Ampère's Law - AP Physics C Electricity and Magnetism

1. 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?
   Answer: Zero, because the magnetic fields from the two solenoids cancel each other completely 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 $\mu_0 n I$ but opposite directions, so they cancel completely, giving zero net field.
2. Which of the following best describes why Ampère's law is particularly useful for calculating magnetic fields in highly symmetric current distributions?
   Answer: The symmetry allows the magnetic field to be factored out of the circulation integral as a constant 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 $d\vec{l}$. This allows $\vec{B}$ to be factored out as a constant, making the integral $\oint \vec{B} \cdot d\vec{l} = B \oint dl$ easily solvable.
3. In applying Ampère's law, the line integral $\oint \vec{B} \cdot d\vec{l}$ represents which physical quantity?
   Answer: The circulation of the magnetic field around the chosen closed path or Amperian loop The line integral $\oint \vec{B} \cdot d\vec{l}$ is the circulation of the magnetic field - it measures how much the field 'circulates' or 'curls' around the chosen path. This circulation equals $\mu_0$ times the enclosed current according to Ampère's law.
4. A long air-core solenoid has $n=500\,\text{turns/m}$ and carries a steady current $I=3.0\,\text{A}$. The solenoid radius is $R=2.0\,\text{cm}$ and its length is $L=0.80\,\text{m}$, so $L\gg R$ 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 $\ell$ inside the solenoid and the return side outside. According to Ampère's Law, $\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$ only the inside segment contributes significantly so $\oint \vec B\cdot d\vec \ell\approx B\ell$, and the enclosed current is $I_{\text{enc}}=(n\ell)I$. Take $\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$ and ignore fringing. According to Ampère's Law, calculate the magnetic field inside the solenoid.
   Answer: $1.9\times 10^{-3}\,\text{T}$ 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).
5. In which of the following situations would Ampère's law be most directly applicable for calculating the magnetic field?
   Answer: A current configuration with high symmetry where the magnetic field is constant along chosen paths Ampère's law is most useful when the current configuration has sufficient symmetry that the magnetic field has constant magnitude along portions of the chosen Amperian loop, making the line integral tractable. Without such symmetry, the integral becomes very difficult to evaluate.
6. A finite straight wire segment of length $L$ carries current $I$. Why is Ampère's law not typically used to find the magnetic field at a nearby point?
   Answer: The current configuration lacks sufficient symmetry to make the Amperian loop integral tractable For a finite wire segment, the magnetic field varies both in magnitude and direction around any reasonable Amperian loop, making the line integral $\oint \vec{B} \cdot d\vec{l}$ very difficult to evaluate. The Biot-Savart law is more appropriate for such geometries.
7. Ampère's law relates the circulation of the magnetic field around a closed path to which of the following quantities?
   Answer: The total electric current enclosed by the path multiplied by the permeability of free space Ampère's law states that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$, 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.
8. A coaxial cable has a solid inner conductor of radius $a=1.0\,\text{mm}$ and a thin outer conductor of inner radius $b=5.0\,\text{mm}$. The inner conductor carries a steady current $I=6.0\,\text{A}$ in the $+z$ direction, and the outer conductor carries $-6.0\,\text{A}$ as a return current on its surface. Assume vacuum between conductors, negligible end effects, and perfect cylindrical symmetry so $\vec B$ is azimuthal and depends only on radius $r$. A circular Amperian loop of radius $r=3.0\,\text{mm}$ lies between the conductors ($a<r<b$). According to Ampère's Law, $\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$ the enclosed current is $I_{\text{enc}}=I$ for this loop, and $\oint \vec B\cdot d\vec \ell=B(2\pi r)$. Use $\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$. According to Ampère's Law, find the magnetic field between the conductors.
   Answer: $2.0\times 10^{-4}\,\text{T}$ 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.
9. A toroidal solenoid has $N$ total turns and carries current $I$. The magnetic field inside the torus at radius $r$ from the center is given by which expression?
   Answer: $\frac{\mu_0 N I}{2\pi r}$ For a toroidal solenoid, using a circular Amperian loop of radius $r$ inside the torus, Ampère's law gives $B(2\pi r) = \mu_0 N I$, where $N$ is the total number of turns. Therefore, $B = \frac{\mu_0 N I}{2\pi r}$.
10. Two parallel wires separated by distance $d$ carry currents $I_1$ and $I_2$ in opposite directions. At the midpoint between the wires, what is the magnitude of the total magnetic field?
    Answer: $\frac{\mu_0(I_1 + I_2)}{\pi d}$ At the midpoint, each wire is at distance $d/2$. The magnetic field from each wire has magnitude $\frac{\mu_0 I}{2\pi(d/2)} = \frac{\mu_0 I}{\pi d}$. Since the currents are in opposite directions, the fields point in the same direction at the midpoint, so they add: $B_{total} = \frac{\mu_0 I_1}{\pi d} + \frac{\mu_0 I_2}{\pi d} = \frac{\mu_0(I_1 + I_2)}{\pi d}$.
11. A current $I$ 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?
    Answer: The magnetic field is zero everywhere because the radial current distribution encloses no net current 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, $\oint \vec{B} \cdot d\vec{l} = \mu_0 \cdot 0 = 0$, which combined with symmetry arguments shows $\vec{B} = 0$.
12. Two coaxial solenoids with the same length $L$ have $N_1$ and $N_2$ turns respectively, carrying currents $I_1$ and $I_2$ in the same direction. What is the magnetic field inside both solenoids?
    Answer: $\mu_0 \frac{(N_1 I_1 + N_2 I_2)}{L}$ The magnetic fields from both solenoids point in the same direction inside the coaxial arrangement. By superposition, the total field is the sum: $B = \mu_0 n_1 I_1 + \mu_0 n_2 I_2 = \mu_0 \frac{N_1 I_1}{L} + \mu_0 \frac{N_2 I_2}{L} = \mu_0 \frac{(N_1 I_1 + N_2 I_2)}{L}$.
13. A square current loop with side length $a$ carries current $I$. Why is Ampère's law not the preferred method to find the magnetic field at the center of the loop?
    Answer: The geometry lacks the rotational symmetry needed to choose an Amperian loop where the field is constant A square loop does not have the rotational symmetry of a circular loop. There is no simple Amperian loop around the center where the magnetic field would be constant in magnitude and direction, making Ampère's law impractical. The Biot-Savart law is more suitable for this geometry.
14. A toroidal inductor in air has $N=300$ turns and carries a steady current $I=2.0\,\text{A}$. The inner and outer radii are $a=4.0\,\text{cm}$ and $b=9.0\,\text{cm}$, respectively. Assume the magnetic field is negligible outside the windings and is approximately circular and tangent to a circle of radius $r$ inside the core region $a<r<b$. Consider an Amperian loop that is a circle of radius $r=5.0\,\text{cm}$ centered on the toroid axis. According to Ampère's Law, $\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$ and by symmetry $\oint \vec B\cdot d\vec \ell=B(2\pi r)$ while $I_{\text{enc}}=NI$ because the loop links all turns. Use $\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$ and ignore fringing. According to Ampère's Law, determine the magnetic field within the toroid at a radius of $5.0\,\text{cm}$.
    Answer: $1.5\times 10^{-3}\,\text{T}$ 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.
15. A hollow cylindrical conductor of inner radius $a$ and outer radius $b$ carries a uniformly distributed current $I$. What is the magnetic field at radius $r$ where $a < r < b$?
    Answer: $\frac{\mu_0 I (r^2 - a^2)}{2\pi r (b^2 - a^2)}$ The current density is $J = \frac{I}{\pi(b^2 - a^2)}$. The current enclosed by radius $r$ is $I_{enc} = J \cdot \pi(r^2 - a^2) = \frac{I(r^2 - a^2)}{b^2 - a^2}$. Applying Ampère's law: $B(2\pi r) = \mu_0 I_{enc}$, giving $B = \frac{\mu_0 I (r^2 - a^2)}{2\pi r (b^2 - a^2)}$.
16. A tightly wound solenoid in air has $n=800\,\text{turns/m}$ and carries a steady current $I=2.5\,\text{A}$. The solenoid is long compared with its radius (length $L=0.50\,\text{m}$, radius $R=1.5\,\text{cm}$), 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 $\ell=0.20\,\text{m}$ inside the solenoid parallel to the axis and the opposite long side outside where $B\approx 0$. According to Ampère's Law, $\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$ with $\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$. Using the diagram, only the inside segment contributes significantly, so $\oint \vec B\cdot d\vec \ell\approx B\ell$. The enclosed current equals the number of turns pierced by the loop times $I$, so $I_{\text{enc}}=(n\ell)I$. Assume steady current, negligible fringing, and no ferromagnetic core. Using the diagram, calculate the magnetic field inside the solenoid.
    Answer: $2.5\times 10^{-3}\,\text{T}$ 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.
17. A long, straight wire carries a steady current $I=25.0\,\text{A}$ upward along the $+y$ axis in vacuum. Assume cylindrical symmetry so the magnetic field circles the wire and has magnitude $B(r)$ that depends only on the radial distance $r$ from the wire. A circular Amperian loop of radius $r=2.0\,\text{cm}$ is centered on the wire in a plane perpendicular to the wire. According to Ampère's Law, $\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$ where $\mu_0=4\pi\times10^{-7}\,\text{T}\cdot\text{m/A}$. The loop encloses the entire current, and $\vec B$ is tangent and uniform in magnitude along the loop, so $\oint \vec B\cdot d\vec \ell=B(2\pi r)$. Ignore external fields and any magnetic materials. According to Ampère's Law, what is the magnetic field strength at a distance of $2.0\,\text{cm}$ from the wire?
    Answer: $2.5\times 10^{-4}\,\text{T}$ 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.
18. A long, straight wire carries a steady current $I$. Using Ampère's law with a circular Amperian loop of radius $r$ centered on the wire, what is the magnitude of the magnetic field at distance $r$ from the wire?
    Answer: $\frac{\mu_0 I}{2\pi r}$ Applying Ampère's law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$. For a circular path, $B$ is constant and tangent to the circle, so $B(2\pi r) = \mu_0 I$, giving $B = \frac{\mu_0 I}{2\pi r}$.
19. Which of the following statements about Amperian loops is correct when applying Ampère's law?
    Answer: The Amperian loop must be chosen to exploit the symmetry of the current distribution for practical calculations The key to successfully applying Ampère's law is choosing an Amperian loop that takes advantage of the symmetry of the problem, allowing the magnetic field to be constant along portions of the path and making the integral tractable.
20. A solid cylindrical conductor of radius $R$ carries current $I$ uniformly distributed over its cross-section. What is the magnetic field at radius $r < R$ from the center?
    Answer: $\frac{\mu_0 I r}{2\pi R^2}$ The current density is $J = \frac{I}{\pi R^2}$. The current enclosed by radius $r$ is $I_{enc} = J \cdot \pi r^2 = \frac{I r^2}{R^2}$. Applying Ampère's law with a circular loop: $B(2\pi r) = \mu_0 I_{enc}$, so $B = \frac{\mu_0 I r}{2\pi R^2}$.

Start this quiz