AP Physics C Electricity and Magnetism Quiz: Amperes Law
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Amperes LawQuestion 1 of 20

A long air-core solenoid has turn density n=1200 m−1n=1200\,\text{m}^{-1} and carries a steady current I=0.80 AI=0.80\,\text{A}. The solenoid length is much greater than its radius, so the magnetic field inside is approximately uniform and axial, and the magnetic field outside is approximately zero. Choose a rectangular Amperian loop with length ℓ\ell inside the solenoid parallel to the axis and the return path outside. According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, with μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}. Using the symmetry assumptions, the integral reduces to BℓB\ell for the inside segment, and Ienc=(nℓ)II_{\text{enc}}=(n\ell)I because nℓn\ell turns are enclosed. Ignore fringing and assume the current is steady.

According to Ampère's Law, calculate the magnetic field inside the solenoid.

1.2×10−3 T1.2\times 10^{-3}\,\text{T}
9.6×10−4 T9.6\times 10^{-4}\,\text{T}
3.0×10−7 T3.0\times 10^{-7}\,\text{T}
3.8×10−3 T3.8\times 10^{-3}\,\text{T}
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AP Physics C Electricity and Magnetism Quiz

AP Physics C Electricity and Magnetism Quiz: Amperes Law

Practice Amperes Law 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.

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

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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.

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Question 1

A long air-core solenoid has turn density n=1200 m−1n=1200\,\text{m}^{-1} and carries a steady current I=0.80 AI=0.80\,\text{A}. The solenoid length is much greater than its radius, so the magnetic field inside is approximately uniform and axial, and the magnetic field outside is approximately zero. Choose a rectangular Amperian loop with length ℓ\ell inside the solenoid parallel to the axis and the return path outside. According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, with μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}. Using the symmetry assumptions, the integral reduces to BℓB\ell for the inside segment, and Ienc=(nℓ)II_{\text{enc}}=(n\ell)I because nℓn\ell turns are enclosed. Ignore fringing and assume the current is steady.

According to Ampère's Law, calculate the magnetic field inside the solenoid.

  1. 1.2×10−3 T1.2\times 10^{-3}\,\text{T}
  2. 9.6×10−4 T9.6\times 10^{-4}\,\text{T} (correct answer)
  3. 3.0×10−7 T3.0\times 10^{-7}\,\text{T}
  4. 3.8×10−3 T3.8\times 10^{-3}\,\text{T}
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to find the magnetic field inside a long solenoid. Ampère's Law is ideal for solenoids because the interior field is uniform and axial while the exterior field is negligible, allowing us to use a rectangular Amperian loop with one side inside and one outside. In this scenario, the solenoid has turn density n = 1200 m⁻¹ and carries current I = 0.80 A, with the rectangular loop having length ℓ inside where the field contributes to the line integral. Choice B is correct because applying Ampère's Law gives Bℓ = μ₀(nℓ)I, which simplifies to B = μ₀nI = (4π×10⁻⁷)(1200)(0.80) = 9.6×10⁻⁴ T. Choice A would result from using 1.0 A instead of 0.80 A, while choice D might come from calculation errors or unit confusion. To help students: Stress that the magnetic field inside an ideal solenoid depends only on μ₀, n, and I, not on the solenoid's radius or the specific loop chosen. Practice recognizing when approximations (like neglecting fringing fields) are valid and ensure proper unit handling with turn density.

Question 2

A long, straight wire carries a steady current II. Using Ampère's law with a circular Amperian loop of radius rr centered on the wire, what is the magnitude of the magnetic field at distance rr from the wire?

  1. μ0I2πr\frac{\mu_0 I}{2\pi r} (correct answer)
  2. μ0I4πr\frac{\mu_0 I}{4\pi r}
  3. μ0Iπr2\frac{\mu_0 I}{\pi r^2}
  4. μ0I2πr2\frac{\mu_0 I}{2\pi r^2}
Explanation: Applying Ampère's law: ∮B⃗⋅dl⃗=μ0I\oint \vec{B} \cdot d\vec{l} = \mu_0 I. For a circular path, BB is constant and tangent to the circle, so B(2πr)=μ0IB(2\pi r) = \mu_0 I, giving B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}.

Question 3

Two coaxial solenoids with the same length LL have N1N_1 and N2N_2 turns respectively, carrying currents I1I_1 and I2I_2 in the same direction. What is the magnetic field inside both solenoids?

  1. μ0(N1I1+N2I2)L\mu_0 \frac{(N_1 I_1 + N_2 I_2)}{L} (correct answer)
  2. μ0(N1I1−N2I2)L\mu_0 \frac{(N_1 I_1 - N_2 I_2)}{L}
  3. μ0(N1I1)2+(N2I2)2L\mu_0 \frac{\sqrt{(N_1 I_1)^2 + (N_2 I_2)^2}}{L}
  4. μ0(N1+N2)(I1+I2)L\mu_0 \frac{(N_1 + N_2)(I_1 + I_2)}{L}
Explanation: The magnetic fields from both solenoids point in the same direction inside the coaxial arrangement. By superposition, the total field is the sum: B=μ0n1I1+μ0n2I2=μ0N1I1L+μ0N2I2L=μ0(N1I1+N2I2)LB = \mu_0 n_1 I_1 + \mu_0 n_2 I_2 = \mu_0 \frac{N_1 I_1}{L} + \mu_0 \frac{N_2 I_2}{L} = \mu_0 \frac{(N_1 I_1 + N_2 I_2)}{L}.

Question 4

The magnetic field outside a long solenoid is approximately zero. What does this imply about the circulation integral ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} for an Amperian loop entirely outside the solenoid?

  1. The circulation integral equals zero because the loop encloses no current from the solenoid windings (correct answer)
  2. The circulation integral equals μ0I\mu_0 I where II is the current in the solenoid windings
  3. The circulation integral is undefined because the magnetic field is zero everywhere on the path
  4. The circulation integral equals the total magnetic flux through the area bounded by the loop
Explanation: An Amperian loop entirely outside the solenoid encloses no net current (the current goes into the solenoid on one side and out on the other, giving zero net enclosed current). Therefore, by Ampère's law, ∮B⃗⋅dl⃗=μ0⋅0=0\oint \vec{B} \cdot d\vec{l} = \mu_0 \cdot 0 = 0, consistent with B⃗≈0\vec{B} \approx 0 outside.

Question 5

An infinite current sheet with surface current density KK produces a magnetic field of magnitude μ0K2\frac{\mu_0 K}{2} on either side. Which Amperian loop geometry is most appropriate for this derivation?

  1. A rectangular loop with sides parallel and perpendicular to the current sheet surface (correct answer)
  2. A circular loop centered on a point in the current sheet with radius perpendicular to the surface
  3. A triangular loop with one vertex touching the current sheet and base parallel to the surface
  4. A helical loop that spirals around the current sheet with increasing radius from the surface
Explanation: A rectangular Amperian loop with one side on each side of the current sheet exploits the symmetry: the field is parallel to the sheet and equal in magnitude on both sides. The contributions from the sides perpendicular to the sheet cancel, leaving only the parallel contributions.

Question 6

A current II flows radially outward from the center of a large conducting disk. Using Ampère's law with a circular Amperian loop centered on the disk, what can be concluded about the magnetic field?

  1. The magnetic field is zero everywhere because the radial current distribution encloses no net current (correct answer)
  2. The magnetic field has magnitude μ0I2πr\frac{\mu_0 I}{2\pi r} and is tangent to circles centered on the disk
  3. The magnetic field has magnitude μ0Iπr2\frac{\mu_0 I}{\pi r^2} and is directed perpendicular to the disk
  4. The magnetic field cannot be determined using Ampère's law due to the radial current geometry
Explanation: For a circular Amperian loop centered on the disk, the radial currents pass through the loop symmetrically - equal amounts of current cross the loop going inward and outward, so the net enclosed current is zero. By Ampère's law, ∮B⃗⋅dl⃗=μ0⋅0=0\oint \vec{B} \cdot d\vec{l} = \mu_0 \cdot 0 = 0, which combined with symmetry arguments shows B⃗=0\vec{B} = 0.

Question 7

Two parallel wires separated by distance dd carry currents I1I_1 and I2I_2 in opposite directions. At the midpoint between the wires, what is the magnitude of the total magnetic field?

  1. μ0(I1+I2)πd\frac{\mu_0(I_1 + I_2)}{\pi d} (correct answer)
  2. μ0(I1−I2)πd\frac{\mu_0(I_1 - I_2)}{\pi d}
  3. μ0I12+I22πd\frac{\mu_0\sqrt{I_1^2 + I_2^2}}{\pi d}
  4. μ0(I1+I2)2πd\frac{\mu_0(I_1 + I_2)}{2\pi d}
Explanation: At the midpoint, each wire is at distance d/2d/2. The magnetic field from each wire has magnitude μ0I2π(d/2)=μ0Iπd\frac{\mu_0 I}{2\pi(d/2)} = \frac{\mu_0 I}{\pi d}. Since the currents are in opposite directions, the fields point in the same direction at the midpoint, so they add: Btotal=μ0I1πd+μ0I2πd=μ0(I1+I2)πdB_{total} = \frac{\mu_0 I_1}{\pi d} + \frac{\mu_0 I_2}{\pi d} = \frac{\mu_0(I_1 + I_2)}{\pi d}.

Question 8

A very long solenoid with nn turns per unit length carries current II. Using Ampère's law, what is the magnitude of the magnetic field inside the solenoid?

  1. μ0nI\mu_0 n I (correct answer)
  2. μ0nI2\frac{\mu_0 n I}{2}
  3. μ0I2πn\frac{\mu_0 I}{2\pi n}
  4. μ0nI2π\frac{\mu_0 n I}{2\pi}
Explanation: For a long solenoid, the magnetic field is uniform inside and zero outside. Using a rectangular Amperian loop with one side inside the solenoid of length LL, Ampère's law gives BL=μ0(nL)IBL = \mu_0(nL)I, so B=μ0nIB = \mu_0 n I.

Question 9

A long air-core solenoid has n=500 turns/mn=500\,\text{turns/m} and carries a steady current I=3.0 AI=3.0\,\text{A}. The solenoid radius is R=2.0 cmR=2.0\,\text{cm} and its length is L=0.80 mL=0.80\,\text{m}, so L≫RL\gg R and the interior magnetic field is approximately uniform and parallel to the axis, while the exterior field is negligible. Use an Amperian rectangle with one side of length ℓ\ell inside the solenoid and the return side outside. According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, only the inside segment contributes significantly so ∮B⃗⋅dℓ⃗≈Bℓ\oint \vec B\cdot d\vec \ell\approx B\ell, and the enclosed current is Ienc=(nℓ)II_{\text{enc}}=(n\ell)I. Take μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A} and ignore fringing.

According to Ampère's Law, calculate the magnetic field inside the solenoid.

  1. 1.9×10−3 T1.9\times 10^{-3}\,\text{T} (correct answer)
  2. 6.0×10−3 T6.0\times 10^{-3}\,\text{T}
  3. 3.8×10−4 T3.8\times 10^{-4}\,\text{T}
  4. 1.2×10−6 T1.2\times 10^{-6}\,\text{T}
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to determine the magnetic field inside a long solenoid. Ampère's Law is particularly powerful for solenoids because the field inside is approximately uniform and axial for L >> R, while the field outside is negligible, making the choice of a rectangular Amperian loop straightforward. In this scenario, the solenoid has 500 turns/m and carries 3.0 A, with the rectangular loop having one side inside where B contributes and one outside where B ≈ 0. Choice A is correct because the enclosed current is (nℓ)I and the line integral is Bℓ, giving B = μ₀nI = (4π×10⁻⁷)(500)(3.0) = 1.9×10⁻³ T. Choice B might result from calculation errors, while other choices could come from incorrect application of the formula or unit mistakes. To help students: Reinforce that the magnetic field inside an ideal solenoid is B = μ₀nI, independent of the solenoid's radius. Practice problems with different solenoid parameters and emphasize the distinction between turn density (n) and total turns (N).

Question 10

A long, straight wire carries a steady current I=25.0 AI=25.0\,\text{A} upward along the +y+y axis in vacuum. Assume cylindrical symmetry so the magnetic field circles the wire and has magnitude B(r)B(r) that depends only on the radial distance rr from the wire. A circular Amperian loop of radius r=2.0 cmr=2.0\,\text{cm} is centered on the wire in a plane perpendicular to the wire. According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, where μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}. The loop encloses the entire current, and B⃗\vec B is tangent and uniform in magnitude along the loop, so ∮B⃗⋅dℓ⃗=B(2πr)\oint \vec B\cdot d\vec \ell=B(2\pi r). Ignore external fields and any magnetic materials.

According to Ampère's Law, what is the magnetic field strength at a distance of 2.0 cm2.0\,\text{cm} from the wire?

  1. 2.5×10−4 T2.5\times 10^{-4}\,\text{T} (correct answer)
  2. 1.3×10−5 T1.3\times 10^{-5}\,\text{T}
  3. 3.1×10−6 T3.1\times 10^{-6}\,\text{T}
  4. 8.0×10−5 T8.0\times 10^{-5}\,\text{T}
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to find the magnetic field around a straight current-carrying wire. Ampère's Law relates the circulation of the magnetic field around a closed path to the current enclosed by that path, making it ideal for problems with cylindrical symmetry. In this scenario, a straight wire carries 25.0 A upward, and we use a circular Amperian loop of radius 2.0 cm perpendicular to the wire, where the magnetic field is tangent to the loop and has constant magnitude. Choice A is correct because applying Ampère's Law gives B(2πr) = μ₀I, yielding B = μ₀I/(2πr) = (4π×10⁻⁷)(25.0)/(2π×0.02) = 2.5×10⁻⁴ T. Choice D might result from using diameter instead of radius, while choices B and C represent various calculation errors. To help students: Stress the importance of recognizing cylindrical symmetry and choosing circular Amperian loops. Practice converting between different units (cm to m) and ensure students understand that the field magnitude depends inversely on distance from the wire.

Question 11

A tightly wound solenoid in air has n=800 turns/mn=800\,\text{turns/m} and carries a steady current I=2.5 AI=2.5\,\text{A}. The solenoid is long compared with its radius (length L=0.50 mL=0.50\,\text{m}, radius R=1.5 cmR=1.5\,\text{cm}), so the magnetic field inside is approximately uniform and parallel to the solenoid axis, while the field outside is negligible. An Amperian loop is chosen as a rectangle with one long side of length ℓ=0.20 m\ell=0.20\,\text{m} inside the solenoid parallel to the axis and the opposite long side outside where B≈0B\approx 0. According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, with μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}. Using the diagram, only the inside segment contributes significantly, so ∮B⃗⋅dℓ⃗≈Bℓ\oint \vec B\cdot d\vec \ell\approx B\ell. The enclosed current equals the number of turns pierced by the loop times II, so Ienc=(nℓ)II_{\text{enc}}=(n\ell)I. Assume steady current, negligible fringing, and no ferromagnetic core.

Using the diagram, calculate the magnetic field inside the solenoid.

  1. 2.5×10−3 T2.5\times 10^{-3}\,\text{T} (correct answer)
  2. 8.0×10−4 T8.0\times 10^{-4}\,\text{T}
  3. 1.3×10−2 T1.3\times 10^{-2}\,\text{T}
  4. 6.3×10−6 T6.3\times 10^{-6}\,\text{T}
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to calculate the magnetic field inside a solenoid. Ampère's Law is particularly powerful for solenoids because the field inside is uniform and parallel to the axis, while the field outside is negligible, creating an ideal situation for choosing a rectangular Amperian loop. In this scenario, the solenoid has 800 turns/m and carries 2.5 A, with the rectangular loop having one side of length 0.20 m inside the solenoid where B is uniform, and the opposite side outside where B ≈ 0. Choice A is correct because the line integral reduces to Bℓ on the inside segment, and the enclosed current is (nℓ)I = (800)(0.20)(2.5) = 400 A, giving B = μ₀nI = (4π×10⁻⁷)(800)(2.5) = 2.5×10⁻³ T. Choice B might result from using turns instead of turn density, while other choices represent calculation errors. To help students: Emphasize that for solenoids, the field inside is B = μ₀nI regardless of the loop size chosen. Practice identifying when to use turn density (n) versus total turns (N), and watch for unit consistency when working with turns per meter.

Question 12

A toroidal inductor in air has N=300N=300 turns and carries a steady current I=2.0 AI=2.0\,\text{A}. The inner and outer radii are a=4.0 cma=4.0\,\text{cm} and b=9.0 cmb=9.0\,\text{cm}, respectively. Assume the magnetic field is negligible outside the windings and is approximately circular and tangent to a circle of radius rr inside the core region a<r<ba<r<b. Consider an Amperian loop that is a circle of radius r=5.0 cmr=5.0\,\text{cm} centered on the toroid axis. According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, and by symmetry ∮B⃗⋅dℓ⃗=B(2πr)\oint \vec B\cdot d\vec \ell=B(2\pi r) while Ienc=NII_{\text{enc}}=NI because the loop links all turns. Use μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A} and ignore fringing.

According to Ampère's Law, determine the magnetic field within the toroid at a radius of 5.0 cm5.0\,\text{cm}.

  1. 2.4×10−3 T2.4\times 10^{-3}\,\text{T}
  2. 7.5×10−4 T7.5\times 10^{-4}\,\text{T}
  3. 1.5×10−3 T1.5\times 10^{-3}\,\text{T} (correct answer)
  4. 4.8×10−4 T4.8\times 10^{-4}\,\text{T}
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to calculate the magnetic field inside a toroidal inductor. Ampère's Law is ideal for toroids due to their circular symmetry, where the magnetic field lines are circles centered on the toroid axis and confined to the core region between inner and outer radii. In this scenario, the toroid has 300 turns carrying 2.0 A, with inner radius 4.0 cm and outer radius 9.0 cm, and we evaluate the field at r = 5.0 cm using a circular Amperian loop that links all turns. Choice C is correct because applying Ampère's Law gives B(2πr) = μ₀NI, so B = μ₀NI/(2πr) = (4π×10⁻⁷)(300)(2.0)/(2π×0.05) = 1.5×10⁻³ T. Choice A would result from using a different radius, while choices B and D represent various calculation errors. To help students: Stress that the magnetic field in a toroid varies inversely with radius within the core region. Practice setting up Ampère's Law for different toroidal geometries and emphasize that the enclosed current includes contributions from all N turns when the loop is inside the toroid.

Question 13

A coaxial cable has a solid inner conductor of radius a=1.0 mma=1.0\,\text{mm} and a thin outer conductor of inner radius b=5.0 mmb=5.0\,\text{mm}. The inner conductor carries a steady current I=6.0 AI=6.0\,\text{A} in the +z+z direction, and the outer conductor carries −6.0 A-6.0\,\text{A} as a return current on its surface. Assume vacuum between conductors, negligible end effects, and perfect cylindrical symmetry so B⃗\vec B is azimuthal and depends only on radius rr. A circular Amperian loop of radius r=3.0 mmr=3.0\,\text{mm} lies between the conductors (a<r<ba<r<b). According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, the enclosed current is Ienc=II_{\text{enc}}=I for this loop, and ∮B⃗⋅dℓ⃗=B(2πr)\oint \vec B\cdot d\vec \ell=B(2\pi r). Use μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}.

According to Ampère's Law, find the magnetic field between the conductors.

  1. 4.0×10−4 T4.0\times 10^{-4}\,\text{T}
  2. 1.3×10−4 T1.3\times 10^{-4}\,\text{T}
  3. 0 T0\,\text{T}
  4. 2.0×10−4 T2.0\times 10^{-4}\,\text{T} (correct answer)
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to find the magnetic field in a coaxial cable between the inner and outer conductors. Ampère's Law is ideal for this geometry due to cylindrical symmetry, where the magnetic field forms circles around the central axis and has magnitude depending only on radius. In this scenario, the inner conductor (radius 1.0 mm) carries 6.0 A forward while the outer conductor (at radius 5.0 mm) carries 6.0 A return current, and we calculate the field at r = 3.0 mm where the Amperian loop encloses only the inner conductor's current. Choice D is correct because applying Ampère's Law gives B(2πr) = μ₀I, so B = μ₀I/(2πr) = (4π×10⁻⁷)(6.0)/(2π×0.003) = 2.0×10⁻⁴ T. Choice C (zero) would apply outside the outer conductor where both currents cancel, not between the conductors. To help students: Stress the importance of identifying which currents are enclosed at different radial positions. Practice sketching the field configuration in coaxial cables and understanding why the field is zero outside when currents are equal and opposite.

Question 14

A toroidal coil (air core) has N=500N=500 tightly wound turns carrying a steady current I=1.50 AI=1.50\,\text{A}. The toroid has inner radius a=6.0 cma=6.0\,\text{cm} and outer radius b=10.0 cmb=10.0\,\text{cm}. Assume the windings are uniform, the field is confined primarily within the core region a<r<ba<r<b, and the magnetic field lines are circular and tangent to circles centered on the toroid axis. Choose a circular Amperian loop of radius r=8.0 cmr=8.0\,\text{cm} centered on the toroid axis (so it lies within the core). According to Ampère's Law, ∮B⃗⋅dℓ⃗=μ0Ienc,\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}}, with μ0=4π×10−7 T⋅m/A\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}. Using the diagram, B⃗\vec B is tangent and approximately constant in magnitude along the loop, so ∮B⃗⋅dℓ⃗=B(2πr)\oint \vec B\cdot d\vec \ell=B(2\pi r). The enclosed current equals NINI because the loop links all NN turns. Ignore fringing fields and any magnetic materials.

Using the diagram, determine the magnetic field within the toroid at a radius of 8.0 cm8.0\,\text{cm}.

  1. 1.9×10−3 T1.9\times 10^{-3}\,\text{T}
  2. 5.9×10−4 T5.9\times 10^{-4}\,\text{T} (correct answer)
  3. 9.4×10−3 T9.4\times 10^{-3}\,\text{T}
  4. 2.4×10−4 T2.4\times 10^{-4}\,\text{T}
Explanation: This question tests AP Physics C understanding of Ampère's Law, specifically its application to toroidal geometries where the magnetic field forms circular loops inside the toroid. Ampère's Law is particularly useful for toroids because of their azimuthal symmetry, allowing us to choose circular Amperian loops where B is tangent and constant in magnitude. In this scenario, a toroid with 500 turns carries 1.50 A, and we use a circular loop at radius r = 8.0 cm (between inner radius 6.0 cm and outer radius 10.0 cm), which links all N turns of the toroid. Choice B is correct because applying Ampère's Law gives B(2πr) = μ₀NI, yielding B = μ₀NI/(2πr) = (4π×10⁻⁷)(500)(1.50)/(2π×0.08) = 5.9×10⁻⁴ T. Choice A might result from using the inner radius instead of the actual loop radius, while choice C could come from calculation errors. To help students: Emphasize that for toroids, the field varies with radius as 1/r within the core region. Practice distinguishing between the loop radius (where we calculate B) and the toroid's inner/outer radii, and ensure students understand that all turns contribute to the enclosed current.

Question 15

Ampère's law relates the circulation of the magnetic field around a closed path to which of the following quantities?

  1. The total electric current enclosed by the path multiplied by the permeability of free space (correct answer)
  2. The total electric charge enclosed by the path multiplied by the permeability of free space
  3. The total magnetic flux passing through the surface bounded by the path
  4. The rate of change of electric flux passing through the surface bounded by the path
Explanation: Ampère's law states that ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}, where the circulation of the magnetic field around a closed path equals the permeability of free space times the enclosed current. This is the fundamental statement of Ampère's law.

Question 16

Which of the following best describes why Ampère's law is particularly useful for calculating magnetic fields in highly symmetric current distributions?

  1. The symmetry allows the magnetic field to be factored out of the circulation integral as a constant (correct answer)
  2. Symmetric current distributions always produce uniform magnetic fields throughout all regions of space
  3. The mathematical complexity of the Biot-Savart law becomes undefined for symmetric geometries
  4. Ampère's law automatically accounts for the vector nature of magnetic fields in symmetric cases
Explanation: In highly symmetric situations, we can choose Amperian loops where the magnetic field has constant magnitude along portions of the path and is either parallel or perpendicular to the path element dl⃗d\vec{l}. This allows B⃗\vec{B} to be factored out as a constant, making the integral ∮B⃗⋅dl⃗=B∮dl\oint \vec{B} \cdot d\vec{l} = B \oint dl easily solvable.

Question 17

Two identical long solenoids are placed end-to-end with their axes aligned, carrying currents in opposite directions. What is the magnetic field in the region where the solenoids overlap?

  1. Zero, because the magnetic fields from the two solenoids cancel each other completely (correct answer)
  2. μ0nI\mu_0 n I, because the magnetic fields from the two solenoids add constructively
  3. 2μ0nI2\mu_0 n I, because the magnetic fields from both solenoids contribute equally
  4. μ0nI2\frac{\mu_0 n I}{2}, because the effective turn density is reduced in the overlap region
Explanation: When the currents are in opposite directions, the magnetic fields produced by each solenoid point in opposite directions along the axis. In the overlap region, the fields have equal magnitude μ0nI\mu_0 n I but opposite directions, so they cancel completely, giving zero net field.

Question 18

Maxwell's modification to Ampère's law includes which additional term to account for changing electric fields?

  1. μ0ε0dΦEdt\mu_0 \varepsilon_0 \frac{d\Phi_E}{dt} where ΦE\Phi_E is the electric flux through the Amperian loop (correct answer)
  2. 1μ0ε0dΦEdt\frac{1}{\mu_0 \varepsilon_0} \frac{d\Phi_E}{dt} where ΦE\Phi_E is the electric flux through the Amperian loop
  3. μ0dΦBdt\mu_0 \frac{d\Phi_B}{dt} where ΦB\Phi_B is the magnetic flux through the Amperian loop
  4. ε0dEdt\varepsilon_0 \frac{dE}{dt} where EE is the electric field magnitude at the loop location
Explanation: Maxwell's modification to Ampère's law adds the displacement current term: ∮B⃗⋅dl⃗=μ0I+μ0ε0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 I + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}. This accounts for the magnetic field produced by changing electric flux, completing the electromagnetic field equations.

Question 19

A current II flows in a helical coil (like a spring) with nn turns per unit length. If the coil has a large length compared to its radius, what is the magnetic field inside the coil?

  1. μ0nI\mu_0 n I directed along the axis of the coil in the direction given by the right-hand rule (correct answer)
  2. μ0nI2\frac{\mu_0 n I}{2} directed along the axis of the coil in the direction given by the right-hand rule
  3. μ0I2πn\frac{\mu_0 I}{2\pi n} directed perpendicular to the axis of the coil at each point
  4. μ0nI\mu_0 n I directed in circles around the axis of the coil at each interior point
Explanation: A helical coil with large length-to-radius ratio behaves like a solenoid. The magnetic field inside is uniform and parallel to the axis with magnitude B=μ0nIB = \mu_0 n I, where nn is the number of turns per unit length. The direction follows the right-hand rule.

Question 20

A solid cylindrical conductor of radius RR carries current II uniformly distributed over its cross-section. What is the magnetic field at radius r<Rr < R from the center?

  1. μ0Ir2πR2\frac{\mu_0 I r}{2\pi R^2} (correct answer)
  2. μ0I2πr\frac{\mu_0 I}{2\pi r}
  3. μ0Ir22πR2\frac{\mu_0 I r^2}{2\pi R^2}
  4. μ0IR2πr2\frac{\mu_0 I R}{2\pi r^2}
Explanation: The current density is J=IπR2J = \frac{I}{\pi R^2}. The current enclosed by radius rr is Ienc=J⋅πr2=Ir2R2I_{enc} = J \cdot \pi r^2 = \frac{I r^2}{R^2}. Applying Ampère's law with a circular loop: B(2πr)=μ0IencB(2\pi r) = \mu_0 I_{enc}, so B=μ0Ir2πR2B = \frac{\mu_0 I r}{2\pi R^2}.