Ampere loops for symmetric currents

Science · Electricity & Magnetism · ages 19-21

Name ______________________ Date ____________

In the ideal infinite-solenoid model in vacuum, the magnetic field is zero outside the windings.
Circle one: True False
For steady currents in vacuum, what does Ampere's law state?
- Closed-path circulation of B equals mu0 times signed enclosed current
- Closed-surface electric flux equals mu0 times total enclosed electric charge
- Closed-path induced voltage equals mu0 times magnetic flux without any time change
Which closed path lets Ampere's law directly determine the field around an infinitely long straight wire?
- A spherical surface centred on a point along the wire
- A centred circle in a plane perpendicular to the wire
- A rectangular path placed far away on one side of the wire
An ideal toroid has N uniformly distributed turns carrying current I, with vacuum in its winding region. At radius r from its central axis within that region, what is B?
- mu0 N I/(2 pi r), tangent to a centred circle
- mu0 N I/(2 pi r^2), tangent to a centred circle
- mu0 N I/(2 pi r), directed along the central axis
A long circular wire of radius a carries current uniformly across its cross-section. What fraction of its total current lies inside a centred circle of radius a/2? Give a decimal.
Answer: ______________
The right-hand normal of an Amperian path defines positive current. Its surface is crossed by 9 A with that normal and 7 A against it. What is Ienc?
- 16 A, adding the magnitudes of both crossings
- 9 A, keeping the positive crossing and discarding the negative crossing
- 2 A, adding the two crossings with their signs
An ideal solenoid in vacuum has an interior field of 2.0 mT. Its turns per metre triple and its current halves, with the ideal model unchanged. Find the new field magnitude in mT.
Answer: ______________
A single circular wire loop carries steady current in vacuum. Why does Ampere's law not directly give B at an axial point?
- Its finite radius changes vacuum permeability, so the law contains an unknown magnetic constant
- No useful symmetry path reduces the integral to that B times a known length
- Its steady current necessarily creates a changing electric field, so the steady-current law fails

Answer key

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True · The uniformly wound ideal infinite solenoid has zero exterior field. This is not an exact statement about every finite coil.
Closed-path circulation of B equals mu0 times signed enclosed current · Ampere's law states that the closed-path integral of B dot dl equals mu0 Ienc. The current is counted with signs.
A centred circle in a plane perpendicular to the wire · On the centred perpendicular circle, B is tangent and constant in magnitude. Its circulation is B times 2 pi r.
mu0 N I/(2 pi r), tangent to a centred circle · The circular path encloses N current crossings. B(2 pi r) = mu0 N I gives B = mu0 N I/(2 pi r), tangent to the circle.
0.25 · Uniform current density makes the current fraction equal to the area fraction: (a/2)^2/a^2 = 1/4 = 0.25.
2 A, adding the two crossings with their signs · The first current is +9 A and the second is -7 A. Their signed sum is 2 A.
3 · B follows n I. The factor is 3 times 1/2 = 1.5, so the new field is 2.0 mT times 1.5 = 3.0 mT.
No useful symmetry path reduces the integral to that B times a known length · The law still gives circulation. The loop's symmetry does not isolate its axial field as B times a known path length; use Biot-Savart for that field.