AP Physics C Electricity and Magnetism Quiz: Magnetism And Moving Charges
18 questions · exam conditions
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Magnetism And Moving ChargesQuestion 1 of 18

A rectangular current loop (dimensions 0.10 m×0.20 m0.10\,\text{m}\times0.20\,\text{m}) carries current I=2.0 AI=2.0\,\text{A}, meaning charges move around the loop and create a magnetic dipole moment μ⃗=IA n^\vec\mu=IA\,\hat n. The loop is in a uniform magnetic field B⃗=0.30 T\vec B=0.30\,\text{T} to the right. The torque on the loop is τ⃗=μ⃗×B⃗\vec\tau=\vec\mu\times\vec B; equilibrium occurs when μ⃗\vec\mu is parallel or antiparallel to B⃗\vec B. Refer to the scenario above. Determine the net force on the loop in the uniform magnetic field.

Net force is zero
Net force is IABIAB
Net force is qvBqvB
Net force is μB\mu B toward B⃗\vec B
Net force is μ0I/(2πr)\mu_0 I/(2\pi r)
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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Magnetism And Moving Charges

Practice Magnetism And Moving Charges in AP Physics C Electricity and Magnetism with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Magnetism And Moving Charges, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Electricity and Magnetism.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A rectangular current loop (dimensions 0.10 m×0.20 m0.10\,\text{m}\times0.20\,\text{m}) carries current I=2.0 AI=2.0\,\text{A}, meaning charges move around the loop and create a magnetic dipole moment μ⃗=IA n^\vec\mu=IA\,\hat n. The loop is in a uniform magnetic field B⃗=0.30 T\vec B=0.30\,\text{T} to the right. The torque on the loop is τ⃗=μ⃗×B⃗\vec\tau=\vec\mu\times\vec B; equilibrium occurs when μ⃗\vec\mu is parallel or antiparallel to B⃗\vec B. Refer to the scenario above. Determine the net force on the loop in the uniform magnetic field.

  1. Net force is zero (correct answer)
  2. Net force is IABIAB
  3. Net force is qvBqvB
  4. Net force is μB\mu B toward B⃗\vec B
  5. Net force is μ0I/(2πr)\mu_0 I/(2\pi r)
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. In a uniform magnetic field, the net force on a current loop is zero because forces on opposite sides cancel - each side experiences F = ILB, but opposite sides have opposite force directions. The scenario describes a rectangular loop in a uniform field, where the magnetic forces on the four sides form two canceling pairs. While the loop experiences torque (τ = μ×B) that tends to align its magnetic moment with the field, the net translational force is zero. Choice A is correct because uniform fields exert zero net force on current loops, only torque. Choice B is incorrect because it confuses torque with net force - IAB relates to torque magnitude, not force. To help students: Distinguish between force (causes translation) and torque (causes rotation), analyze forces on each loop segment systematically, and emphasize that non-uniform fields are needed for net force. Use demonstrations with wire loops in magnetic fields.

Question 2

In a uniform magnetic field region, B⃗=0.25 T\vec B=0.25\,\text{T} directed out of the page. A positive charge moves upward at v=2.0×105 m/sv=2.0\times10^5\,\text{m/s}. Moving charges create magnetic fields, and in an external field the charge experiences the Lorentz force F⃗=q v⃗×B⃗\vec F=q\,\vec v\times\vec B, which is perpendicular to both vectors. Refer to the scenario above. What is the direction of the magnetic force on the positive charge?

  1. To the right
  2. Upward
  3. To the left (correct answer)
  4. Out of the page
  5. Into the page
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic force on a moving charge is F = qv×B, with direction determined by the right-hand rule for positive charges. In this scenario, the positive charge moves upward, and the magnetic field points out of the page, so using the right-hand rule: fingers point up (velocity), curl out of page (field), and thumb points left. Choice C is correct because the cross product of upward velocity and out-of-page field gives leftward force for a positive charge. Choice A is incorrect because it gives the opposite direction, which would be correct for a negative charge but not a positive one. To help students: Emphasize that the right-hand rule applies directly to positive charges (use left hand for negative charges), practice with 3D visualization tools, and create mnemonics. Always specify charge sign and verify force direction makes physical sense.

Question 3

A straight horizontal wire lies along the +x+x direction and carries conventional current I=3.0 AI=3.0\,\text{A} (moving charges). It is placed in a uniform magnetic field B⃗=0.50 T\vec B=0.50\,\text{T} directed upward (+y+y). The magnetic force on a current-carrying wire is F⃗=I L⃗×B⃗\vec F=I\,\vec L\times\vec B (from the same right-hand rule used for qv⃗×B⃗q\vec v\times\vec B). The wire segment in the field has length L=0.20 mL=0.20\,\text{m}. Refer to the scenario above. What is the direction of the magnetic force on the wire segment?

  1. Into the page (−z-z)
  2. Along +x+x (to the right)
  3. Upward (+y+y)
  4. Out of the page (+z+z) (correct answer)
  5. Along −x-x (to the left)
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic force on a current-carrying wire is F = IL×B, where the direction follows the same right-hand rule as for moving charges. In this scenario, the current flows along +x and the magnetic field points along +y, so using the right-hand rule: fingers point along +x (current direction), curl toward +y (field direction), and the thumb points along +z (out of the page). Choice D is correct because the cross product of +x and +y gives +z direction (out of the page). Choice A is incorrect because it gives the opposite direction, suggesting confusion with the right-hand rule or treating the current as negative. To help students: Practice the right-hand rule systematically with coordinate axes, emphasize that conventional current direction determines the force, and use physical demonstrations with wires and magnets. Remind students that x×y = z in right-handed coordinate systems.

Question 4

A charged particle in a uniform magnetic field illustrates how moving charges and magnetic fields interact. An electron (∣q∣=1.60×10−19 C|q|=1.60\times10^{-19}\,\text{C}, m=9.11×10−31 kgm=9.11\times10^{-31}\,\text{kg}) enters a region with B⃗=2.0×10−3 T\vec B=2.0\times10^{-3}\,\text{T} directed upward. Its velocity is v=4.0×106 m/sv=4.0\times10^6\,\text{m/s} to the east, so θ=90∘\theta=90^\circ and the Lorentz force magnitude is F=∣q∣vBF=|q|vB. The force is perpendicular to v⃗\vec v, so the electron moves in a circle at constant speed with radius r=mv/(∣q∣B)r=mv/(|q|B). Refer to the scenario above. Calculate the radius of curvature of the electron's path.

  1. 1.1×10−2 m1.1\times10^{-2}\,\text{m} (correct answer)
  2. 1.1×10−3 m1.1\times10^{-3}\,\text{m}
  3. 1.1×10−1 m1.1\times10^{-1}\,\text{m}
  4. 1.1×10−2 T1.1\times10^{-2}\,\text{T}
  5. 2.3×10−2 m2.3\times10^{-2}\,\text{m}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. When a charged particle moves perpendicular to a magnetic field, the magnetic force provides centripetal force: |q|vB = mv²/r, which gives r = mv/(|q|B). In this scenario, we have an electron with m = 9.11 × 10⁻³¹ kg, |q| = 1.60 × 10⁻¹⁹ C, v = 4.0 × 10⁶ m/s, and B = 2.0 × 10⁻³ T. Calculating: r = (9.11 × 10⁻³¹ × 4.0 × 10⁶)/(1.60 × 10⁻¹⁹ × 2.0 × 10⁻³) = 1.14 × 10⁻² m ≈ 1.1 × 10⁻² m. Choice A is correct because it properly applies the radius formula with correct values and units. Choice B is incorrect by a factor of 10, suggesting a calculation or power-of-ten error. To help students: Practice balancing centripetal and magnetic forces, ensure familiarity with electron mass and charge values, and emphasize checking units and order of magnitude. Use dimensional analysis to verify the formula gives length units.

Question 5

A uniform magnetic field exists between pole faces, B⃗=0.40 T\vec B=0.40\,\text{T} directed into the page. A positive ion (q=+3.2×10−19 Cq=+3.2\times10^{-19}\,\text{C}, m=6.4×10−27 kgm=6.4\times10^{-27}\,\text{kg}) enters with velocity v⃗\vec v to the right, perpendicular to B⃗\vec B. Moving charges create magnetic fields, and the force on a moving charge is F⃗=q v⃗×B⃗\vec F=q\,\vec v\times\vec B, always perpendicular to v⃗\vec v, so the speed stays constant while direction changes. Refer to the scenario above. Explain why the charged particle follows a circular path in the magnetic field.

  1. Because F⃗\vec F is always parallel to v⃗\vec v, increasing speed uniformly
  2. Because F⃗=qv⃗×B⃗\vec F=q\vec v\times\vec B stays perpendicular to v⃗\vec v, providing centripetal force (correct answer)
  3. Because the magnetic field does work, increasing kinetic energy each second
  4. Because the ion is attracted to the north pole like a small magnet
  5. Because F⃗=qE⃗\vec F=q\vec E dominates even though E⃗=0\vec E=0
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic force on a moving charge is F = qv×B, which is always perpendicular to the velocity vector, meaning it cannot change the particle's speed, only its direction. In this scenario, the ion enters perpendicular to the magnetic field, so the force is maximum (sin θ = 1) and perpendicular to both v and B. Since the force is always perpendicular to velocity, it acts as a centripetal force, continuously changing the particle's direction while maintaining constant speed, resulting in circular motion. Choice B is correct because it identifies that the perpendicular force provides centripetal acceleration for circular motion. Choice C is incorrect because magnetic forces do no work (F⊥v means W = F·d = 0), so kinetic energy remains constant. To help students: Emphasize that perpendicular forces cannot change speed, only direction. Use analogies like a ball on a string or planetary orbits to reinforce centripetal force concepts.

Question 6

A long straight wire carries conventional current I=12 AI=12\,\text{A} to the east. Moving charges in the wire create a magnetic field that forms concentric circles around the wire; use the right-hand rule (thumb along current, fingers give B⃗\vec B). At a point r=3.0 cmr=3.0\,\text{cm} directly above the wire, the field magnitude is B=μ0I/(2πr)B=\mu_0 I/(2\pi r). Refer to the scenario above. Calculate the magnetic field strength at the point above the wire.

  1. 8.0×10−5 T8.0\times10^{-5}\,\text{T} (correct answer)
  2. 8.0×10−6 T8.0\times10^{-6}\,\text{T}
  3. 8.0×10−4 T8.0\times10^{-4}\,\text{T}
  4. 8.0×10−5 m/s8.0\times10^{-5}\,\text{m/s}
  5. 2.4×10−4 T2.4\times10^{-4}\,\text{T}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic field around a straight current-carrying wire is given by B = μ₀I/(2πr), where μ₀ = 4π × 10⁻⁷ T·m/A. In this scenario, we need to calculate the field strength at r = 3.0 cm = 0.03 m from a wire carrying I = 12 A. Substituting values: B = (4π × 10⁻⁷)(12)/(2π × 0.03) = 8.0 × 10⁻⁵ T. Choice A is correct because it properly applies the formula with correct unit conversions and calculations. Choice B is incorrect because it represents an error of a factor of 10, likely from incorrect conversion of centimeters to meters or calculation error. To help students: Emphasize careful unit conversion (cm to m), practice using μ₀ = 4π × 10⁻⁷ T·m/A, and verify dimensions in the formula. Create a systematic approach: identify given values, convert units, substitute carefully, and check order of magnitude.

Question 7

A proton (q=+1.60×10−19 Cq=+1.60\times10^{-19}\,\text{C}) moves through a uniform magnetic field B⃗=0.80 T\vec B=0.80\,\text{T} while its velocity makes a 30∘30^\circ angle with B⃗\vec B. Moving charges produce magnetic fields, and the force on a moving charge in a magnetic field is F=qvBsin⁡θF=qvB\sin\theta. Refer to the scenario above. Calculate the magnetic force magnitude if v=1.0×106 m/sv=1.0\times10^6\,\text{m/s}.

  1. 6.4×10−14 N6.4\times10^{-14}\,\text{N} (correct answer)
  2. 1.3×10−13 N1.3\times10^{-13}\,\text{N}
  3. 6.4×10−14 T6.4\times10^{-14}\,\text{T}
  4. 2.6×10−13 N2.6\times10^{-13}\,\text{N}
  5. 0 N0\,\text{N}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic force magnitude on a moving charge at angle θ to the field is F = qvB sin θ, accounting for the component of velocity perpendicular to B. In this scenario, q = 1.60 × 10⁻¹⁹ C, v = 1.0 × 10⁶ m/s, B = 0.80 T, and θ = 30°, so sin 30° = 0.5. Calculating: F = (1.60 × 10⁻¹⁹)(1.0 × 10⁶)(0.80)(0.5) = 6.4 × 10⁻¹⁴ N. Choice A is correct because it properly includes the sin θ factor for the 30° angle. Choice B is incorrect because it appears to use sin 30° = 1 or makes a calculation error, giving twice the correct value. To help students: Emphasize that only the velocity component perpendicular to B contributes to force, practice identifying angles in 3D scenarios, and remember special angle values (sin 30° = 0.5). Draw vector diagrams showing v, B, and the perpendicular component.

Question 8

A long straight wire carries I=12 AI=12\,\text{A} to the right. Moving charges in the wire generate a magnetic field circling the wire; apply the right-hand rule to set B⃗\vec B direction. A point is located r=0.060 mr=0.060\,\text{m} above the wire in air. Refer to the scenario above. Calculate the magnetic field strength at that point.

  1. 4.0×10−5 T4.0\times10^{-5}\,\text{T}
  2. 2.0×10−5 T2.0\times10^{-5}\,\text{T} (correct answer)
  3. 4.0×10−5 N4.0\times10^{-5}\,\text{N}
  4. 1.3×10−5 T1.3\times10^{-5}\,\text{T}
  5. 8.0×10−5 T8.0\times10^{-5}\,\text{T}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. A straight current-carrying wire creates a magnetic field with magnitude B = μ₀I/(2πr) at distance r, forming concentric circles around the wire. In this scenario, a wire carries 12 A to the right, and we calculate the field 0.060 m above it. Choice B is correct because B = (4π×10⁻⁷)(12)/(2π×0.060) = 48π×10⁻⁷/(0.12π) = 48×10⁻⁷/0.12 = 4.0×10⁻⁵/2 = 2.0×10⁻⁵ T. Choice A represents double the correct value, a common error from omitting the 2π factor in the denominator. To help students: memorize B = μ₀I/(2πr) for straight wires, practice dimensional analysis to catch formula errors, and use the right-hand rule to find field direction (thumb along current, fingers curl in B direction).

Question 9

A positive particle (q=+2.0×10−6 Cq=+2.0\times10^{-6}\,\text{C}) moves at v=1500 (m/s)v=1500\,(\text{m/s}) north through a uniform magnetic field B=0.80 TB=0.80\,\text{T} directed east. Moving charges create magnetic fields, and the right-hand rule gives the direction of v⃗×B⃗\vec v\times\vec B; the magnetic force magnitude is F=qvBsin⁡θF=qvB\sin\theta. Refer to the scenario above. What is the direction of the magnetic force on a positive charge moving as described?

  1. East
  2. Upward
  3. Downward (correct answer)
  4. North
  5. West
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic force on a moving charge is F = qv×B, with direction found using the right-hand rule for the cross product, considering the charge sign. In this scenario, a positive charge moves north while the magnetic field points east, requiring careful application of the right-hand rule. Choice C is correct because using the right-hand rule: point fingers north (velocity), curl them east (field), and the thumb points downward (force direction for positive charge). Choice B would result from reversing the order of the cross product or misapplying the right-hand rule. To help students: practice the right-hand rule with orthogonal vectors, emphasize that v×B ≠ B×v (order matters), and use coordinate systems to verify directions (north×east = down in standard orientation).

Question 10

A solenoid of length 0.30 m0.30\,\text{m} has N=900N=900 turns and carries current I=2.0 AI=2.0\,\text{A}. The moving charges in the coils create a magnetic field; inside a long solenoid, B≈μ0nIB\approx\mu_0 n I with n=N/Ln=N/L, and the right-hand rule (curl fingers with current, thumb gives B⃗\vec B inside) sets direction. Refer to the scenario above. Calculate the magnetic field strength inside the solenoid.

  1. 7.5×10−3 T7.5\times10^{-3}\,\text{T} (correct answer)
  2. 2.4×10−2 T2.4\times10^{-2}\,\text{T}
  3. 7.5×10−3 m/s7.5\times10^{-3}\,\text{m/s}
  4. 3.8×10−3 T3.8\times10^{-3}\,\text{T}
  5. 1.5×10−2 T1.5\times10^{-2}\,\text{T}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. A solenoid creates a nearly uniform magnetic field inside with magnitude B = μ₀nI, where n = N/L is the turn density (turns per unit length). In this scenario, a solenoid has 900 turns over 0.30 m length carrying 2.0 A current. Choice A is correct because n = 900/0.30 = 3000 turns/m, so B = (4π×10⁻⁷)(3000)(2.0) = 24π×10⁻⁴ = 7.54×10⁻³ ≈ 7.5×10⁻³ T. Choice E represents double the correct value, possibly from using diameter instead of length or misapplying the formula. To help students: emphasize that n = N/L is turns per unit length, practice unit analysis to ensure consistency, and use the right-hand rule for solenoids (curl fingers with current, thumb points along B inside).

Question 11

A proton (q=+eq=+e, m=1.67×10−27 kgm=1.67\times10^{-27}\,\text{kg}) enters a uniform magnetic field B=0.30 TB=0.30\,\text{T} with velocity perpendicular to B⃗\vec B. Moving charges create magnetic fields, but here the external field causes a Lorentz force F=qvBF=qvB that bends the path without changing speed. The proton speed is v=2.4×106 m/sv=2.4\times10^{6}\,\text{m/s}. Refer to the scenario above. Explain why the charged particle follows a circular path in the magnetic field.

  1. The force is always perpendicular to velocity, acting as centripetal force (correct answer)
  2. The force is zero because magnetic fields only affect stationary charges
  3. The force is parallel to velocity, so the particle spirals outward faster
  4. The magnetic field does positive work each cycle to keep the circle closed
  5. The particle curves because electric attraction to the magnet provides centripetal force
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic force F = qvB acts perpendicular to both velocity and magnetic field, providing centripetal force for circular motion when v⊥B. In this scenario, a proton enters a magnetic field with velocity perpendicular to B, creating conditions for uniform circular motion. Choice A is correct because the Lorentz force is always perpendicular to velocity (from the cross product v×B), providing the centripetal force mv²/r = qvB needed for circular motion at constant speed. Choice D is incorrect because magnetic forces do no work (W = F·d = 0 when F⊥v), so energy cannot be transferred to maintain the motion. To help students: use vector diagrams showing F, v, and B at multiple points on the circle, emphasize the work-energy theorem showing magnetic forces do no work, and practice setting centripetal force equal to magnetic force.

Question 12

A rectangular current loop (height 0.10 m0.10\,\text{m}, width 0.060 m0.060\,\text{m}) lies in a uniform magnetic field B=0.40 TB=0.40\,\text{T} to the right. Current I=3.0 AI=3.0\,\text{A} flows clockwise as viewed from above; moving charges in the wire create magnetic fields, and the loop experiences forces from the external field via F⃗=I L⃗×B⃗\vec F=I\,\vec L\times\vec B. The loop's plane is vertical, so its area vector points north. Refer to the scenario above. Determine the net force on the loop in the given magnetic field.

  1. Nonzero, upward
  2. Nonzero, to the right
  3. Zero, forces cancel pairwise (correct answer)
  4. Nonzero, into the page
  5. Zero, because BB is uniform only inside the loop
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. A current loop in a uniform magnetic field experiences forces on each segment given by F = IL×B, where the net force depends on the loop's orientation and field uniformity. In this scenario, a rectangular loop with clockwise current sits in a uniform field pointing right, with the loop's plane vertical. Choice C is correct because in a uniform field, the forces on opposite sides of the loop are equal in magnitude but opposite in direction, causing them to cancel pairwise - the top and bottom segments have opposing forces, as do the left and right segments. Choice A is incorrect because it ignores the cancellation of forces on opposite segments in a uniform field. To help students: draw force vectors on each segment using F = IL×B, show how opposite sides experience opposite forces, and emphasize that uniform fields produce zero net force but can create torque.

Question 13

A positive ion (q=+1.60×10−19 Cq=+1.60\times10^{-19}\,\text{C}) enters a uniform magnetic field B⃗=0.50 T\vec B=0.50\,\text{T} directed into the page. Its velocity is v=2.0×106 m/sv=2.0\times10^{6}\,\text{m/s} to the right, so moving charge produces no B⃗\vec B here but feels Lorentz force F⃗=q v⃗×B⃗\vec F=q\,\vec v\times\vec B. Use the right-hand rule for v⃗×B⃗\vec v\times\vec B. Refer to the scenario above. What is the direction of the magnetic force on a positive charge moving as described?

  1. Upward (correct answer)
  2. Downward
  3. To the right
  4. To the left
  5. Into the page
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The Lorentz force on a moving charge is F = qv×B, with direction determined by the right-hand rule for the cross product v×B, then considering the sign of the charge. In this scenario, a positive ion moves to the right while the magnetic field points into the page. Choice A is correct because using the right-hand rule: point fingers right (velocity), curl them into the page (field), and the thumb points upward (force direction for positive charge). Choice B would be correct for a negative charge, showing a common confusion about charge sign effects. To help students: practice the right-hand rule systematically - fingers along v, curl toward B, thumb shows F for positive charges; emphasize that negative charges experience force opposite to the right-hand rule result.

Question 14

A long straight wire produces a magnetic field because its charges move; field lines circle the wire by the right-hand rule. A student measures the field at a point r=5.0 cmr=5.0\,\text{cm} from the wire and uses B=μ0I/(2πr)B=\mu_0 I/(2\pi r). If the current is reversed (same magnitude), the magnetic field direction reverses but its magnitude at the point is unchanged. Refer to the scenario above. Calculate the magnetic field strength when I=10 AI=10\,\text{A}.

  1. 4.0×10−5 T4.0\times10^{-5}\,\text{T} (correct answer)
  2. 4.0×10−6 T4.0\times10^{-6}\,\text{T}
  3. 2.0×10−5 T2.0\times10^{-5}\,\text{T}
  4. 4.0×10−5 N4.0\times10^{-5}\,\text{N}
  5. 8.0×10−5 T8.0\times10^{-5}\,\text{T}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic field around a straight wire carrying current I is B = μ₀I/(2πr), where μ₀ = 4π × 10⁻⁷ T·m/A. In this scenario, I = 10 A and r = 5.0 cm = 0.05 m, so B = (4π × 10⁻⁷ × 10)/(2π × 0.05) = 4.0 × 10⁻⁵ T. The problem notes that reversing current reverses field direction but not magnitude, reinforcing that B depends on |I|. Choice A is correct because it properly applies the formula with correct unit conversion and calculation. Choice B is incorrect by a factor of 10, likely from a calculation error or incorrect unit conversion. To help students: Emphasize systematic unit conversion (always convert cm to m), practice using μ₀ in calculations, and verify results have correct units (Tesla). Create a checklist: identify formula, list given values with units, convert to SI units, calculate carefully.

Question 15

In a lab, students note that moving charges create magnetic fields: a long straight wire carries conventional current I=5.0 AI=5.0\,\text{A} upward, so the magnetic field circles the wire by the right-hand rule (thumb along II, curled fingers give B⃗\vec B). A proton (q=+1.60×10−19 Cq=+1.60\times10^{-19}\,\text{C}) passes r=2.0 cmr=2.0\,\text{cm} east of the wire moving due north at v=3.0×106 m/sv=3.0\times10^6\,\text{m/s}. The magnetic force on a moving charge is the Lorentz force F⃗=q v⃗×B⃗\vec F=q\,\vec v\times\vec B (here sin⁡θ=1\sin\theta=1). Refer to the scenario above. What is the direction of the magnetic force on the proton?

  1. Upward (parallel to the wire)
  2. West (toward the wire) (correct answer)
  3. South (opposite the velocity)
  4. East (away from the wire)
  5. Downward (into the floor)
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic field from a current-carrying wire circles the wire according to the right-hand rule, and the force on a moving charge is given by F = qv×B. In this scenario, the wire carries current upward, creating a magnetic field that circles the wire - at a point east of the wire, the field points south. The proton moves north, so v×B points west (toward the wire) using the right-hand rule. Choice B is correct because the force direction is determined by the cross product v×B, which points west when v is north and B is south. Choice D is incorrect because students might confuse the force direction with repulsion from the wire, not recognizing that magnetic forces follow the cross product rule. To help students: Practice the right-hand rule systematically - first find B direction around the wire, then apply v×B for the force. Use 3D models or animations to visualize field lines circling wires and practice determining force directions from various positions.

Question 16

A wire segment of length L=0.25 mL=0.25\,\text{m} carries I=6.0 AI=6.0\,\text{A} upward through a uniform magnetic field B=0.30 TB=0.30\,\text{T} directed to the right. Moving charges in the wire create magnetic fields, but the external field exerts a magnetic force on the current: F⃗=I L⃗×B⃗\vec F=I\,\vec L\times\vec B. Use the right-hand rule for L⃗×B⃗\vec L\times\vec B. Refer to the scenario above. What is the direction of the magnetic force on a positive charge moving as described?

  1. Upward
  2. Into the page (correct answer)
  3. To the right
  4. Out of the page
  5. Downward
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. A current-carrying wire in a magnetic field experiences force F = IL×B, where L points in the current direction and the cross product determines force direction. In this scenario, current flows upward (L upward) through a field pointing right, requiring the right-hand rule for L×B. Choice B is correct because using the right-hand rule: point fingers upward (current/L direction), curl them to the right (field), and the thumb points into the page (force direction). Choice D would result from confusing the cross product order or misapplying the right-hand rule. To help students: practice F = IL×B systematically, emphasize that L points along current flow, and verify directions using coordinate systems (up×right = into page in standard orientation).

Question 17

A solenoid (length L=0.40 mL=0.40\,\text{m}, N=800N=800 turns) carries current I=1.5 AI=1.5\,\text{A}. The moving charges in its windings create a magnetic field that is approximately uniform inside and much weaker outside; the right-hand rule (fingers follow current, thumb gives B⃗\vec B inside) sets direction. The field magnitude inside is B≈μ0nIB\approx\mu_0 n I where n=N/Ln=N/L. Refer to the scenario above. Calculate the magnetic field strength inside the solenoid.

  1. 3.8×10−3 T3.8\times10^{-3}\,\text{T} (correct answer)
  2. 3.8×10−2 T3.8\times10^{-2}\,\text{T}
  3. 3.8×10−4 T3.8\times10^{-4}\,\text{T}
  4. 3.8×10−3 N3.8\times10^{-3}\,\text{N}
  5. 1.9×10−3 T1.9\times10^{-3}\,\text{T}
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic field inside a solenoid is approximately uniform and given by B = μ₀nI, where n = N/L is the turn density and μ₀ = 4π × 10⁻⁷ T·m/A. In this scenario, N = 800 turns, L = 0.40 m, so n = 800/0.40 = 2000 turns/m, and I = 1.5 A. Calculating: B = (4π × 10⁻⁷)(2000)(1.5) = 3.77 × 10⁻³ T ≈ 3.8 × 10⁻³ T. Choice A is correct because it properly calculates turn density and applies the solenoid field formula. Choice C is incorrect by a factor of 10, suggesting an error in calculating n or in the final multiplication. To help students: Emphasize that n is turns per unit length (not total turns), practice the solenoid formula B = μ₀nI, and verify units work out to Tesla. Draw solenoids showing how closely spaced turns create strong internal fields.

Question 18

In a lab, a straight wire carries I=8.0 AI=8.0\,\text{A} upward. Moving charges in the wire create circular magnetic field lines around the wire; use the right-hand rule (thumb along current, curled fingers give B⃗\vec B direction). A point r=0.040 mr=0.040\,\text{m} east of the wire lies in air (μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}). Refer to the scenario above. Calculate the magnetic field strength at that point.

  1. 4.0×10−5 T4.0\times10^{-5}\,\text{T}
  2. 1.0×10−5 T1.0\times10^{-5}\,\text{T}
  3. 4.0×10−5 m/s4.0\times10^{-5}\,\text{m/s}
  4. 8.0×10−5 T8.0\times10^{-5}\,\text{T}
  5. 2.0×10−5 T2.0\times10^{-5}\,\text{T} (correct answer)
Explanation: This question tests AP Physics C: Electricity and Magnetism skills, specifically understanding of magnetism and moving charges. The magnetic field around a straight current-carrying wire is given by B = μ₀I/(2πr), where the field forms concentric circles around the wire with direction determined by the right-hand rule. In this scenario, a wire carries 8.0 A upward, and we need to find the field strength at a point 0.040 m east of the wire. Choice E is correct because B = (4π×10⁻⁷)(8.0)/(2π×0.040) = 32×10⁻⁷/0.080 = 4.0×10⁻⁵/2 = 2.0×10⁻⁵ T. Choice A is incorrect as it represents double the correct value, likely from forgetting the factor of 2π in the denominator. To help students: emphasize the formula B = μ₀I/(2πr) for straight wires, practice unit analysis to ensure dimensional consistency, and use the right-hand rule to determine field direction (thumb along current, fingers curl in field direction).