetic energy: eV = ½m_ev², so v = √(2 × 1.60 × 10⁻¹⁹ C × 500 V ÷ 9.11 × 10⁻³¹ kg) = 1.3 × 10⁷ m s⁻¹. The plate separation does not affect the final speed.
2.7 × 10⁵ m s⁻¹ — A student who takes the field strength as V × d = 10 V m⁻¹ finds a gain in kinetic energy of eEd = 3.2 × 10⁻²⁰ J. The field is V/d, so eEd = eV = 8.0 × 10⁻¹⁷ J and v = 1.3 × 10⁷ m s⁻¹.
Working Work done by the field = eV = ΔE_k. eV = ½m_ev² ⇒ v = √(2eV/m_e) = √(2 × 1.60 × 10⁻¹⁹ C × 500 V / 9.11 × 10⁻³¹ kg) = √(1.76 × 10¹⁴ m² s⁻²) = 1.33 × 10⁷ m s⁻¹ ≈ 1.3 × 10⁷ m s⁻¹. (Equivalently E = V/d = 2.5 × 10⁴ V m⁻¹, a = eE/m_e, v² = 2ad: the separation cancels.)
2 A proton moves in a circle in a uniform magnetic field that is perpendicular to its velocity. A second proton moves at twice the speed of the first, perpendicular to the same field. Compared with the first proton, what are the radius of the second proton's circle and the time it takes to complete one revolution?
Answer and reasoning
Radius halved; time for one revolution reduced to a quarter — A student who reasons that the larger force at higher speed pulls the proton into a tighter circle picks this. The force qvB doubles, but the centripetal force needed, mv²/r, would quadruple at the same radius, so the radius must double: r = mv/(qB).
Radius doubled; time for one revolution unchanged — From qvB = mv²/r, r = mv/(qB), so doubling v doubles r. The time for one revolution is T = 2πr/v = 2πm/(qB): the path is twice as long but covered at twice the speed, so T is unchanged.
Radius unchanged; time for one revolution halved — A student who thinks the field fixes the size of the circle picks this. The radius depends on the speed, r = mv/(qB), so it doubles; the time then stays the same.
Radius doubled; time for one revolution halved as well — A student who assumes a faster particle must complete its circle sooner picks this. The circle is twice as large, so at twice the speed the time 2πr/v is unchanged; T = 2πm/(qB) does not depend on speed.
3 An electron moves towards the right of the page, in the plane of the page, through a uniform magnetic field directed into the page. What is the direction of the magnetic force on the electron at that instant?
Answer and reasoning
Towards the top of the page — A student who ignores the sign of the charge picks this; it is the direction of the force on a proton with the same velocity. The electron is negative, so the force on it is reversed: towards the bottom of the page.
Towards the bottom of the page — The force is perpendicular to both the velocity (right) and the field (into the page), so it lies along the top–bottom line of the page. A positive charge would be pushed towards the top of the page; the electron is negative, so the force on it is towards the bottom.
Towards the left of the page — A student who thinks the magnetic field resists motion like friction picks this. The magnetic force is perpendicular to the velocity, so it has no component opposing the motion.
Into the plane of the page — A student who thinks the magnetic force acts along the field lines picks this. The force is perpendicular to the field, so it cannot point into the page when the field does.
4 A straight wire 0.40 m long carries a current of 3.0 A. Only 0.25 m of the wire lies inside a uniform magnetic field of strength 0.080 T, and the wire makes an angle of 30° with the direction of the field. What is the magnitude of the force on the wire?
Answer and reasoning
4.8 × 10⁻² N — A student who uses the whole length of the wire picks this. Only the 0.25 m inside the field experiences a force: F = BIL sin θ = 0.080 × 3.0 × 0.25 × sin 30° = 3.0 × 10⁻² N.
5.2 × 10⁻² N — A student who uses the component of the wire along the field uses cos 30°. The force depends on the component perpendicular to the field, so sin θ is used: F = 3.0 × 10⁻² N.
6.0 × 10⁻² N — A student who ignores the angle calculates BIL, which applies only when the wire is perpendicular to the field. Here sin 30° = 0.50, so F = 3.0 × 10⁻² N.
3.0 × 10⁻² N — F = BIL sin θ, with L the length inside the field and θ the angle between the wire and the field: F = 0.080 T × 3.0 A × 0.25 m × sin 30° = 3.0 × 10⁻² N.
Working L = length in the field = 0.25 m; θ = 30°. F = BIL sin θ = (0.080 T)(3.0 A)(0.25 m)(sin 30°) = 0.060 N × 0.50 = 0.030 N = 3.0 × 10⁻² N.
5 Two long straight parallel wires in air are 4.0 cm apart and carry currents of 5.0 A and 8.0 A. What is the magnitude of the magnetic force on a 1.5 m length of the wire carrying 8.0 A? Take μ₀ = 4π × 10⁻⁷ T m A⁻¹ for air.
Answer and reasoning
3.0 × 10⁻⁴ N — F/L = μ₀I₁I₂/(2πr) = (4π × 10⁻⁷ × 5.0 × 8.0) ÷ (2π × 0.040) = 2.0 × 10⁻⁴ N m⁻¹, so the force on 1.5 m is 2.0 × 10⁻⁴ × 1.5 = 3.0 × 10⁻⁴ N.
2.0 × 10⁻⁴ N — A student who stops at μ₀I₁I₂/(2πr) has the force per unit length, 2.0 × 10⁻⁴ N m⁻¹. The force on 1.5 m of wire is 1.5 times this: 3.0 × 10⁻⁴ N.
7.5 × 10⁻³ N — A student who assumes an inverse-square law divides by r². For long parallel wires F/L = μ₀I₁I₂/(2πr), with r to the first power, giving 3.0 × 10⁻⁴ N.
1.5 × 10⁻⁴ N — A student who thinks the calculated force is shared between the two wires halves it. F/L = μ₀I₁I₂/(2πr) gives the force per unit length on each wire: the wires exert equal and opposite forces on each other, each of magnitude 2.0 × 10⁻⁴ × 1.5 = 3.0 × 10⁻⁴ N.
Working F/L = μ₀I₁I₂/(2πr) = (4π × 10⁻⁷ T m A⁻¹)(5.0 A)(8.0 A)/(2π × 0.040 m) = (2 × 10⁻⁷ × 40)/0.040 N m⁻¹ = 2.0 × 10⁻⁴ N m⁻¹. F = (F/L) × L = 2.0 × 10⁻⁴ N m⁻¹ × 1.5 m = 3.0 × 10⁻⁴ N.
6 Two long straight parallel copper wires are 5.0 cm apart in air. One carries a current of 2.0 A and the other a current of 6.0 A, in the same direction. Which statement about the magnetic forces between the wires is correct?
Answer and reasoning
The wires push each other apart, as their currents are in the same direction. — A student who applies 'like repels like' from charges and poles picks this. The field of each wire at the other, combined with F = BIL, gives a force towards the other wire: currents in the same direction attract.
Each wire is pulled towards the other by a force of the same magnitude. — Each wire lies in the magnetic field of the other, and for currents in the same direction the force F = BIL on each points towards the other wire. The forces form a Newton's third law pair: F/L = μ₀I₁I₂/(2πr) is the same for both wires even though the currents differ.
Neither wire feels a force, as copper is not a magnetic material. — A student who thinks magnetic forces act only on iron-like materials picks this. Magnetic forces act on moving charges: each current sits in the field of the other and experiences a force F = BIL.
The 6.0 A wire pulls the 2.0 A wire harder than the 2.0 A wire pulls it. — A student who thinks the larger current exerts the larger force picks this. The 6.0 A wire produces the stronger field, but the 2.0 A current sits in it; the product I₁I₂ is the same for both, and the forces are an equal and opposite third-law pair.
7 The diagram shows a proton moving through a region of uniform magnetic field B, which is represented by the dots. In which direction is the magnetic force on the proton at the instant shown?
Answer and reasoning
Towards the right of the page — Dots show a field out of the page. For a positive charge moving towards the top of the page, F = qv × B (Fleming's left-hand rule with the current towards the top) gives a force towards the right of the page.
Towards the left of the page — To the left is the force if the dots showed a field into the page. A dot is the point of an arrow coming towards you, so the field is out of the page; with v towards the top and B out of the page the force on a positive charge is towards the right.
Towards the bottom of the page — The magnetic force is not a drag opposing the motion. It is always perpendicular to the velocity, so it cannot act towards the bottom of the page when the proton moves towards the top; here it acts to the right.
Out of the page towards you — The magnetic force is perpendicular to the field, not along it. The dots show the field out of the page, so the force lies in the plane of the page, at right angles to both the field and the velocity: to the right.
8 The diagram shows the path of a charged particle that enters a region of uniform magnetic field at O, moving to the right, and strikes a detector. The dots represent the field. No other forces act on the particle. Which statement about the particle is correct?
Answer and reasoning
It carries a negative charge. — The dots show the field out of the page. At O a positive charge moving to the right would feel a force towards the bottom of the page (F = qv × B) and curve downwards. This particle curves upwards, so the force on it is reversed: its charge is negative.
It is positively charged. — A positive charge would curve upwards only if the field were into the page. Dots are the points of arrows coming towards you, so the field here is out of the page, and a positive charge moving right at O would curve downwards. The upward path means the charge is negative.
It slows down along the arc. — The magnetic force is always perpendicular to the velocity, so it changes the direction of motion but not the speed. The particle moves at constant speed round the semicircle, which is why the arc has a single radius.
Its kinetic energy gets larger. — A force perpendicular to the velocity does no work, so although the magnetic force accelerates the particle (by changing its direction) it cannot change its kinetic energy. The kinetic energy stays constant.
9 The diagram shows an electron entering, at O, the region between two parallel plates in a vacuum. The potential difference between the plates is 150 V. The electron enters midway between the plates at speed v, parallel to them, and leaves at the far end of the plates having been deflected by the distance marked. Gravity is negligible. What is v? Take e = 1.60 × 10⁻¹⁹ C and m_e = 9.11 × 10⁻³¹ kg.
Answer and reasoning
3.4 × 10⁶ m s⁻¹ — This uses the potential difference, 150 V, as if it were the field strength. The field is E = V/d = 150/0.020 = 7500 V m⁻¹, which makes the acceleration 50 times larger and v about 7 times larger: 2.4 × 10⁷ m s⁻¹.
2.4 × 10⁷ m s⁻¹ — E = V/d = 150/0.020 = 7500 V m⁻¹, so a = eE/m_e = 1.32 × 10¹⁵ m s⁻². The horizontal speed is constant, so t = L/v and y = ½at² gives v = L√(a/2y) = 0.060 × √(1.32 × 10¹⁵/0.0080) = 2.4 × 10⁷ m s⁻¹.
4.9 × 10⁵ m s⁻¹ — This takes E = Vd = 150 × 0.020 = 3.0 V m⁻¹. Field strength is potential difference per unit separation, E = V/d = 7500 V m⁻¹, and the speed works out at 2.4 × 10⁷ m s⁻¹.
7.7 × 10⁶ m s⁻¹ — This substitutes 2.0, 6.0 and 4.0 as read from the diagram without converting to metres. With d = 0.020 m, L = 0.060 m and y = 0.0040 m, v = 2.4 × 10⁷ m s⁻¹.
Working From the diagram: plate separation d = 2.0 cm = 0.020 m, plate length L = 6.0 cm = 0.060 m, deflection y = 4.0 mm = 0.0040 m. Field between the plates E = V/d = 150/0.020 = 7500 V m⁻¹. Acceleration towards the positive plate a = eE/m_e = (1.60 × 10⁻¹⁹ × 7500)/(9.11 × 10⁻³¹) = 1.32 × 10¹⁵ m s⁻². The horizontal velocity is unchanged, so the time between the plates is t = L/v, and the deflection is y = ½at² = ½a(L/v)². Hence v = L√(a/2y) = 0.060 × √(1.32 × 10¹⁵/(2 × 0.0040)) = 0.060 × 4.06 × 10⁸ = 2.4 × 10⁷ m s⁻¹.
10 A straight wire of length 25 cm carries a current I and lies at 30° to a uniform magnetic field of strength B. The graph shows how the magnetic force F on the wire varies with I. What is B?
Answer and reasoning
120 mT — This takes the gradient to be BL, ignoring the angle. Only the component of the field perpendicular to the wire produces a force, so the gradient is BL sin θ and B = 0.030/(0.25 × 0.50) = 240 mT.
240 mT — The gradient of the F–I line is BL sin θ. Reading the graph, gradient = 0.120/4.0 = 0.030 N A⁻¹, so B = 0.030/(0.25 × sin 30°) = 0.030/0.125 = 0.24 T = 240 mT.
139 mT — This divides by cos 30° instead of sin 30°. The force depends on the component of the wire perpendicular to the field, F = BIL sin θ, so the gradient is BL sin θ and B = 0.030/(0.25 × 0.50) = 240 mT.
2.4 mT — This substitutes the length as 25 rather than 0.25 m. In SI units, B = 0.030/(0.25 × 0.50) = 0.24 T = 240 mT.
Working F = BIL sin θ, so a graph of F against I is a straight line through the origin with gradient BL sin θ. From the graph, gradient = 0.120 N/4.0 A = 0.030 N A⁻¹. With L = 0.25 m and θ = 30°: B = gradient/(L sin θ) = 0.030/(0.25 × 0.50) = 0.24 T = 240 mT.
11 The diagram shows three long straight parallel wires P, Q and R lying in the same plane, equally spaced, each carrying a current of the same magnitude I in the direction shown. What is the direction of the resultant magnetic force on wire Q?
Answer and reasoning
To the right, towards R — Unlike like charges and like poles, which repel, currents in the same direction attract. P pulls Q to the left and R, with the opposite current, pushes Q to the left as well, so the resultant is towards P.
Upwards, along the wire Q — The magnetic force on a wire is always perpendicular to the wire, F = BIL with B at right angles to the current, so it cannot act along Q. Both forces on Q act sideways, towards P.
To the left, towards P — P and Q carry currents in the same direction, so they attract: the force on Q from P is towards P. R carries a current opposite to Q, so R and Q repel: the force on Q from R is away from R, which is also towards P. The two forces are equal in size (same currents, same separation) and both point to the left.
Zero, as the forces cancel — The forces would cancel only if both neighbours attracted Q. R carries the opposite current to Q, so it repels Q, pushing it towards P, the same way that P pulls it. The forces add to give a resultant to the left.
12 The diagram shows a velocity selector. Positive ions enter the region between two charged plates moving to the right at speed v, midway between the plates; the crosses represent a uniform magnetic field B. Ions with speed v₀ pass straight through undeflected. Gravity is negligible. Which describes the path of an ion whose speed is greater than v₀?
Answer and reasoning
It curves towards the positive plate at the top — The electric force on a positive ion is towards the negative plate, downwards, and is the same for every speed. The magnetic force, from the crosses (field into the page) and the velocity to the right, is towards the top of the page and grows with speed, qvB. For v > v₀ the magnetic force wins, so the ion curves upwards, towards the positive plate.
It curves towards the negative plate below — A faster ion does not feel a smaller magnetic force: F = qvB grows with speed. Above v₀ the magnetic force (towards the top of the page) exceeds the electric force, so the ion is deflected towards the positive plate, not the negative one.
It continues straight through undeflected — Only at v₀ do the forces balance. The magnetic force qvB depends on speed while the electric force qE does not, so a faster ion feels a net force towards the top of the page and curves towards the positive plate.
It curves into the page, along the field — The magnetic force is perpendicular to the field, not along it. With the field into the page and the velocity to the right, the force lies in the plane of the page, towards the top, and the faster ion curves towards the positive plate.
That was your twenty minutes. Real practice on D.3 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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