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.

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.

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.

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.

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).