The total current $$I$$ is found by integrating the current density $$J$$ over the cross-sectional area $$A$$. We use an infinitesimal ring of area $$dA = 2\pi r dr$$. The integral is $$I = \int_A J dA = \int_0^R J_0 (1 - r/R) (2\pi r dr) = 2\pi J_0 \int_0^R (r - r^2/R) dr = 2\pi J_0 [\frac{r^2}{2} - \frac{r^3}{3R}]_0^R = 2\pi J_0 (\frac{R^2}{2} - \frac{R^3}{3R}) = 2\pi J_0 (\frac{R^2}{2} - \frac{R^2}{3}) = 2\pi J_0 (\frac{R^2}{6}) = \frac{1}{3} \pi R^2 J_0$$.

Once the particle enters the wire, it becomes a charge carrier. Its motion inside the wire is characterized by frequent collisions with the lattice, resulting in a very small average drift velocity, which determines the current ($$I=nqAv_d$$). The high speed gained during initial acceleration is quickly lost to collisions, and the particle settles into the slow drift that constitutes the current.

The current density $$J$$ is the total current $$I$$ divided by the cross-sectional area $$A$$ through which the current flows. For a hollow cylinder, the cross-sectional area is the area of the outer circle minus the area of the inner circle: $$A = \pi R_2^2 - \pi R_1^2 = \pi(R_2^2 - R_1^2)$$. Therefore, the current density is $$J = I / A = I / (\pi(R_2^2 - R_1^2))$$.

Current is $$I = V/R$$, and resistance is $$R = \rho L/A$$. So, $$I = VA/(\rho L)$$. Drift speed is related to current by $$I = nqAv_d$$, so $$v_d = I/(nqA)$$. Substituting the expression for $$I$$, we get $$v_d = (VA/(\rho L))/(nqA) = V/(nq\rho L)$$. Since $$V$$ and $$L$$ are the same for both wires, $$v_d \propto 1/(n\rho)$$. Thus, $$ \frac{v_{d,N}}{v_{d,A}} = \frac{n_A \rho_A}{n_N \rho_N} = \frac{n_A \rho_A}{(0.5 n_A)(60 \rho_A)} = \frac{1}{30}$$. So, $$v_{d,N} \approx \frac{1}{30} v_{d,A}$$.

The relationship between electric field, resistivity, and current density is $$E = \rho J$$. The current density is $$J = I/A$$. Since both wires have the same current $$I$$ and area $$A$$, their current densities are equal. Therefore, the electric field is directly proportional to the resistivity ($$E \propto \rho$$). Since aluminum has a higher resistivity than copper ($$\rho_{Al} > \rho_{Cu}$$), the electric field in the aluminum wire is greater than in the copper wire ($$E_{Al} > E_{Cu}$$).

For a current to flow, there must be a net force on the charge carriers. This force is provided by an electric field inside the conductor. The electric field exerts a force $$F = qE$$ on the positive charge carriers, causing them to drift and create a current. Thus, the electric field points in the same direction as the conventional current. The condition of zero electric field inside a conductor only applies to electrostatic equilibrium.

This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current calculation using Ohm's Law. Electric current is the flow of electric charge in a circuit, calculated as the ratio of voltage to resistance. In this problem, we have a voltage of 9 V across a resistance of 3 Ω, requiring straightforward application of I = V/R. Choice D is correct because I = V/R = 9 V / 3 Ω = 3.0 A, properly applying Ohm's Law to find the current. Choice C is incorrect because it multiplies voltage and resistance (9 × 3 = 27), which is a common algebraic error when students forget the correct relationship or confuse multiplication with division. To help students: Create a triangle diagram with V at the top and I and R at the bottom to visualize the relationships. Watch for: Students who multiply quantities instead of dividing, or who forget to check their units (amperes) for reasonableness.

This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current behavior in parallel circuits. Electric current in parallel circuits divides among branches, with the total current being the sum of individual branch currents. When adding a resistor in parallel, you create an additional path for current flow, which affects the circuit's total resistance and current. Choice B is correct because adding a parallel resistor decreases the total circuit resistance (1/R_total = 1/R₁ + 1/R₂), and with constant voltage, decreased resistance leads to increased total current according to Ohm's Law. Choice A is incorrect because it applies series circuit logic to a parallel situation, assuming that adding components always increases total resistance, which is a fundamental misconception about parallel circuits. To help students: Use the analogy of multiple lanes on a highway - more lanes (parallel paths) allow more traffic (current) flow. Practice calculating equivalent resistance for parallel combinations and emphasize that parallel resistance is always less than the smallest individual resistance.

This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current and circuit analysis using Ohm's Law. Electric current is the flow of electric charge in a circuit, typically measured in amperes, and is calculated using the relationship I = V/R. In this problem, we're given a voltage of 12 V and a resistance of 6 Ω, requiring direct application of Ohm's Law. Choice B is correct because I = V/R = 12 V / 6 Ω = 2.0 A, which properly applies the fundamental relationship between voltage, current, and resistance. Choice D is incorrect because it multiplies voltage and resistance (12 × 6 = 72), which is a common error when students confuse the formula. To help students: Emphasize memorizing Ohm's Law in all three forms (I = V/R, V = IR, R = V/I) and practice unit analysis. Watch for: Students multiplying instead of dividing, or confusing which quantity goes in the numerator versus denominator.

This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current calculation through direct application of Ohm's Law. Electric current is determined by the ratio of applied voltage to circuit resistance, representing the rate of charge flow. Given a voltage of 18 V and a resistance of 9 Ω, we need to calculate the resulting current using I = V/R. Choice E is correct because I = V/R = 18 V / 9 Ω = 2.0 A, which accurately applies the fundamental relationship between these electrical quantities. Choice A is incorrect because it multiplies voltage and resistance (18 × 9 = 162), a common error that suggests the student doesn't understand that resistance opposes current flow. To help students: Create practice problems with simple integer ratios to build confidence before moving to decimals. Watch for: Students who default to multiplication when unsure, or who don't verify their answers make physical sense.