2023 AP AP Physics C: Electricity and Magnetism 模拟试题及答案详解

(a) Free-Body Diagram
- Downward arrow for gravitational force, labeled $F_g$ (or $mg$ or $W$).
- Horizontal arrow pointing to the right for the electrostatic force, labeled $F_E$ (or $qE$).
- Upward and leftward diagonal arrow pointing along the string for tension, labeled $F_T$ (or $T$).

(b) Derivation
At static equilibrium, applying Newton's second law in two dimensions:
$$\Sigma F_x = F_E - F_T \sin\theta = 0 \implies F_T \sin\theta = F_E$$
$$\Sigma F_y = F_T \cos\theta - F_g = 0 \implies F_T \cos\theta = mg$$
Dividing the two equations:
$$\frac{F_T \sin\theta}{F_T \cos\theta} = \frac{F_E}{mg} \implies \tan\theta = \frac{F_E}{mg}$$
Substituting $F_E = qE = q\left(\frac{V}{d}\right)$:
$$\tan\theta = \left(\frac{q}{mgd}\right)V$$

(c) Data Analysis
i. A straight line of best fit is drawn with approximately equal distribution of points above and below the line, passing through $(0,0)$ and $(1000\text{ V}, 0.39)$.

ii. Calculating slope using two points on the best-fit line, e.g., $(0, 0)$ and $(1000, 0.39)$:
$$\text{Slope} = \frac{0.39 - 0}{1000\text{ V} - 0\text{ V}} = 3.9 \times 10^{-4}\text{ V}^{-1}$$
From the derived equation, $\text{Slope} = \frac{q}{mgd}$:
$$q = \text{Slope} \cdot mgd = (3.9 \times 10^{-4}\text{ V}^{-1})(2.0 \times 10^{-3}\text{ kg})(9.8\text{ m/s}^2)(0.080\text{ m}) = 6.1 \times 10^{-7}\text{ C}$$

iii. At $V = 800\text{ V}$, $\tan\theta = 0.312$, so $\theta = \arctan(0.312) \approx 17.3^\circ$.
$$F_T = \frac{mg}{\cos\theta} = \frac{(2.0 \times 10^{-3}\text{ kg})(9.8\text{ m/s}^2)}{\cos(17.3^\circ)} \approx \frac{0.0196}{0.9548} \approx 0.021\text{ N}$$

(d) Uncharged Conducting Sphere
- Selection: Remains hanging vertically ($\theta = 0$).
- Justification: In a uniform electric field, the conducting sphere polarizes such that negative charges accumulate on the left side and equal positive charges accumulate on the right side. Because the external electric field is uniform, the force to the left on the negative charges equals the force to the right on the positive charges ($F_{\text{net}} = q_+ E - q_- E = 0$). Thus, no net electrostatic force acts on the sphere, and it does not deflect.

(e) Effect of Changing Plate Separation
- Selection: $\theta_2 = \theta_1$
- Justification: When the battery is disconnected, the charge $Q_{\text{plate}}$ on the isolated parallel plates is constant, so the surface charge density $\sigma = \frac{Q_{\text{plate}}}{A}$ is constant. The electric field between large parallel plates is $E = \frac{\sigma}{\varepsilon_0}$, which is independent of the plate separation $d$. Because $E$ remains unchanged, the electrostatic force $F_E = qE$ remains constant, and therefore the equilibrium deflection angle $\theta$ does not change ($\theta_2 = \theta_1$).

Part (a) [2 points total]
- 1 point: For correctly drawing and labeling the electrostatic force pointing horizontally to the right.
- 1 point: For correctly drawing and labeling the gravitational force downward and the tension force pointing upward and to the left (deduct 1 point for any extraneous forces).

Part (b) [2 points total]
- 1 point: For setting up equilibrium conditions equating horizontal and vertical components of tension to electrostatic and gravitational forces ($F_T \sin\theta = F_E$ and $F_T \cos\theta = mg$).
- 1 point: For substituting $F_E = q\frac{V}{d}$ and combining equations to arrive at the target relationship.

Part (c) [5 points total]
- (i) 1 point: For drawing a reasonable single straight line of best fit that follows the trend of the data points.
- (ii) 1 point: For calculating the slope using two points on the drawn best-fit line (not raw data points unless they lie on the line).
- (ii) 1 point: For correctly equating the slope to $\frac{q}{mgd}$ and solving for $q$.
- (ii) 1 point: For a consistent numerical answer with proper units ($5.8 \times 10^{-7}\text{ C}$ to $6.4 \times 10^{-7}\text{ C}$).
- (iii) 1 point: For correctly calculating the tension using $F_T = \frac{mg}{\cos\theta}$ or $F_T = \sqrt{(mg)^2 + F_E^2}$.

Part (d) [2 points total]
- 1 point: For stating that the sphere remains hanging vertically (or does not deflect).
- 1 point: For a correct justification explaining that in a uniform electric field, the induced dipole experiences equal and opposite forces, resulting in zero net electrostatic force.

Part (e) [4 points total]
- 1 point: For selecting '$\theta_2 = \theta_1$'.
- 1 point: For stating that disconnecting the battery keeps the charge $Q$ (or surface charge density $\sigma$) on the plates constant.
- 1 point: For indicating that the electric field $E = \sigma / \varepsilon_0$ between parallel plates is independent of the separation distance $d$.
- 1 point: For connecting the constancy of $E$ to a constant electrostatic force $F_E = qE$, meaning the deflection angle remains unchanged.

(a)(i)
Horizontal forces acting on the rod:
- Tension force \(F_T\) directed to the right (toward the pulley).
- Magnetic/Lorentz force \(F_B\) directed to the left (opposing motion).

(a)(ii)
As the rod moves to the right, the area enclosed by the loop increases, which increases the magnetic flux directed into the page. According to Lenz's law, the induced current must create an induced magnetic field directed out of the page to oppose this increase in flux. By the right-hand rule, an out-of-page magnetic field requires a counterclockwise current in the loop, which means the current flows upward through the rod (from the bottom rail to the top rail).

(b)(i)
The motional EMF induced in the moving rod is:
\[\mathcal{E} = -\frac{d\Phi_B}{dt} = B_0 L v\]
The total resistance in the closed loop is \(R_{\text{total}} = R + r\).
By Ohm's law, the induced current is:
\[I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{B_0 L v}{R + r}\]

(b)(ii)
The magnetic force opposing the rod's motion is:
\[F_B = I L B_0 = \left(\frac{B_0 L v}{R + r}\right) L B_0 = \frac{B_0^2 L^2 v}{R + r}\]
Applying Newton's second law to the combined system of mass \(M + m\):
\[\Sigma F_{\text{ext}} = (M + m)a\]
\[Mg - F_B = (M + m)\frac{dv}{dt}\]
\[(M + m)\frac{dv}{dt} = Mg - \frac{B_0^2 L^2 v}{R + r}\]

(c)
At terminal speed \(v_T\), the acceleration is zero (\(\frac{dv}{dt} = 0\)):
\[0 = Mg - \frac{B_0^2 L^2 v_T}{R + r}\]
\[v_T = \frac{Mg(R + r)}{B_0^2 L^2}\]

(d)
The speed-time graph starts at \((0,0)\) with initial slope \(a_0 = \frac{Mg}{M+m}\), is concave down, and asymptotically approaches the horizontal line \(v = v_T\).

(e)(i)
\(v_{T,\text{new}} < v_T\)

(e)(ii)
From the derived equation, terminal speed is inversely proportional to the square of the magnetic field (\(v_T \propto \frac{1}{B_0^2}\)). Physically, a stronger magnetic field produces a larger induced EMF and a stronger magnetic braking force at any given speed. Therefore, the magnetic force balances the gravitational pulling force \(Mg\) at a lower velocity. Specifically, when \(B_0\) is doubled to \(2B_0\), the new terminal speed becomes \(v_{T,\text{new}} = \frac{1}{4}v_T\).

Part (a): 3 points total
- 1 point for correctly drawing and labeling the tension force directed to the right and the magnetic force directed to the left on the rod, with no extraneous horizontal forces.
- 1 point for correctly stating that the current is directed upward.
- 1 point for a valid justification using Lenz's law and flux change (flux into the page increases, so induced field opposes change by pointing out of the page, leading to CCW current/upward flow through the rod).

Part (b): 4 points total
- 1 point for applying Faraday's law / motional EMF: \(\mathcal{E} = B_0 L v\).
- 1 point for substituting total resistance \(R + r\) into Ohm's law to find \(I = \frac{B_0 L v}{R + r}\).
- 1 point for writing a correct expression for the magnetic force \(F_B = \frac{B_0^2 L^2 v}{R + r}\).
- 1 point for applying Newton's second law to the system to arrive at \((M + m)\frac{dv}{dt} = Mg - \frac{B_0^2 L^2 v}{R + r}\).

Part (c): 2 points total
- 1 point for setting acceleration \(\frac{dv}{dt} = 0\) or setting \(F_B = Mg\).
- 1 point for the correct algebraic expression for \(v_T = \frac{Mg(R+r)}{B_0^2 L^2}\).

Part (d): 2 points total
- 1 point for a curve starting at the origin \((0,0)\) that is monotonically increasing and concave down.
- 1 point for showing clear asymptotic approach to a horizontal asymptote labeled \(v_T\).

Part (e): 4 points total
- 1 point for selecting \(v_{T,\text{new}} < v_T\).
- 1 point for explaining that a larger \(B\) field increases the rate of magnetic braking force per unit velocity.
- 2 points for referencing the inverse-square dependence \(v_T \propto \frac{1}{B_0^2}\) to demonstrate that the new speed is \(\frac{1}{4}v_T\).

(a) When switch $S$ is in Position A, apply Kirchhoff's loop and junction rules:
Let the total current from the battery through $R_1$ be $I_{\text{tot}} = I_{R2} + I_L$.
The potential difference across the parallel branch is $V_{\text{branch}} = V_L = L \frac{dI_L}{dt} = I_{R2} R_2 = 2R I_{R2}$.
Thus, $I_{R2} = \frac{L}{2R} \frac{dI_L}{dt}$, and $I_{\text{tot}} = I_L + \frac{L}{2R} \frac{dI_L}{dt}$.
Applying Kirchhoff's loop rule around the outer loop containing the battery, $R_1$, and inductor $L$:
$$\mathcal{E}_0 - I_{\text{tot}} R_1 - L \frac{dI_L}{dt} = 0$$
$$\mathcal{E}_0 - \left(I_L + \frac{L}{2R}\frac{dI_L}{dt}\right)R - L \frac{dI_L}{dt} = 0$$
$$\mathcal{E}_0 - R I_L - \frac{3}{2} L \frac{dI_L}{dt} = 0$$
Rearranging yields:
$$\frac{dI_L}{dt} = \frac{2\mathcal{E}_0 - 2R I_L}{3L} = \frac{2(\mathcal{E}_0 - R I_L)}{3L}$$

(b)
- **Graph i ($V_L$ vs $t$):** Starts at an initial non-zero maximum value $V_L(0) = \frac{R_2}{R_1+R_2}\mathcal{E}_0 = \frac{2}{3}\mathcal{E}_0$, and decreases monotonically with concave-up curvature, approaching the horizontal asymptote $V_L = 0$ as $t \to \infty$.
- **Graph ii ($I_{R2}$ vs $t$):** Since $V_{R2} = V_L$, the current $I_{R2}(t) = \frac{V_L(t)}{2R}$ begins at $I_{R2}(0) = \frac{\mathcal{E}_0}{3R}$ and decays monotonically with concave-up curvature asymptotically toward $0$.

(c)
- i. At $t = 0^+$, the inductor acts as an open circuit ($I_L = 0$). The circuit is simply $R_1$ and $R_2$ in series:
$$I_{\text{tot}}(0^+) = \frac{\mathcal{E}_0}{R_1 + R_2} = \frac{\mathcal{E}_0}{R + 2R} = \frac{\mathcal{E}_0}{3R}$$
- ii. As $t \to \infty$, the inductor acts as an ideal short circuit ($V_L = 0$), so no current flows through $R_2$. The entire current from the battery passes through $R_1$ and $L$:
$$I_L(\infty) = \frac{\mathcal{E}_0}{R_1} = \frac{\mathcal{E}_0}{R}$$

(d)
- i. Opposite direction. Prior to $t_1$, current through $R_2$ flowed downward (from the upper node to the lower reference rail). Immediately after $t_1$, the inductor maintains its downward current of magnitude $I_0 = \frac{\mathcal{E}_0}{R}$ through its branch, which forces conventional current to circulate upward through resistor $R_2$ in the isolated closed loop.
- ii. Conservation of energy: all the magnetic energy initially stored in the inductor at $t_1$ is converted into thermal energy dissipated in resistor $R_2$ (the only resistive element in the isolated loop):
$$U_B = \frac{1}{2} L I_0^2$$
Substituting $I_0 = \frac{\mathcal{E}_0}{R}$:
$$E_{\text{diss}} = \frac{1}{2} L \left(\frac{\mathcal{E}_0}{R}\right)^2 = \frac{L \mathcal{E}_0^2}{2R^2}$$

(e)
- i. The self-inductance of a solenoid is $L = \mu_0 n^2 A \ell = \mu_0 \frac{N^2}{\ell} A$. Since the number of turns doubles while the geometry remains unchanged, $L \propto N^2$, so:
$$\frac{L_{\text{new}}}{L} = \left(\frac{2N}{N}\right)^2 = 4$$
- ii. Increases. The discharging time constant for an $RL$ loop is $\tau = \frac{L_{\text{new}}}{R_2} = \frac{4L}{2R} = \frac{2L}{R}$, which is $4$ times greater than the original time constant $\tau_0 = \frac{L}{2R}$. Because the half-life $t_{1/2} = \tau \ln 2$ is directly proportional to $\tau$, the time required for current to decay to half of its initial value increases.

Part (a): 2 points
- 1 point: For applying Kirchhoff's loop rule to relate $\mathcal{E}_0$, the voltage across $R_1$, and the induced emf across $L$ ($L \frac{dI_L}{dt}$).
- 1 point: For correctly substituting the total current $I_{\text{tot}} = I_L + \frac{L}{2R}\frac{dI_L}{dt}$ into the loop equation to obtain a valid differential equation in terms of permitted variables.

Part (b): 3 points
- 1 point: For sketching $V_L(t)$ starting at a positive non-zero value, monotonically decreasing, and approaching zero asymptotically.
- 1 point: For sketching $I_{R2}(t)$ starting at a positive non-zero value, monotonically decreasing, and approaching zero asymptotically.
- 1 point: For drawing both curves with concave-up curvature approaching zero slope as $t$ increases.

Part (c): 3 points
- 1 point: For determining the correct initial total current $I_{\text{tot}}(0^+) = \frac{\mathcal{E}_0}{3R}$ by recognizing the inductor acts as an open circuit.
- 2 points: For determining the long-time inductor current $I_L(\infty) = \frac{\mathcal{E}_0}{R}$ (1 point for recognizing inductor acts as a short circuit with zero parallel voltage, 1 point for the correct algebraic expression).

Part (d): 4 points
- 1 point: For stating that the current is in the opposite direction with a correct physical justification using Lenz's law / continuity of inductor current.
- 1 point: For equating total energy dissipated to the initial magnetic energy stored in the inductor ($E_{\text{diss}} = \frac{1}{2}LI^2$).
- 1 point: For using the correct initial current $I_0 = \frac{\mathcal{E}_0}{R}$ at $t = t_1$.
- 1 point: For the correct algebraic derivation leading to $E_{\text{diss}} = \frac{L\mathcal{E}_0^2}{2R^2}$.

Part (e): 3 points
- 1 point: For calculating the correct inductance ratio $\frac{L_{\text{new}}}{L} = 4$ based on $L \propto N^2$.
- 1 point: For selecting "Increases".
- 1 point: For a justification relating the increase in time to the larger time constant $\tau = L/R$.