4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A proton is first held at rest at a point O. Later it moves through O with constant velocity. Which statement correctly describes the fields the proton produces at a nearby point P in each case?
Answer and reasoning
AAt rest and moving, an electric field alone; magnetic fields come from magnets A student who thinks only magnets and magnetic materials produce magnetic fields picks this. A single moving charged particle produces a magnetic field; the fields of magnets themselves come from moving charges inside them.
BAt rest, an electric field by itself; in motion, electric and magnetic fieldsCorrect A charged particle always produces an electric field. A magnetic field is produced only by moving charge, so the moving proton produces both fields at P, while the proton at rest produces only the electric field. Motion adds the magnetic field; it does not remove the electric one.
CAt rest, an electric field alone; when it is moving, a magnetic field in its place A student who thinks electric effects belong to charges at rest and magnetic effects to moving charges picks this. The moving proton still has its electric field; its motion adds a magnetic field.
DAt rest and when moving, both an electric field and a magnetic field A student who treats charges as magnetic poles picks this. A proton at rest produces no magnetic field; only moving charge produces one.
The diagram shows a proton, its velocity v⃗ and a uniform magnetic field B⃗ at one instant. What is the direction of the magnetic force on the proton at that instant?
Answer and reasoning
ADown the page, perpendicular to the field A student who takes the product in the wrong order, B⃗ × v⃗, or uses the wrong hand gets the opposite direction. For a positive charge the force is along v⃗ × B⃗, which points up the page here.
BTo the right, along the field lines A student who thinks the magnetic force acts along the field lines, as the electric force does, picks this. The magnetic force is perpendicular to B⃗.
COut of the page, along the proton's velocity A student who thinks the magnetic force pushes a charge along its direction of motion picks this. The magnetic force is perpendicular to v⃗; it turns the proton without speeding it up.
DUp the page, perpendicular to the fieldCorrect F⃗B = q(v⃗ × B⃗) with q > 0. v⃗ points out of the page and B⃗ to the right: (out of page) × (right) = up the page (ẑ × x̂ = ŷ). The force is perpendicular to both the velocity and the field.
Working v⃗ = v ẑ (out of page), B⃗ = B x̂. F⃗ = e v B (ẑ × x̂) = evB ŷ: up the page.
A particle with charge −q (where q > 0) moves with speed v in the +x-direction through a region containing a uniform electric field of magnitude E in the +y-direction and a uniform magnetic field of magnitude B in the +z-direction. What is the y-component of the net force on the particle at that instant? Neglect gravity.
Answer and reasoning
A−q(E + vB) A student who reverses the electric force for the negative charge but takes the magnetic force straight from the right-hand rule, as for a positive charge, gets −qvB ŷ for it, and then −q(E + vB). The negative charge reverses v⃗ × B⃗ too, giving +qvB ŷ.
B−qE A student who thinks the magnetic force acts along the magnetic field, here along z, finds no magnetic contribution to Fy. The magnetic force is perpendicular to B⃗; here it is along y.
Cq(vB − E)Correct The two fields exert independent forces. Electric: (−q)E ŷ = −qE ŷ. Magnetic: (−q)(v x̂ × B ẑ) = (−q)(−vB ŷ) = +qvB ŷ, because x̂ × ẑ = −ŷ. Adding the y-components: Fy = qvB − qE = q(vB − E).
DqE − qvB A student who drops the minus sign of the charge and finds both forces as for a positive charge takes the electric force along E⃗, +qE ŷ, and the magnetic force along v⃗ × B⃗ = −vB ŷ, giving qE − qvB. The particle's charge is negative, so both forces are reversed: Fy = −qE + qvB = q(vB − E).
Working F⃗E = (−q)E ŷ. F⃗B = (−q)(v x̂ × B ẑ) = (−q)vB(−ŷ) = qvB ŷ. Fy = qvB − qE = q(vB − E). (Positive-charge rule for the magnetic part only: −q(E + vB); magnetic force along z: −qE; both forces taken as for a positive charge: qE − qvB.)
The diagram shows a flat metal strip, seen face-on, carrying a current I in a uniform magnetic field B⃗ directed into the page. The charge carriers in the metal are electrons. Which statement correctly describes the Hall potential difference in the strip?
Answer and reasoning
AThe top edge is at a higher potential than the edge at the bottom. A student who treats the carriers as positive charges moving with the conventional current finds them pushed toward the top edge and makes that edge positive. The carriers pushed toward the top edge are electrons, so the top edge is the negative one.
BThe potential differs between the front and back faces of the strip. A student who thinks the magnetic force pushes the carriers along the field lines, into or out of the page, picks this. The magnetic force is perpendicular to B⃗, so it pushes the carriers toward the top or bottom edge.
CThe potential differs only along the length of the strip, not across it. A student who thinks the only potential difference in a conductor is along the current picks this. The magnetic force on the drifting carriers produces an additional potential difference across the strip, between its top and bottom edges.
DThe bottom edge is at a higher potential than the top edge.Correct The electrons drift to the left, opposite to I. The force on each is (−e)(v⃗ × B⃗): v⃗ × B⃗ = (left) × (into page) points down the page, and the negative charge reverses it, so the electrons are pushed toward the top edge. The top edge becomes negative and the bottom edge, left with a deficit of electrons, is at the higher potential.
Working Electrons: v⃗ = −v x̂, B⃗ = −B ẑ. v⃗ × B⃗ = vB(x̂ × ẑ) = −vB ŷ; F⃗ = (−e)(−vB ŷ) = +evB ŷ: electrons pushed to the top edge. Top negative, bottom at higher potential.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
12.2.A.1 Magnetic field of a moving charged particle Fix
Magnetic field of a moving charged particle
A charged particle moving with velocity v⃗ produces a magnetic field in the space around it, in addition to its electric field. For speeds much less than the speed of light, B⃗ = (μ₀/4π)q(v⃗ × r̂)/r², which is the Biot–Savart law with qv⃗ in place of I dℓ⃗; r is the distance from the particle to the point and r̂ the unit vector from the particle toward the point. SI unit of B: tesla (T).
Vacuum permeability, μ₀
The constant that sets the strength of magnetic fields produced by moving charges and currents in empty space: μ₀ = 4π × 10⁻⁷ T·m/A.
Dependence on velocity and distance
In a given direction from a moving charged particle, the magnitude of its magnetic field is proportional to the particle's speed and inversely proportional to the square of the distance from the particle. A particle at rest produces no magnetic field.
Direction of the field of a moving charge
The magnetic field at a point is perpendicular to both the particle's velocity and the position vector from the particle to the point. For a positive charge it points along v⃗ × r̂, found with the right-hand rule; for a negative charge it points the opposite way. The field lines are circles around the line of motion.
Maximum field direction
Because B ∝ v sin θ, where θ is the angle between v⃗ and the position vector to the point, the field at a given distance is largest where θ = 90° and zero at points on the particle's line of motion (θ = 0° or 180°).
Students often think Only magnets and magnetic materials produce magnetic fields; a charged particle, whether moving or at rest, produces only an electric field. In fact A single charged particle produces a magnetic field whenever it moves. Magnets are not the only sources: moving charges are, and the fields of magnets themselves come from the circular or rotational motion of charges, such as electrons, inside them.
Students often think Electric charges act like magnetic poles, so a charged particle produces a magnetic field and interacts with magnets even when it is at rest. In fact No. A charged particle at rest produces an electric field but no magnetic field, and a magnetic field exerts no force on it. Electric charges are not magnetic poles.
12.2.B.1 Magnetic force on a moving charge, F⃗B = q(v⃗ × B⃗) Fix
Magnetic force on a moving charge, F⃗B = q(v⃗ × B⃗)
The force a magnetic field exerts on a charged particle moving through it. Magnitude |q|vB sin θ, where θ is the angle between v⃗ and B⃗; direction perpendicular to both v⃗ and B⃗, along v⃗ × B⃗ for a positive charge and opposite for a negative charge. It is zero for a charge at rest or moving along the field. SI unit: newton (N).
Tesla, T
The SI unit of magnetic field: 1 T is the field that exerts a force of 1 N on a charge of 1 C moving at 1 m/s perpendicular to it, so 1 T = 1 N·s/(C·m) = 1 N/(A·m).
Magnetic force does no work
Because the magnetic force is always perpendicular to the velocity, it does no work on the particle: it changes the direction of motion but not the speed or kinetic energy.
Circular and helical motion in a uniform field
A charged particle moving perpendicular to a uniform magnetic field moves in a circle: |q|vB = mv²/r gives r = mv/(|q|B) and period T = 2πm/(|q|B), independent of speed. A velocity component along the field is unaffected, so the general path is a helix whose pitch is v∥T.
Students often think The order of the vectors in the cross product does not matter, so the direction of the magnetic force can be found from B⃗ × v⃗ as well as from v⃗ × B⃗, or with either hand. In fact Along v⃗ × B⃗: with the fingers of the right hand along v⃗ and curled toward B⃗, the thumb points along the force. Reversing the order of the vectors, B⃗ × v⃗, gives the opposite direction.
Students often think The magnetic force on a moving charge acts along (or against) the magnetic field lines, as the electric force acts along the electric field. In fact No. The magnetic force q(v⃗ × B⃗) is perpendicular to the magnetic field as well as to the velocity. A charge moving along the field lines feels no magnetic force at all.
12.2.B.2 Independent electric and magnetic forces Fix
Independent electric and magnetic forces
In a region containing both fields, a moving charged particle feels the electric force qE⃗, which does not depend on its velocity, and the magnetic force q(v⃗ × B⃗) independently; the net force is their vector sum, F⃗ = qE⃗ + q(v⃗ × B⃗).
Crossed fields (velocity selector)
Uniform electric and magnetic fields perpendicular to each other and to a particle's velocity, arranged so that the two forces are opposite. A particle passes undeflected only if qE = qvB, that is, v = E/B, whatever its charge or mass.
Students often think Electric fields and forces belong to charges at rest and magnetic fields and forces to moving charges, so when a charge moves, its electric field and the electric force on it are replaced by magnetic ones (or weaken as… In fact Yes. A charged particle always produces an electric field and always feels the force qE⃗ in an electric field, whether it is at rest or moving. Motion adds a magnetic field and a magnetic force; it does not replace or weaken the electric ones.
Students often think The forces on a charged particle can be found as if its charge were positive: the electric force points along E⃗ and the magnetic force along v⃗ × B⃗, whatever the sign of the charge. In fact Yes. F⃗E = qE⃗ and F⃗B = q(v⃗ × B⃗) both contain q, so for a negative charge each force points opposite to E⃗ and to v⃗ × B⃗ respectively.
12.2.B.3 Hall effect Fix
Hall effect
When a conductor carrying a current is in a magnetic field with a component perpendicular to the current, the magnetic force pushes the moving charge carriers toward one side. Charge builds up until the transverse electric field it creates balances the magnetic force, leaving a potential difference across the conductor, perpendicular to both the current and the field.
Hall potential difference, ΔVH
The potential difference across a conductor produced by the Hall effect. For carriers of charge q, density n and drift speed vd in a strip of width w (across which ΔVH is measured) and thickness t (along B⃗): EH = vdB and ΔVH = vdBw = IB/(n|q|t). SI unit: volt (V). Its sign shows whether the carriers are positive or negative.
Students often think The charge carriers in a metal move in the direction of the conventional current, as positive charges would, so the charge that collects on one side of a conductor in a magnetic field is positive. In fact No. In metals the charge carriers are electrons, which drift opposite to the conventional current. The magnetic force on an electron drifting opposite to I points the same way as the force on a positive carrier moving with I, but the charge that collects on that side is negative.
Students often think The only potential difference in a current-carrying conductor is along the direction of the current, so any effect of a magnetic field appears as a potential difference along the conductor's length. In fact Yes. The magnetic force pushes the moving carriers toward one side of the conductor, and the charge that builds up there creates a potential difference across the conductor, perpendicular to both the current and the field: the Hall potential difference. It is separate from the potential difference along the conductor that drives the current.
12 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 12
A particle with charge +q moves with speed v. The magnetic field produced by a charged particle moving with velocity v⃗ is B⃗ = (μ₀/4π)q(v⃗ × r̂)/r², where r is the distance from the particle to the point and r̂ is the unit vector from the particle toward the point. At one instant, point P is a distance r from the particle in a direction perpendicular to the particle's velocity, and the magnetic field produced by the particle at P has magnitude B₀. A second particle, also with charge +q, moves with speed 2v. At one instant, point Q is a distance 2r from this particle, and the position vector from the particle to Q makes an angle of 37° with its velocity. What is the magnitude of the magnetic field produced by the second particle at Q? (sin 37° = 0.60)
Answer and reasoning
A0.50 B₀ A student who thinks the field has the same magnitude in every direction at a given distance leaves out the angle: 2 × (1/4) = 0.50. At a given distance the field depends on the direction through sin θ, which is 0.60 here, not 1.
B0.15 B₀ A student who treats the speed as an on-off switch for the field, rather than a factor in the given expression, gets (1/4) × 0.60 = 0.15. The field is proportional to v, so doubling v doubles it.
C0.60 B₀ A student who carries over the 1/r dependence of a long straight wire, instead of the r² in the given expression, gets 2 × (1/2) × 0.60 = 0.60. Doubling the distance divides this field by 4.
D0.30 B₀Correct From the field given in the stem, B = (μ₀/4π)|q|v sin θ/r². Relative to P, the speed is doubled (×2), the distance is doubled (×1/4) and sin θ changes from 1 to 0.60 (×0.60): 2 × (1/4) × 0.60 = 0.30, so B = 0.30 B₀.
Working B = (μ₀/4π)|q|v sin θ/r². At P: B₀ = (μ₀/4π)qv(1)/r². At Q: B = (μ₀/4π)q(2v)(0.60)/(2r)² = (2 × 0.60/4)B₀ = 0.30B₀. (Angle omitted: 0.50B₀; speed omitted: 0.15B₀; 1/r: 0.60B₀.)
The diagram shows an electron moving with velocity v⃗ and a point P at one instant. What is the direction of the magnetic field produced by the electron at P at that instant?
Answer and reasoning
AInto the page, away from you A student who applies the right-hand rule to v⃗ × r̂ and forgets to reverse it for a negative charge picks this. The field is q(v⃗ × r̂) times a positive factor, and q is negative for an electron, so the field points out of the page.
BUp the page, toward the electron A student who draws the magnetic field like the electric field of a negative charge, pointing toward the charge, picks this. The magnetic field is perpendicular to the position vector from the electron to P, so it has no component toward the electron.
COut of the page, toward youCorrect The position vector from the electron to P points down the page. With v⃗ to the right, v⃗ × r̂ = (right) × (down) points into the page. The electron's charge is negative, so its field is opposite to v⃗ × r̂: out of the page at P.
DRightward, along its velocity A student who pictures the field as carried along with the motion picks this. The magnetic field of a moving charge is perpendicular to its velocity; here it is perpendicular to the page.
Working r̂ from electron to P = −ŷ; v⃗ = +x̂. v⃗ × r̂ = x̂ × (−ŷ) = −ẑ (into page). q < 0 reverses it: B⃗ at P along +ẑ, out of the page.
The diagram shows a positively charged particle moving with velocity v⃗ and three points, P₁, P₂ and P₃, at the distances and angle shown, at one instant. The magnetic field produced by a charged particle moving with velocity v⃗ is B⃗ = (μ₀/4π)q(v⃗ × r̂)/r², where r is the distance from the particle to the point and r̂ is the unit vector from the particle toward the point. B₁, B₂ and B₃ are the magnitudes of the magnetic field produced by the particle at P₁, P₂ and P₃ at that instant. Which ranking is correct?
Answer and reasoning
AB₃ > B₂ > B₁Correct From the field given in the stem, B ∝ sin θ/r², where θ is the angle between v⃗ and the position vector to the point. P₁ is on the line of motion (θ = 0°), so B₁ = 0. P₂: sin 90°/(2d)² = 1/(4d²). P₃: sin 30°/d² = 1/(2d²). So B₃ > B₂ > B₁.
BB₁ = B₃ > B₂ A student who thinks the field is equally strong in every direction at a given distance ranks by distance alone: P₁ and P₃ are both at d, P₂ at 2d. The field depends on direction through sin θ: it is zero at P₁, on the line of motion.
CB₂ = B₃ > B₁ A student who carries over the long wire's 1/r dependence instead of the given 1/r² gets 1/(2d) at P₂ and sin 30°/d = 1/(2d) at P₃. A single moving charge produces a field proportional to 1/r², so P₂, at twice the distance, has only 1/(4d²), less than P₃'s 1/(2d²).
DB₁ > B₃ > B₂ A student who uses the velocity component toward the point (cos θ) finds the field largest directly ahead (P₁), smaller at 30° (P₃) and zero at 90° (P₂). The field depends on sin θ: it is zero ahead of the particle and largest perpendicular to its velocity.
Working B = (μ₀/4π)qv sin θ/r². P₁: θ = 0 → 0. P₂: θ = 90°, r = 2d → 0.25 (in units of μ₀qv/(4πd²)). P₃: θ = 30°, r = d → 0.50. B₃ > B₂ > B₁. (No angle: 1, 0.25, 1; 1/r: 0, 0.5, 0.5; cos θ: 1, 0, 0.87.)
A proton with kinetic energy 1.0 × 10⁻¹⁵ J moves in a uniform magnetic field of magnitude 0.20 T. The angle between the proton's velocity and the field is 30°. What is the magnitude of the magnetic force on the proton? Use e = 1.60 × 10⁻¹⁹ C and mp = 1.67 × 10⁻²⁷ kg.
Answer and reasoning
A3.0 × 10⁻¹⁴ N A student who uses the velocity component along the field, cos 30°, gets evB cos 30° = 3.0 × 10⁻¹⁴ N. The magnetic force depends on the component perpendicular to the field: sin 30°.
B3.5 × 10⁻¹⁴ N A student who uses F = evB whatever the angle gets 3.5 × 10⁻¹⁴ N. The force is evB sin θ; at 30° it is half of evB.
C1.8 × 10⁻¹⁴ NCorrect The proton's speed is v = √(2K/mp) = √(2(1.0 × 10⁻¹⁵ J)/(1.67 × 10⁻²⁷ kg)) = 1.09 × 10⁶ m/s. The magnetic force has magnitude evB sin θ = (1.60 × 10⁻¹⁹ C)(1.09 × 10⁶ m/s)(0.20 T)(sin 30°) = 1.8 × 10⁻¹⁴ N.
D1.6 × 10⁻²⁰ N A student who writes the magnetic force as eB sin θ, by analogy with qE, leaves out the speed and gets (1.60 × 10⁻¹⁹ C)(0.20 T)(0.50) = 1.6 × 10⁻²⁰. The force is proportional to the speed: F = evB sin θ.
Working v = √(2K/mp) = √(2 × 1.0e-15/1.67e-27) = 1.094 × 10⁶ m/s. F = evB sin 30° = 1.60e-19 × 1.094e6 × 0.20 × 0.50 = 1.75 × 10⁻¹⁴ N ≈ 1.8 × 10⁻¹⁴ N. (cos 30°: 3.0 × 10⁻¹⁴ N; no angle: 3.5 × 10⁻¹⁴ N; no speed: 1.6 × 10⁻²⁰.)
A charged particle moves in a circle of radius r₀ in a uniform magnetic field, with its velocity perpendicular to the field. The particle's kinetic energy is then increased by a factor of 4, and it continues to move perpendicular to the same field. What is the radius of its new circular path?
Answer and reasoning
AThe radius doubles, to 2r₀Correct From |q|vB = mv²/r, r = mv/(|q|B), so r is proportional to the speed. Kinetic energy (1/2)mv² is proportional to v², so four times the kinetic energy means twice the speed, and the radius becomes 2r₀.
BThe radius becomes 4r₀ A student who takes the radius to be proportional to the kinetic energy picks this. The radius is proportional to the momentum mv; the speed only doubles when the kinetic energy is multiplied by 4.
CThe radius falls to r₀/2 A student who notes that the magnetic force doubles with the speed, and concludes that the particle is bent more tightly, picks this. The force needed for circular motion, mv²/r, grows as v², faster than |q|vB, so the circle gets larger.
DThe radius remains r₀ A student who carries over the speed-independence of the period to the radius picks this. The period 2πm/(|q|B) does not depend on speed, but the radius mv/(|q|B) does.
Working r = mv/(|q|B) ∝ v. K ∝ v²: K × 4 → v × 2 → r × 2 = 2r₀. (r ∝ K: 4r₀; force-only reasoning: r₀/2; r independent of v: r₀.)
The diagram shows three particles, 1, 2 and 3, each with charge +q, moving with the velocities shown in a uniform magnetic field B⃗. F₁, F₂ and F₃ are the magnitudes of the magnetic forces on particles 1, 2 and 3. Which ranking is correct?
Answer and reasoning
AF₂ > F₁ = F₃ A student who uses F = qvB whatever the direction of the velocity ranks by speed alone: 2qvB for particle 2 and qvB for particles 1 and 3. The force depends on sin θ: it is zero for particle 1, which moves along the field.
BF₂ = F₃ > F₁Correct F = qvB sin θ, where θ is the angle between the velocity and the field. Particle 1 moves along the field: F₁ = 0. Particle 2: q(2v)B sin 30° = qvB. Particle 3 moves perpendicular to the field: F₃ = qvB. So F₂ = F₃ > F₁.
CF₂ > F₁ > F₃ A student who uses the velocity component along the field, cos θ, gets 2v cos 30° ≈ 1.7v for particle 2, v for particle 1 and zero for particle 3. The force depends on the component perpendicular to the field, sin θ.
DF₃ > F₂ > F₁ A student who leaves the speed out of the magnetic force, as for F = qE, ranks by sin θ alone: 1 for particle 3, 0.5 for particle 2 and 0 for particle 1. The force is proportional to the speed, so particle 2's doubled speed makes up for its sin 30°.
Working F = qvB sin θ. 1: θ = 0 → 0. 2: 2v, 30° → qvB. 3: v, 90° → qvB. F₂ = F₃ > F₁. (No angle: 1, 2, 1; cos: 1, 1.73, 0; no speed: 0, 0.5, 1.)
A proton moves with velocity 3.0 × 10⁵ m/s in the +x-direction through a region containing a uniform electric field of 4.0 × 10⁴ N/C in the +y-direction and a uniform magnetic field of 0.10 T, also in the +y-direction. What is the magnitude of the net force on the proton? Use e = 1.60 × 10⁻¹⁹ C and neglect gravity.
Answer and reasoning
A8.0 × 10⁻¹⁵ NCorrect Electric force: eE = (1.60 × 10⁻¹⁹ C)(4.0 × 10⁴ N/C) = 6.4 × 10⁻¹⁵ N along +y. Magnetic force: e(v⃗ × B⃗) with v⃗ along +x and B⃗ along +y gives evB = 4.8 × 10⁻¹⁵ N along +z. The forces are perpendicular, so the net force is √(6.4² + 4.8²) × 10⁻¹⁵ N = 8.0 × 10⁻¹⁵ N.
B1.1 × 10⁻¹⁴ N A student who thinks the magnetic force acts along the magnetic field, here along +y like the electric force, adds the two: 6.4 × 10⁻¹⁵ N + 4.8 × 10⁻¹⁵ N = 1.1 × 10⁻¹⁴ N. The magnetic force is along +z, perpendicular to the electric force.
C4.8 × 10⁻¹⁵ N A student who thinks a moving charge feels only the magnetic force keeps evB alone. The electric force eE acts on the proton whatever its velocity.
D6.4 × 10⁻¹⁵ N A student who thinks the magnetic force depends on the velocity component along the field finds zero magnetic force, because v⃗ is perpendicular to B⃗, and keeps eE alone. Perpendicular velocity gives the largest magnetic force, evB.
Working FE = 1.60e-19 × 4.0e4 = 6.4 × 10⁻¹⁵ N (+y). FB = 1.60e-19 × 3.0e5 × 0.10 = 4.8 × 10⁻¹⁵ N (x̂ × ŷ = ẑ: +z). |F| = √(6.4² + 4.8²) × 10⁻¹⁵ = 8.0 × 10⁻¹⁵ N.
A beam of protons moving in the +x-direction passes undeflected through a region containing a uniform electric field and a uniform magnetic field that are perpendicular to each other and to the beam. A student claims that protons entering the region with a greater speed will be deflected. Which reasoning correctly supports the student's claim?
Answer and reasoning
AThe magnetic field does work on the faster protons, so their kinetic energy increases and their paths bend. A student who thinks the magnetic force can do work picks this. The magnetic force is perpendicular to the velocity and does no work; the deflection comes from the unbalanced forces, not from any change in kinetic energy caused by the magnetic field.
BThe electric force does not depend on speed, but the magnetic force is proportional to it, so the forces no longer cancel.Correct For the beam to pass undeflected, the electric force qE and the magnetic force qvB must be equal and opposite, which happens only at v = E/B. The electric force on a proton is the same at any speed, but the magnetic force increases with speed, so a faster proton feels a net force toward the magnetic force's side and is deflected.
CThe magnetic force on a faster proton is larger and acts along the magnetic field, pushing it sideways. A student who thinks the magnetic force acts along the magnetic field picks this. The magnetic force is perpendicular to B⃗; in the undeflected beam it is directly opposite to the electric force.
DA faster proton is less affected by the electric field, since electric forces act mainly on charges at rest. A student who thinks electric forces act on charges at rest and magnetic forces on moving ones picks this. The electric force qE on a proton is the same whatever its speed; it is the magnetic force that changes.
The diagram, not drawn to scale, shows a thin copper strip carrying a current of 12 A in a uniform magnetic field of 1.5 T directed perpendicular to the strip's broad face. The charge carriers are electrons, with a density of 8.5 × 10²⁸ electrons per cubic meter. What is the magnitude of the Hall potential difference across the strip? Use e = 1.60 × 10⁻¹⁹ C.
Answer and reasoning
A8.8 × 10⁻⁸ V A student who thinks the magnetic force pushes the carriers along the field, so that the charge builds up on the broad faces, uses the thickness t = 0.20 mm instead of the width: vdBt = 8.8 × 10⁻⁸ V. The force is perpendicular to B⃗, across the width w.
B4.4 × 10⁻⁴ V A student who takes the Hall field, EH = vdB = 4.4 × 10⁻⁴ V/m, as the potential difference picks this. The potential difference across the strip is the field times the width it acts across: EHw.
C6.6 × 10⁻⁶ VCorrect The drift speed is vd = I/(neA) with A = wt: vd = 12 A/[(8.5 × 10²⁸ m⁻³)(1.60 × 10⁻¹⁹ C)(0.015 m)(2.0 × 10⁻⁴ m)] = 2.9 × 10⁻⁴ m/s. The magnetic force pushes the electrons across the width w, so the Hall field is EH = vdB = 4.4 × 10⁻⁴ V/m and the Hall potential difference is EHw = 6.6 × 10⁻⁶ V. Equivalently, ΔVH = IB/(net).
D2.2 × 10⁻⁵ V A student who thinks the magnetic field's effect appears along the length of the strip uses ℓ = 5.0 cm: vdBℓ = 2.2 × 10⁻⁵ V. The Hall potential difference is across the strip, perpendicular to the current, over the width w.
Working A = wt = (0.015 m)(2.0e-4 m) = 3.0e-6 m². vd = I/(neA) = 12/(8.5e28 × 1.60e-19 × 3.0e-6) = 2.94e-4 m/s. EH = vdB = 4.41e-4 V/m. ΔVH = EHw = 6.6 × 10⁻⁶ V (= IB/(net)).
A particle of mass m and charge q > 0 enters a region of uniform magnetic field of magnitude B with speed v. Its velocity has a component v∥ parallel to the field and a component v⊥ perpendicular to the field, and it moves along a helix. What distance does the particle move along the field direction during one revolution?
Answer and reasoning
A2πmv/(qB) A student who takes the distance moved along the field in one revolution to be the whole distance traveled along the helical path in one period, vT, picks this. Only the parallel component carries the particle along the field; the distance along B⃗ per revolution is v∥T.
Bmv⊥/(qB) A student who takes the distance along the field per revolution to be the radius of the helix picks this. mv⊥/(qB) is the radius of the circular part of the motion; the advance along the field is v∥ times the period.
C2πmv∥v⊥/(qB) A student who writes the magnetic force as qB, without the speed, gets qB = mv⊥²/r, r = mv⊥²/(qB) and T = 2πmv⊥/(qB), so v∥T = 2πmv∥v⊥/(qB). The force is qv⊥B, which makes the period independent of speed; the units of this expression, m²/s, also show it is not a distance.
D2πmv∥/(qB)Correct The perpendicular component gives circular motion: qv⊥B = mv⊥²/r, so r = mv⊥/(qB) and the period is T = 2πr/v⊥ = 2πm/(qB). The magnetic force has no component along B⃗, so v∥ stays constant, and in one period the particle moves v∥T = 2πmv∥/(qB) along the field.
Working qv⊥B = mv⊥²/r → r = mv⊥/(qB); T = 2πr/v⊥ = 2πm/(qB); pitch p = v∥T = 2πmv∥/(qB). Units: kg·(m/s)/(C·T) = m. (Path length vT: 2πmv/(qB); radius: mv⊥/(qB); F = qB: 2πmv∥v⊥/(qB), units m²/s.)
A particle of mass m and charge q > 0 moves with speed v in a circle in a uniform magnetic field of magnitude B. Its velocity is perpendicular to the field, and no other force acts on it. What is the magnitude of the average magnetic force on the particle (its change in momentum divided by the time taken) while it travels half-way around the circle, from one point to the point diametrically opposite?
Answer and reasoning
AqvB A student who averages the force's magnitude, which stays qvB, picks this. The force turns through 180° during the half revolution, so its components partly cancel, and the average force vector, Δp⃗/Δt, is smaller than qvB.
B0 A student who reasons that a force doing no work cannot change momentum picks this. The magnetic force does no work, so the speed is constant, but it turns the velocity: the momentum changes from mv⃗ to −mv⃗, a change of magnitude 2mv.
C(2/π)qvBCorrect The field does no work, so the speed stays v and after half a revolution the velocity is reversed: |Δp⃗| = 2mv. From qvB = mv²/r, r = mv/(qB), so the half revolution takes t = πr/v = πm/(qB). The average force is 2mv/t = (2/π)qvB.
D4qvB A student who takes the period of the circular motion as 1/ω = m/(qB) gets a half-revolution time of m/(2qB) and an average force of 2mv ÷ m/(2qB) = 4qvB. One revolution is 2π radians, so the half revolution takes π/ω = πm/(qB).
Working The magnetic force does no work, so the speed stays v; after half a revolution the velocity is reversed, so |Δp⃗| = 2mv. From qvB = mv²/r, r = mv/(qB); the half revolution takes t = πr/v = πm/(qB). Favg = 2mv/t = 2mv·qB/(πm) = (2/π)qvB. Distractors: average of the magnitude → qvB; no work, so no momentum change → 0; period taken as 1/ω = m/(qB), so the half revolution takes m/(2qB) → Favg = 4qvB.
A particle of mass m and charge q > 0 is in a region containing a uniform magnetic field of magnitude B in the +z-direction and a uniform electric field of magnitude E, also in the +z-direction. At time t = 0 the particle's velocity has magnitude v₀ and is perpendicular to the z-axis. Neglect gravity. How far does the particle move in the z-direction during its first complete revolution about the z-direction (as seen looking along the z-axis)?
Answer and reasoning
AmE/(2qB²) A student who takes the period of the circular motion as 1/ω = m/(qB) picks this: ½(qE/m)(m/(qB))². One revolution is 2π radians, so it takes T = 2π/ω = 2πm/(qB), and the distance is 4π² times larger.
B2π²mE/(qB²)Correct The fields act independently. Seen along z, the magnetic force keeps the particle on a circle with period T = 2πm/(qB), whatever vz is, since vz is parallel to B⃗. Along z only the electric force acts, so the particle starts with vz = 0, accelerates at qE/m, and in one period moves z = ½(qE/m)T² = 2π²mE/(qB²).
C2πmv₀/(qB) A student who takes the distance along the field in one revolution to be the whole path length at the initial speed, v₀T = 2πmv₀/(qB), picks this. The motion at speed v₀ is around the circle, perpendicular to z; the motion along z starts from rest and comes only from the electric force.
Dmv₀/(qB) A student who takes the distance along the field in one revolution to be the radius of the circle, mv₀/(qB), picks this. The radius describes the motion perpendicular to z; the distance along z comes from the electric force acting for one period.
Working The electric force qE ẑ is along z; the magnetic force q(v⃗ × B⃗) depends only on the velocity component perpendicular to B⃗ and is itself perpendicular to z. So the motion seen along z is a uniform circle at speed v₀, with qv₀B = mv₀²/r and period T = 2πr/v₀ = 2πm/(qB), unaffected by E or by vz. Along z the particle starts with vz = 0 and has acceleration qE/m, so in one period z = ½(qE/m)T² = ½(qE/m)(2πm/(qB))² = 2π²mE/(qB²). Distractors: T taken as 1/ω = m/(qB) → ½(qE/m)(m/(qB))² = mE/(2qB²); distance along the field taken as the path length at the initial speed, v₀T = 2πmv₀/(qB); distance along the field taken as the radius, mv₀/(qB).
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
