These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This set is for the calculus-based course. Data for every question: μ₀ = 4π × 10⁻⁷ T·m/A, so μ₀/(2π) = 2 × 10⁻⁷ T·m/A. “Long” wires, cylinders and solenoids, and “large” slabs, are much bigger than the distances involved. A scientific calculator is assumed; round only at the end.
Question 1 (multiple choice · foundation)
Three long straight wires are perpendicular to the page. Wire A carries 4.0 A out of the page and wire B carries 7.0 A into the page; both pass through a circular Amperian loop drawn on the page. Wire C carries 2.0 A out of the page and lies outside the loop. The loop is traversed anticlockwise. What is ∮B·dℓ?
- (A) −3.8 × 10⁻⁶ T·m
- (B) +3.8 × 10⁻⁶ T·m
- (C) −1.3 × 10⁻⁶ T·m
- (D) +1.4 × 10⁻⁵ T·m
Answer and explanation
Answer: (A). With the loop traversed anticlockwise, the right-hand rule makes currents out of the page positive. I_enc = +4.0 − 7.0 = −3.0 A, so ∮B·dℓ = μ₀I_enc = (4π × 10⁻⁷)(−3.0) = −3.8 × 10⁻⁶ T·m.
- (B) has the right size but the wrong sign: it treats currents into the page as positive for an anticlockwise loop.
- (C) includes wire C (4.0 − 7.0 + 2.0 = −1.0 A). Currents outside the loop add nothing to ∮B·dℓ, even though they change B on the loop.
- (D) adds the sizes of the two enclosed currents (4.0 + 7.0 = 11 A). Currents in opposite directions must subtract.
Question 2 (multiple choice · core)
An ideal solenoid has its number of turns doubled and its radius doubled. Its length and current stay the same. By what factor does the field inside change?
- (A) 2
- (B) 1
- (C) 4
- (D) 8
Answer and explanation
Answer: (A). B = μ₀nI with n = N/L. Doubling N at fixed L doubles n, so B doubles. The radius does not appear: the Amperian rectangle gives Bℓ = μ₀nℓI wherever its inside side sits.
- (B) treats the solenoid like a single loop, B ∝ N/R, so that the two factors cancel.
- (C) multiplies by the radius factor as well, as if B ∝ NR.
- (D) multiplies by the change in cross-sectional area (4) as well as N. That would affect flux (Unit 13), not B.
Question 3 (multiple choice · core)
A long coaxial cable has a thin inner wire carrying current I along the axis and a thin outer cylindrical shell carrying the same current I in the opposite direction. What is the size of B at a distance r outside the cable?
- (A) 0
- (B) μ₀I/(2πr)
- (C) μ₀I/(πr)
- (D) μ₀I/(4πr)
Answer and explanation
Answer: (A). A coaxial circle of radius r outside the cable encloses I − I = 0. By symmetry B is the same size all round the circle and tangent to it, so B(2πr) = 0 and B = 0.
- (B) counts only the inner wire. That is the field between the wire and the shell.
- (C) adds the two currents instead of subtracting them.
- (D) has no basis in the geometry; it halves the single-wire result.
Question 4 (calculation · core)
A long hollow metal tube has inner radius 2.0 mm and outer radius 4.0 mm. It carries 9.0 A along its length, spread uniformly over the metal. Find B at (a) 1.0 mm, (b) 3.0 mm and (c) 8.0 mm from the axis.
Worked solution
Use coaxial circles; by symmetry B is tangent to each and constant on it, so B(2πr) = μ₀I_enc.
- (a) r = 1.0 mm is in the hollow space. I_enc = 0, so B = 0.
- (b) r = 3.0 mm is in the metal. The current fraction enclosed is the area fraction: (r² − a²)/(b² − a²) = (9 − 4)/(16 − 4) = 5/12. I_enc = 9.0 × 5/12 = 3.75 A. B = (2 × 10⁻⁷)(3.75) ÷ (3.0 × 10⁻³) = 2.5 × 10⁻⁴ T.
- (c) r = 8.0 mm is outside. I_enc = 9.0 A. B = (2 × 10⁻⁷)(9.0) ÷ (8.0 × 10⁻³) = 2.25 × 10⁻⁴ T (2.3 × 10⁻⁴ T).
Suggested mark points (4): 1 for B = 0 in the hollow with a reason; 1 for the enclosed fraction (r² − a²)/(b² − a²); 1 for 2.5 × 10⁻⁴ T; 1 for 2.25 × 10⁻⁴ T outside.
Common error: using the solid-wire result μ₀Ir/(2πb²) in the metal. That gives 3.4 × 10⁻⁴ T, because it wrongly counts current in the hollow core.
Question 5 (calculation · core)
A large flat conducting slab is 4.0 cm thick. It carries a uniform current density J = 5.0 × 10⁴ A/m² in the +x direction, parallel to its faces. The mid-plane of the slab is z = 0, so the slab fills −0.020 m ≤ z ≤ 0.020 m.
(a) Find B at z = 0.010 m. (b) Find B at z = 0.050 m and at z = 0.10 m. (c) State the direction of B above the slab (z > 0.020 m).
Worked solution
Use a rectangle of length ℓ along y, placed symmetrically about the mid-plane, with its long sides at +z and −z. By symmetry B is parallel to y, equal in size and opposite in direction at ±z. The short sides are along z, perpendicular to B, so ∮B·dℓ = 2Bℓ.
- (a) Inside: I_enc = J(2z)ℓ, so B = μ₀Jz = (4π × 10⁻⁷)(5.0 × 10⁴)(0.010) = 6.3 × 10⁻⁴ T.
- (b) Outside: I_enc = Jtℓ, so B = μ₀Jt/2 = (4π × 10⁻⁷)(5.0 × 10⁴)(0.040)/2 = 1.26 × 10⁻³ T at both distances.
- (c) dℓ along +x and r̂ with a +z component give x̂ × ẑ = −ŷ. Above the slab, B points in −y (and in +y below it).
Suggested mark points (4): 1 for a symmetric rectangle with ∮B·dℓ = 2Bℓ; 1 for 6.3 × 10⁻⁴ T inside; 1 for 1.26 × 10⁻³ T outside with the statement that it does not depend on distance; 1 for the direction −y above.
Common error: using the symmetric rectangle but writing ∮B·dℓ = Bℓ, counting only one long side. Both long sides lie in the field, so this doubles the answers.
Question 6 (constructed response · core)
A long cylindrical conductor of radius R carries current along its axis with current density J(r) = α/r for 0 < r ≤ R, where α is a positive constant.
(a) Show that the total current is I = 2παR. (b) Use Ampère’s law to derive B(r) for r ≤ R and for r > R. Justify your Amperian loop. (c) Show that your two results agree at r = R. (d) For R = 3.0 mm and I = 6.0 A, find α and B at r = R/2 and at r = 2R. (e) Compare B at r = R/2 with a conductor of the same R and I carrying a uniform current density.
Worked solution and suggested Marlbridge rubric
(a) I_enc(r) = ∫₀ʳ (α/r′) 2πr′ dr′ = 2πα ∫₀ʳ dr′ = 2παr. At r = R, I = 2παR.
(b) Use a coaxial circle of radius r. By symmetry B is tangent to the circle and has the same size at every point on it, so ∮B·dℓ = B(2πr). Inside: B(2πr) = μ₀(2παr), so B = μ₀α, the same at every r inside. Outside: B(2πr) = μ₀I, so B = μ₀I/(2πr).
(c) At r = R, the outside result is μ₀(2παR)/(2πR) = μ₀α, the same as inside.
(d) α = I/(2πR) = 6.0 ÷ (2π × 3.0 × 10⁻³) = 318 A/m. Inside, B = μ₀α = μ₀I/(2πR) = (2 × 10⁻⁷)(6.0) ÷ (3.0 × 10⁻³) = 4.0 × 10⁻⁴ T at r = R/2. At r = 2R, B = 2.0 × 10⁻⁴ T.
(e) Uniform J: B = μ₀Ir/(2πR²), which is half the surface value at R/2, so 2.0 × 10⁻⁴ T. The α/r conductor gives twice this, because its current is concentrated near the axis, so more current is enclosed by small circles.
| Point | What earns it |
|---|---|
| 1 | Sets up I_enc = ∫J 2πr′ dr′ and gets 2παr, so I = 2παR |
| 1 | Coaxial circle justified: B tangent and constant by symmetry |
| 1 | Inside result B = μ₀α (constant) |
| 1 | Outside result μ₀I/(2πr) and continuity at r = R shown |
| 1 | α = 318 A/m and B values 4.0 × 10⁻⁴ T and 2.0 × 10⁻⁴ T |
| 1 | Comparison: uniform case gives 2.0 × 10⁻⁴ T at R/2, with a reason based on where the current flows |
Accept α in A/m or equivalent units. Carry forward an error in I_enc once.
Question 7 (constructed response · stretch)
A long solenoid is 0.25 m long, has 400 turns and carries 0.80 A.
(a) Draw a suitable Amperian loop on a sketch of a section through the solenoid, and use it to derive B = μ₀nI. State each assumption you make. (b) Calculate B inside. (c) A student says: “If I move the inside side of my loop closer to the windings, the enclosed current is the same, so B there must be the same.” Is the student right? Explain. (d) The solenoid is replaced by a parallel-plate capacitor that is being charged. A circular loop is drawn in the gap between the plates, parallel to them, and no charge crosses any flat surface bounded by it. Explain why there is nevertheless a magnetic field around this loop.
Worked solution and suggested Marlbridge rubric
(a) A rectangle with one side of length ℓ inside the solenoid, parallel to the axis, and the opposite side outside. Assumptions: the solenoid is ideal, so B inside is uniform and parallel to the axis, and B outside is negligible. Inside side: Bℓ. The two short sides are perpendicular to B inside and in zero field outside: 0. Outside side: 0. Enclosed current: nℓ turns each carrying I. So Bℓ = μ₀nℓI and B = μ₀nI.
(b) n = 400/0.25 = 1600 m⁻¹. B = (4π × 10⁻⁷)(1600)(0.80) = 1.6 × 10⁻³ T (1.61 × 10⁻³ T).
(c) Yes. Any position of the inside side encloses the same nℓ turns, and gives Bℓ = μ₀nℓI, so B is the same everywhere inside an ideal solenoid. This is why the field inside is uniform.
(d) As the capacitor charges, the electric field between the plates grows, so the electric flux through the loop changes. Maxwell’s addition to Ampère’s law says a changing electric field produces a magnetic field, just as a current does. So a magnetic field circles the gap even though no charge flows across it.
| Point | What earns it |
|---|---|
| 1 | Rectangular loop with one side inside along the axis and one outside, drawn or described |
| 1 | Ideal-solenoid assumptions stated (uniform B inside, negligible outside) |
| 1 | Each side’s contribution justified, giving Bℓ = μ₀nℓI |
| 1 | n = 1600 m⁻¹ and B = 1.6 × 10⁻³ T |
| 1 | Agrees with the student, linking the same enclosed current to the same B (uniform field) |
| 1 | A changing electric field (changing electric flux) produces a magnetic field |
Do not require the Maxwell equation in symbols for (d); a clear statement of the idea earns the point. A common error in (b) is using L = 25 cm, which gives n = 16 and a field 100 times too small.
How did you do?
- Q1 wrong: re-read the sign rule in “From the long-wire field to Ampère’s law” in the study guide.
- Q2 or Q7 wrong: revisit “Long solenoids” and Figure 2.
- Q3 or Q4 wrong: revisit “Wires and cylinders”, especially hollow cylinders and coaxial cables.
- Q5 wrong: revisit “Slabs and current sheets”.
- Q6 incomplete: work through Worked example 2 again, step by step.
Then tick off the topic checklist.