12.2 Magnetism and Moving Charges

1. ADirectly out of the page
   A student who applies the right-hand rule without allowing for the electron's negative charge picks this. Out of the page is the field of a positive charge moving to the right; for an electron the direction is reversed.
2. BDown to the electron
   A student who gives the magnetic field the pattern of the electron's electric field, which points toward a negative charge, picks this. The magnetic field at P is perpendicular to the position vector from the electron to P, so it cannot point along the line to the electron.
3. CDirectly into the page Correct
   For a positive charge moving to the right, the right-hand rule (thumb along the velocity, fingers curling around the line of motion) gives a field out of the page at points above it. The electron's charge is negative, so its field is reversed: into the page at P. This direction is perpendicular to both the velocity and the position vector from the electron to P.
4. DRight, along its velocity
   A student who thinks a moving charge's magnetic field points along its motion picks this. The field is perpendicular to the velocity at every point; at P it points into the page.

CED 12.2.A.1.ii · Read this in Fix

1. AA magnetic force is exerted on the proton at rest, not on the other.
   A student who thinks a magnetic field exerts a force on a charge whatever its velocity picks this. Proton 1's magnetic field does reach proton 2, but proton 2 is at rest, and the magnetic force qvB sin θ is zero when v = 0.
2. BNeither of the protons exerts a magnetic force on the other. Correct
   Proton 1 moves, so it produces a magnetic field at proton 2, but proton 2 is at rest and a magnetic field exerts no force on a charge at rest (F_B = qvB sin θ with v = 0). Proton 2 is at rest, so it produces no magnetic field and exerts no magnetic force on proton 1. Only their electric forces remain.
3. CA magnetic force is exerted on the moving proton, not on the other.
   A student who thinks a charge at rest produces a magnetic field picks this, reasoning that proton 2's field pushes on the moving proton 1. Proton 2 is at rest, so it produces an electric field but no magnetic field, and no magnetic force is exerted on proton 1.
4. DEach exerts a magnetic force on the other, just as with electric forces.
   A student who thinks magnetic forces act between any two charges, as electric forces do, picks this. The protons exert electric forces on each other, but magnetic forces describe interactions between moving charges, and here proton 2 is at rest.

CED 12.2.B.1 · Read this in Fix

1. AConstant, as the net force exerted on it is zero
   A student who thinks constant speed means zero net force picks this. The proton's velocity keeps changing direction, so a net force, the magnetic force toward the center of the circle, is exerted on it; the speed is constant because that force is perpendicular to the velocity.
2. BConstant, as the force is perpendicular to its motion Correct
   The magnetic force is perpendicular to the velocity at every instant, so it does no work on the proton (W = Fd cos 90° = 0). With no work done, the kinetic energy, and so the speed, stays constant; the force changes only the direction of motion, which is why the path curves.
3. CDecreasing, as the magnetic force is against its motion
   A student who thinks the magnetic force acts like friction, against the motion, picks this. The magnetic force is perpendicular to the velocity, not opposite to it, so it does no work and does not slow the proton.
4. DIncreasing, as an unbalanced force is exerted on it
   A student who thinks any net force must change an object's speed picks this. A net force perpendicular to the velocity changes only the direction of motion; the magnetic force does no work, so the speed stays constant.

CED 12.2.B.2.ii · Read this in Fix

1. A3.2 × 10⁻¹⁶ N Correct
   The two fields exert independent forces. The electric force is eE = 4.8 × 10⁻¹⁶ N toward the bottom of the page. The magnetic force is evB = 1.6 × 10⁻¹⁶ N toward the top of the page (right-hand rule for a positive charge moving right in a field into the page). They are opposite, so the net force is 4.8 × 10⁻¹⁶ N − 1.6 × 10⁻¹⁶ N = 3.2 × 10⁻¹⁶ N.
2. B6.4 × 10⁻¹⁶ N
   A student who adds the two force magnitudes without regard to direction picks this: 4.8 × 10⁻¹⁶ N + 1.6 × 10⁻¹⁶ N. The electric force points toward the bottom of the page and the magnetic force toward the top, so their magnitudes must be subtracted.
3. C4.8 × 10⁻¹⁶ N
   A student who uses F_B = qvB cos θ picks this: with θ = 90° the magnetic force comes out zero, leaving only the electric force. The magnetic force depends on sin θ, which is 1 at 90°, so F_B = 1.6 × 10⁻¹⁶ N and it must be included.
4. D1.6 × 10⁻¹⁶ N
   A student who thinks an electric field exerts forces only on charges at rest, so that a moving proton feels only the magnetic force, picks this. An electric field exerts the force qE on a charge whether or not it moves, independently of the magnetic force.

Working Electric force: F_E = eE = (1.60 × 10⁻¹⁹ C)(3.0 × 10³ N/C) = 4.8 × 10⁻¹⁶ N, toward the bottom of the page (along E⃗ for a positive charge). Magnetic force: F_B = evB sin 90° = (1.60 × 10⁻¹⁹ C)(2.0 × 10⁴ m/s)(0.050 T) = 1.6 × 10⁻¹⁶ N; the right-hand rule (v⃗ to the right, B⃗ into the page, positive charge) gives toward the top of the page. The forces are independent and opposite: F_net = 4.8 × 10⁻¹⁶ N − 1.6 × 10⁻¹⁶ N = 3.2 × 10⁻¹⁶ N, toward the bottom of the page.

CED 12.2.B.3 · Read this in Fix

1. AThe edge toward the bottom of the page
   A student who treats the moving charges as positive charges moving with the current picks this: positive carriers moving right would be pushed to the top edge, leaving the bottom edge negative. In a metal the carriers are electrons moving left, and they are pushed to the top edge, making it negative.
2. BThe face on the far side, into the page
   A student who thinks the magnetic force is exerted along the field picks this, sending the electrons into the page. The magnetic force is perpendicular to the field, so the charges separate between the top and bottom edges, not the front and back faces.
3. CThe right-hand end of the strip
   A student who thinks the magnetic force drags against the electrons' motion picks this, piling them up at the right end. The magnetic force is perpendicular to the electrons' velocity, so it pushes them sideways, across the strip, not back along it.
4. DThe edge toward the top of the page Correct
   The electrons move opposite to the current, toward the left. For a positive charge moving left in a field into the page, the right-hand rule gives a force toward the bottom of the page; the electrons are negative, so they are pushed toward the top edge, which becomes negatively charged, leaving the bottom edge positive. The resulting potential difference is across the strip, perpendicular to both the current and the field.

CED 12.2.B.4 · Read this in Fix

1. AElectrons produce a magnetic field whether or not they are moving.
   A student who thinks charged objects produce magnetic fields even at rest picks this. A field that did not depend on the electrons' motion could not reverse when the beam's direction reverses; the reversal shows that the field depends on the electrons' velocity.
2. BMoving electrons produce a magnetic field directed along their velocity.
   A student who thinks a moving charge's magnetic field points along its motion picks this. A field along the beam would keep the needle pointing along the tube, north–south; instead the needle swings toward a direction at right angles to the beam.
3. CThe electrons' charge attracts one pole of the compass needle.
   A student who pictures magnetic poles as charges picks this. An attraction between a charge and a pole would not depend on which way the electrons travel, yet reversing the beam reverses the needle's swing. The needle responds to the magnetic field of the moving electrons, not to their charge as such.
4. DMoving electrons produce a magnetic field perpendicular to their velocity. Correct
   The needle lines up with the magnetic field where it sits. It turns only when electrons move along the tube, it turns toward a direction at right angles to the beam, and it turns the other way when the electrons' velocity is reversed. So the moving electrons produce a magnetic field perpendicular to their velocity, whose direction depends on the direction of that velocity.

CED 12.2.A.1 · Read this in Fix

1. AB₂ > B₁ > B₃ > 0
   A student who thinks a charged object at rest still produces a magnetic field picks this. The proton in Case 3 produces an electric field at P₃ but no magnetic field; only moving charges produce magnetic fields, so B₃ = 0.
2. BB₂ > B₁ > B₃ = 0 Correct
   Each point is the same distance d from its proton, in the same direction relative to the velocity, so only the velocity differs. A faster proton produces a stronger magnetic field at the same point, so B₂ > B₁. The proton in Case 3 is at rest and produces no magnetic field at all, so B₃ = 0.
3. CB₁ = B₂ = B₃ > 0
   A student who treats the magnetic field like the electric field, set by charge and distance alone, picks this. The magnetic field of a proton depends on its velocity: it is larger for the faster proton and zero for the proton at rest.
4. DB₁ = B₂ = B₃ = 0
   A student who thinks only magnets and wires produce magnetic fields picks this. A single moving proton does produce a magnetic field; only the proton at rest, in Case 3, produces none.

Working Same distance d and same direction (point beside the proton) in every case, so only the velocity differs. The field produced by a moving charge increases with its speed: B₂ (2v) > B₁ (v). A charge at rest produces no magnetic field: B₃ = 0. Ranking: B₂ > B₁ > B₃ = 0.

CED 12.2.A.1.i · Read this in Fix

1. AB_P > B_Q > B_R = 0 Correct
   P and Q are beside the proton, where the position vector is perpendicular to the velocity, so the field there is the largest possible for each distance; P is closer, so B_P > B_Q. R is on the proton's line of motion, where the position vector is parallel to the velocity and the field is zero, even though R is as close as P.
2. BB_P = B_R > B_Q > 0
   A student who thinks the field depends only on distance, as the electric field does, picks this, since P and R are both a distance d away. The magnetic field also depends on direction: it is greatest beside the proton, at P, and zero on its line of motion, at R.
3. CB_R > B_P > B_Q > 0
   A student who thinks a moving charge's field is strongest directly ahead of it picks this. The field is perpendicular to the velocity and is largest beside the proton; at points on its line of motion, such as R, it is zero.
4. DB_P = B_Q = B_R = 0
   A student who thinks a single charged particle produces no magnetic field picks this. A moving proton does produce a magnetic field; it is zero only at points on its line of motion, such as R.

Working P and Q are beside the proton (position vector perpendicular to the velocity), where the field is the largest for a given distance; P (distance d) is closer than Q (distance 2d), so B_P > B_Q > 0. R is on the line of motion (position vector parallel to the velocity), where the field is zero. Ranking: B_P > B_Q > B_R = 0.

CED 12.2.A.1.iii · Read this in Fix

1. AF₁ = F₃ > F₂ = 0
   A student who uses cos θ in place of sin θ picks this, getting the largest force for motion along or against the field and none for motion across it. The magnetic force depends on sin θ: it is zero at 0° and 180° and largest at 90°.
2. BF₁ = F₂ = F₃ > 0
   A student who thinks a magnetic field pushes on a charge as an electric field does, whatever its direction of motion, picks this. The magnetic force depends on the angle between the velocity and the field; it is zero for protons 1 and 3, which move parallel and antiparallel to the field.
3. CF₁ = F₂ = F₃ = 0
   A student who thinks magnetic fields exert forces only on magnets and magnetic materials picks this. A magnetic field exerts a force on a moving charged particle whose velocity is not parallel or antiparallel to the field, such as proton 2.
4. DF₂ > F₁ = F₃ = 0 Correct
   F_B = qvB sin θ. Proton 2 moves perpendicular to the field (θ = 90°, sin θ = 1), so the force on it is the largest possible, qvB. Proton 1 moves along the field (θ = 0°) and proton 3 against it (θ = 180°); sin θ = 0 in both cases, so no magnetic force is exerted on them.

Working F_B = qvB sin θ with the same q, v and B for all three. Proton 1: θ = 0°, sin θ = 0, F₁ = 0. Proton 2: θ = 90°, sin θ = 1, F₂ = qvB. Proton 3: θ = 180°, sin θ = 0, F₃ = 0. Ranking: F₂ > F₁ = F₃ = 0.

CED 12.2.B.2.i · Read this in Fix

1. A½
   A student who reasons that doubling the speed doubles the force, so the proton is pulled into a tighter circle, picks this. The faster proton also needs a larger centripetal force, mv²/r, which grows as v² while the magnetic force grows only as v, so its circle is larger, not smaller.
2. B1
   A student who writes the centripetal force as mv/r instead of mv²/r picks this: qvB = mv/r gives r = m/(qB), with no dependence on speed. The centripetal force is mv²/r, so r = mv/(qB), which doubles when v doubles.
3. C2 Correct
   The magnetic force qvB is the centripetal force mv²/r, so r = mv/(qB). With the same mass, charge and field, r is proportional to v: doubling the speed doubles the force, but the force needed to stay on the old circle would quadruple, so the path widens to a radius of 2r.
4. D4
   A student who treats the magnetic force as qB, independent of speed, picks this: qB = mv²/r gives r = mv²/(qB), so doubling v quadruples r. The magnetic force is qvB; it doubles with the speed, so the radius only doubles.

Working The magnetic force is the centripetal force: qvB = mv²/r, so r = mv/(qB). With m, q and B unchanged, r ∝ v, so doubling v doubles r: k = 2. (Errors: F = qB with no v gives r = mv²/(qB), k = 4; F = mv/r gives r = m/(qB), k = 1; 'twice the force, so a tighter circle' gives k = ½.)

CED 12.2.B.2.i · Read this in Fix

1. AToward the top of the page Correct
   For a positive charge moving to the right in a field out of the page, the right-hand rule gives a force toward the bottom of the page. The electron's charge is negative, so the force on it is reversed: toward the top of the page. It is perpendicular to both the velocity and the field.
2. BToward the bottom of the page
   A student who applies the right-hand rule without reversing it for a negative charge picks this. Toward the bottom of the page is the force on a positive charge; the electron, being negative, is pushed toward the top.
3. COut of the page, along B
   A student who thinks the magnetic force is exerted along the field lines picks this. The magnetic force is perpendicular to the field and to the velocity, so here it lies in the plane of the page.
4. DLeft, opposite its motion
   A student who thinks a magnetic field drags on a moving charge like friction picks this. The magnetic force is perpendicular to the velocity, so it changes the electron's direction of motion, not its speed.

CED 12.2.B.2.ii · Read this in Fix

1. A3.0 × 10⁻⁵ V
   A student who thinks this potential difference forms along the direction of the current picks this, using the length: vBℓ = (2.0 × 10⁻⁴ m/s)(1.5 T)(0.10 m). The magnetic force pushes the electrons sideways, across the strip, so the potential difference forms across the width.
2. B3.0 × 10⁻⁷ V
   A student who thinks the magnetic force pushes the charges along the field, into the page, picks this, using the thickness: vBt = (2.0 × 10⁻⁴ m/s)(1.5 T)(1.0 × 10⁻³ m). The force is perpendicular to the field, so the charges separate across the width, not the thickness.
3. C6.0 × 10⁻⁶ V Correct
   In the steady state eE = evB, so E = vB = (2.0 × 10⁻⁴ m/s)(1.5 T) = 3.0 × 10⁻⁴ V/m. The magnetic force is perpendicular to both the electrons' velocity (along the strip) and the field (perpendicular to the page), so the charges separate across the 0.020 m width. For a uniform field, |ΔV| = E·w = (3.0 × 10⁻⁴ V/m)(0.020 m) = 6.0 × 10⁻⁶ V.
4. D3.0 × 10⁻² V
   A student who writes the magnetic force as eB, without the speed, picks this: eE = eB gives E = B and ΔV = Bw = (1.5)(0.020). The magnetic force is evB; because the electrons drift so slowly, the potential difference is only a few microvolts.

Working Steady state: eE = evB, so E = vB = (2.0 × 10⁻⁴ m/s)(1.5 T) = 3.0 × 10⁻⁴ V/m. The electrons move along the length and the field is perpendicular to the page, so the magnetic force, perpendicular to both, is across the width w: the charges separate between the top and bottom edges. ΔV = Ew = vBw = (3.0 × 10⁻⁴ V/m)(0.020 m) = 6.0 × 10⁻⁶ V.

CED 12.2.B.4 · Read this in Fix

1. A√(mΔV/q/2)/B
   A student who rearranges ½mv² = qΔV as v² = qΔV/(2m), carrying the ½ across without inverting it, picks this: v = √(qΔV/(2m)) and r = mv/(qB) = √(mΔV/(2q))/B = √(mΔV/q/2)/B, half the correct radius. Multiplying both sides by 2 gives v² = 2qΔV/m.
2. Bm/(qB)
   A student who writes the centripetal force as mv/r picks this: qvB = mv/r gives r = m/(qB), with no dependence on speed or on ΔV. The centripetal force is mv²/r, so r = mv/(qB), and the speed from the acceleration must be substituted.
3. C√(2mΔV/q)/B Correct
   Energy conservation during the acceleration gives qΔV = ½mv², so v = √(2qΔV/m). In the field the magnetic force qvB (velocity perpendicular to B) provides the centripetal force: qvB = mv²/r, so r = mv/(qB). Substituting v gives r = √(2mΔV/q)/B.
4. D√(mΔV/q)/B
   A student who writes the kinetic energy as mv², dropping the ½, picks this: qΔV = mv² gives v = √(qΔV/m) and r = mv/(qB) = √(mΔV/q)/B, smaller than the correct radius by a factor of √2. The kinetic energy is ½mv², so v = √(2qΔV/m).

Working Acceleration: the particle's electric potential energy decreases by qΔV and its kinetic energy increases by the same amount: qΔV = ½mv², so v = √(2qΔV/m). In the field the velocity is perpendicular to B (θ = 90°), so F_B = qvB, and this force is the centripetal force: qvB = mv²/r, giving r = mv/(qB). Substituting v: r = (m/(qB))√(2qΔV/m) = √(2mΔV/q)/B. (Errors: dropping the ½, qΔV = mv², gives v = √(qΔV/m) and r = √(mΔV/q)/B; carrying the ½ across uninverted, v² = qΔV/(2m), gives r = √(mΔV/(2q))/B; centripetal force mv/r gives r = m/(qB).)

CED 12.2.B.2.i · Read this in Fix

1. A√(mE/(qD))
   A student who takes the distance D between the entry point and the detector as the radius of the path picks this: D = mE/(qB²) gives B = √(mE/(qD)). After a semicircle the ion is a full diameter from where it entered, so r = D/2.
2. B√(2mE/(qD)) Correct
   In region 1 the ion moves at constant velocity, so qE = qvB and v = E/B. In region 2, qvB = mv²/r gives r = mv/(qB) = mE/(qB²). The detector is a diameter away from the entry point, so r = D/2, giving B² = 2mE/(qD) and B = √(2mE/(qD)).
3. C2m/(qD)
   A student who writes the centripetal force as mv/r picks this: qvB = mv/r gives qB = m/r, so B = m/(q·D/2) = 2m/(qD), and the speed set in region 1 drops out. The centripetal force is mv²/r, so r = mv/(qB) depends on the speed v = E/B.
4. D2mE/(qD)
   A student who solves D/2 = mE/(qB²) for B² and stops there picks this: 2mE/(qD) is B², not B. Taking the square root gives B = √(2mE/(qD)); the units confirm it, since 2mE/(qD) has units of T².

Working Region 1: the ion's velocity is constant, so the electric and magnetic forces, which are exerted independently, are equal in magnitude and opposite in direction: qE = qvB (velocity perpendicular to the field, θ = 90°), so v = E/B. Region 2: only the magnetic force qvB is exerted, and it is the centripetal force: qvB = mv²/r, so r = mv/(qB). The entry point and the detector are the ends of a diameter, so r = D/2. Substituting v = E/B: D/2 = mE/(qB²), so B² = 2mE/(qD) and B = √(2mE/(qD)). (Errors: taking D as the radius gives B = √(mE/(qD)); a centripetal force mv/r gives qB = m/r and B = 2m/(qD), with no dependence on E; stopping at B² gives 2mE/(qD).)

CED 12.2.B.3 · Read this in Fix

1. A2.3 × 10⁻⁸ s
   A student who divides the straight-line distance between the entry and exit points, 2r = 6.8 × 10⁻² m, by the speed picks this: t = 2r/v = 2.3 × 10⁻⁸ s. The electron follows the curved path, whose length is πr, so t = πr/v = 3.6 × 10⁻⁸ s.
2. B7.2 × 10⁻⁸ s
   A student who uses the period of the circular motion, 2πr/v, picks this: 7.2 × 10⁻⁸ s. The electron travels only a semicircle before it leaves the field, so the time is half a period, 3.6 × 10⁻⁸ s.
3. C3.6 × 10⁻⁸ s Correct
   The magnetic force evB is the centripetal force, so r = m_e v/(eB) = 3.4 × 10⁻² m. The electron covers half a circumference, πr, at constant speed: t = πr/v = 3.6 × 10⁻⁸ s.
4. D1.1 × 10⁻¹ s
   A student who takes the magnetic force as eB, independent of the speed, picks this: eB = m_e v²/r gives r = m_e v²/(eB), numerically 1.0 × 10⁵ in SI units (its units are m²/s, not meters, which exposes the error), and t = πr/v = 1.1 × 10⁻¹ s. The magnetic force is evB, so r = m_e v/(eB) = 3.4 × 10⁻² m and t = 3.6 × 10⁻⁸ s.

Working The velocity is perpendicular to the field (θ = 90°), so F_B = evB, and this force is the centripetal force: evB = m_e v²/r, so r = m_e v/(eB) = (9.11 × 10⁻³¹ kg)(3.0 × 10⁶ m/s)/[(1.60 × 10⁻¹⁹ C)(5.0 × 10⁻⁴ T)] = 3.4 × 10⁻² m. The magnetic force does not change the speed, so the electron travels half a circumference, πr, at 3.0 × 10⁶ m/s: t = πr/v = π(3.42 × 10⁻² m)/(3.0 × 10⁶ m/s) = 3.6 × 10⁻⁸ s. Equivalently t = πm_e/(eB), which does not depend on the speed. (Errors: the straight-line distance 2r gives t = 2r/v = 2.3 × 10⁻⁸ s; a full period 2πr/v gives 7.2 × 10⁻⁸ s; a magnetic force eB with no dependence on speed gives r = m_e v²/(eB), numerically 1.0 × 10⁵ in SI units (not a length) and t = πr/v = 1.1 × 10⁻¹ s.)

CED 12.2.B.2.i · Read this in Fix