An original Thinka practice paper modelled on the structure and difficulty of the May 2025 AP AP Physics C: Electricity and Magnetism paper. Not affiliated with or reproduced from AP.
部分 II: Free-Response Questions
Section II contains 4 free-response questions. Show all derivations, equations from the reference sheet, intermediate steps, graphs, and units where appropriate. A scientific or graphing calculator and reference tables are permitted.
4 题目 · 40 分
题目 1 · Free-Response Question
10 分
An isolated electrostatic system consists of a solid nonconducting sphere of radius \(a\) concentric with a thin conducting spherical shell of radius \(b\), where \(b > a\). The solid sphere has a uniform positive volume charge density \(\rho_0\). The thin conducting shell carries a total negative charge \(-Q_0\), where the magnitude of the total charge on the solid sphere equals \(Q_0 = \frac{4}{3}\pi a^3 \rho_0\).
A.
i. Using Gauss's law, derive an expression for the magnitude \(E\) of the electric field as a function of the radial distance \(r\) from the center of the sphere for the region \(a < r < b\). Express your answer in terms of \(a\), \(\rho_0\), \(r\), and physical constants, as appropriate.
ii. Derive an expression for the absolute value \(|\Delta V|\) of the electric potential difference between the surface of the solid sphere at \(r = a\) and the spherical shell at \(r = b\) in terms of \(a\), \(b\), \(\rho_0\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
iii. On a set of axes of electric field magnitude \(E\) versus radial distance \(r\), describe the features of a sketch of \(E(r)\) from \(r = 0\) to a position \(r > b\).
B. A dielectric material of dielectric constant \(\kappa\) is inserted between the solid sphere and the conducting shell, completely filling the region \(a < r < b\).
Derive an expression for the capacitance \(C\) of the spherical capacitor with the dielectric material inserted in terms of \(a\), \(b\), \(\kappa\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
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解题
Part A(i): Apply Gauss's law: \[\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}\] For a spherical Gaussian surface of radius \(r\) where \(a < r < b\): - Surface area: \(A = 4\pi r^2\) - Enclosed charge: \(q_{\text{enc}} = \rho_0 V_{\text{sphere}} = \rho_0 \left(\frac{4}{3}\pi a^3\right) Substituting into Gauss's law: \[E(4\pi r^2) = \frac{\frac{4}{3}\pi a^3 \rho_0}{\varepsilon_0}\] \[E(r) = \frac{\rho_0 a^3}{3\varepsilon_0 r^2}\] --- **Part A(ii):** Relate electric potential difference to electric field: \[\Delta V = -\int_a^b \vec{E} \cdot d\vec{r}\] Taking the magnitude: \[|\Delta V| = \int_a^b \frac{\rho_0 a^3}{3\varepsilon_0 r^2}\, dr = \frac{\rho_0 a^3}{3\varepsilon_0} \left[ -\frac{1}{r} \right]_a^b = \frac{\rho_0 a^3}{3\varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right) = \frac{\rho_0 a^2 (b - a)}{3\varepsilon_0 b}\] --- **Part A(iii):** - For \)0 \le r < a\): \(q_{\text{enc}} = \rho_0 \left(\frac{4}{3}\pi r^3\right) \implies E(r) = \frac{\rho_0 r}{3\varepsilon_0}\), which is a straight line through the origin with positive slope. - For \(a < r < b\): \(E(r) = \frac{\rho_0 a^3}{3\varepsilon_0 r^2}\), which is positive, decreasing, and concave up. - For \(r > b\): The net enclosed charge is \(Q_{\text{net}} = +Q_0 - Q_0 = 0\), so \(E = 0\).
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Part B: Definition of capacitance: \[C = \frac{Q}{|\Delta V|}\] Without dielectric: \[Q = \frac{4}{3}\pi a^3 \rho_0\] \[|\Delta V| = \frac{\rho_0 a^3}{3\varepsilon_0}\left(\frac{b - a}{ab}\right)\] \[C_0 = \frac{\frac{4}{3}\pi a^3 \rho_0}{\frac{\rho_0 a^3 (b - a)}{3\varepsilon_0 ab}} = \frac{4\pi \varepsilon_0 a b}{b - a}\] With the dielectric material of constant \(\kappa\) completely filling the gap: \[C = \kappa C_0 = \frac{4\pi \kappa \varepsilon_0 a b}{b - a}\]
评分标准
Part A(i): (3 points) - Point A1: For stating Gauss's law \(\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}\). - Point A2: For substituting the correct surface area \(4\pi r^2\) for a spherical Gaussian surface. - Point A3: For correctly determining the enclosed charge \(q_{\text{enc}} = \frac{4}{3}\pi a^3 \rho_0\) and solving for \(E(r) = \frac{\rho_0 a^3}{3\varepsilon_0 r^2}\).
Part A(ii): (2 points) - Point A4: For substituting the expression for \(E\) into the integral relation \(|\Delta V| = \int E\, dr\). - Point A5: For evaluating the integral with correct limits from \(a\) to \(b\) to find \(|\Delta V| = \frac{\rho_0 a^3}{3\varepsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right)\).
Part A(iii): (2 points) - Point A6: For describing/sketching a curve that is linear from \(r = 0\) to \(r = a\) and zero for \(r > b\). - Point A7: For describing/sketching a curve that is decreasing and concave up (proportional to \(1/r^2\)) for \(a < r < b\).
Part B: (3 points) - Point B1: For starting from a fundamental definition of capacitance \(C = \frac{Q}{\Delta V}\). - Point B2: For correctly substituting total charge \(Q\) and potential difference \(|\Delta V|\) consistent with Part A. - Point B3: For correctly applying the dielectric constant \(\kappa\) to obtain \(C = \frac{4\pi \kappa \varepsilon_0 ab}{b - a}\).
题目 2 · free-response
12 分
An ideal $LC$ circuit consists of a capacitor of capacitance $C$ connected across an inductor of inductance $L$. The total resistance of the circuit is negligible. At time $t = 0$, the capacitor stores an initial charge $+Q_0$ on its upper plate and the current in the circuit is zero. The system undergoes undamped electromagnetic oscillations with period $T = 2\pi\sqrt{LC}$.
A. The energy stored in the electric field of the capacitor is $U_E$, and the energy stored in the magnetic field of the inductor is $U_B$. At $t = 0$, $U_E = U_{\text{tot}}$ (represented as a bar of height $4\text{ units}$) and $U_B = 0$. In a bar chart, indicate the values of $U_E$ and $U_B$ at times $t = \frac{1}{4}T$, $t = \frac{1}{2}T$, and $t = \frac{3}{4}T$ relative to the value at $t = 0$. If an energy value is zero at any given time, specify "0" for that column.
B. Derive an expression for the magnitude of the maximum current $I_{\max}$ in the inductor in terms of $Q_0$, $L$, and $C$. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
C. On a set of axes displaying $U_B$ on the vertical axis and time $t$ on the horizontal axis, sketch a graph of the magnetic energy $U_B(t)$ stored in the inductor as a function of time $t$ over the interval $0 \le t \le T$.
D. Indicate whether the sketch you drew in part C is or is not consistent with the bars representing $U_B$ in part A. Briefly justify your answer by referencing the functional dependence of $U_B$ on time.
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解题
A. By conservation of energy, the total energy $U_{\text{tot}} = U_E + U_B = \frac{Q_0^2}{2C}$ is constant. - At $t = \frac{1}{4}T$: The capacitor is fully discharged ($q = 0$), so $U_E = 0$. All energy is stored in the inductor's magnetic field, so $U_B = 4\text{ units}$. - At $t = \frac{1}{2}T$: The capacitor is fully charged with opposite polarity ($q = -Q_0$), so $U_E = 4\text{ units}$, and the current is momentarily zero, so $U_B = 0$. - At $t = \frac{3}{4}T$: The capacitor is again fully discharged ($q = 0$), so $U_E = 0$, and $U_B = 4\text{ units}$.
B. Method 1 (Conservation of Energy): $$U_{\text{tot}} = \frac{Q_0^2}{2C} = \frac{1}{2}L I_{\max}^2$$ $$I_{\max}^2 = \frac{Q_0^2}{LC}$$ $$I_{\max} = \frac{Q_0}{\sqrt{LC}}$$
C. The magnetic energy as a function of time is: $$U_B(t) = \frac{1}{2}L [i(t)]^2 = \frac{1}{2}L (I_{\max} \sin(\omega t))^2 = U_{\text{tot}} \sin^2\left(\frac{2\pi t}{T}\right)$$ The graph is a periodic sinusoidal squared wave (always $\ge 0$): - Starts at $(0, 0)$. - Rises smoothly to a peak of $U_{\text{tot}}$ (height 4 units) at $t = \frac{1}{4}T$. - Falls to 0 at $t = \frac{1}{2}T$. - Rises smoothly to a second identical peak of $U_{\text{tot}}$ (height 4 units) at $t = \frac{3}{4}T$. - Falls to 0 at $t = T$. - Shows exactly two full oscillation cycles of energy during one period $T$ of circuit oscillation.
D. Yes, the sketch in part C is consistent with the bar chart in part A. The energy $U_B(t) \propto \sin^2(\omega t)$ demonstrates that $U_B$ reaches zero at $t = 0, \frac{1}{2}T, T$ and reaches its maximum value of $4\text{ units}$ at $t = \frac{1}{4}T$ and $t = \frac{3}{4}T$, aligning with the discrete states represented in the bar chart.
评分标准
Part A (3 points): - 1 point: For correctly identifying and drawing $U_B = 4\text{ units}$ at both $t = \frac{1}{4}T$ and $t = \frac{3}{4}T$. - 1 point: For correctly identifying and drawing $U_E = 4\text{ units}$ at $t = \frac{1}{2}T$. - 1 point: For indicating 0 for $U_E$ at $t = \frac{1}{4}T$ and $\frac{3}{4}T$, and 0 for $U_B$ at $t = \frac{1}{2}T$.
Part B (4 points): - 1 point: For stating a correct fundamental principle (e.g., Conservation of Energy $U_E + U_B = \text{constant}$ or Faraday's/Kirchhoff's loop rule $-L\frac{di}{dt} - \frac{q}{C} = 0$). - 1 point: For equating maximum electric potential energy to maximum magnetic energy ($\frac{1}{2}\frac{Q_0^2}{C} = \frac{1}{2}L I_{\max}^2$) OR finding $i(t) = -\frac{dq}{dt} = Q_0 \omega \sin(\omega t)$. - 1 point: For substituting $\omega = \frac{1}{\sqrt{LC}}$ or correctly manipulating the energy relation. - 1 point: For arriving at the correct final expression $I_{\max} = \frac{Q_0}{\sqrt{LC}}$.
Part C (3 points): - 1 point: For sketching a curve that is concave down at peaks and concave up at troughs (approximate $\sin^2$ shape). - 1 point: For showing exactly two cycles within the interval $0 \le t \le T$. - 1 point: For starting at the origin $(0,0)$ and having minima equal to 0 at $t = 0, \frac{1}{2}T, T$ and equal maxima at $t = \frac{1}{4}T, \frac{3}{4}T$.
Part D (2 points): - 1 point: For stating that the representations are consistent. - 1 point: For a valid justification linking the functional dependence $U_B \propto \sin^2(\omega t)$ or $U_B = \frac{1}{2}Li^2$ to the specific zero and maximum points shown on both the bar chart and graph.
题目 3 · Experimental Design and Analysis (LAB)
10 分
In Experiment 1, a group of students is tasked with designing an investigation to determine the magnetic permeability of free space, \(\mu_0\), using a long solenoid of known length \(L\) and total turns \(N\). The students have access to a variable DC power supply, an ammeter, connecting wires, a magnetic field sensor (Hall probe), and a meterstick.
A. Describe an experimental procedure to collect data that can be used to determine \(\mu_0\), including the circuit setup, the quantities to be measured, and at least one procedure to reduce experimental uncertainty.
B. Explain what quantities should be graphed to yield a linear relationship and how the slope of the resulting graph would be analyzed to calculate \(\mu_0\).
In Experiment 2, another group of students uses a solenoid of length \(L = 0.50\text{ m}\) containing \(N = 400\text{ turns}\) uniformly wound along its length. The magnetic field sensor is positioned at the center of the solenoid along its central axis. The students measure the magnitude of the internal magnetic field \(B\) for several different values of the current \(I\). The collected data are recorded in Table 1 below.
ii. On a coordinate grid, plot the data points from Table 1 according to your chosen axes. Clearly label both axes, including units and appropriate scales.
iii. Draw a best-fit line for the plotted data points.
D. Using the slope of the best-fit line drawn in part C (iii), calculate an experimental value for \(\mu_0\).
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解题
### Part A 1. Connect the variable DC power supply, the ammeter, and the solenoid in a single closed series circuit using connecting wires. 2. Before powering the circuit, zero the magnetic field sensor inside the center of the solenoid to eliminate the effect of the ambient background/Earth's magnetic field. 3. Turn on the power supply and place the active tip of the magnetic field probe at the geometric center of the solenoid, aligned parallel to the central axis. 4. Record the current \(I\) on the ammeter and the magnetic field strength \(B\) on the sensor. 5. Adjust the power supply to at least four additional different current settings, recording the corresponding values of \(I\) and \(B\) for each setting. 6. To reduce experimental uncertainty, repeat the measurements at each current setting for 3 trials and compute average values.
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### Part B The magnetic field inside an ideal long solenoid is given by: \[ B = \mu_0 n I = \mu_0 \left(\frac{N}{L}\right) I \] Graph the magnetic field \(B\) on the vertical axis as a function of current \(I\) on the horizontal axis. Since \(B = \left(\frac{\mu_0 N}{L}\right) I\), the theoretical relationship is linear with a slope of \(m = \frac{\mu_0 N}{L}\). Therefore, \(\mu_0\) is determined from the slope of the best-fit line by: \[ \mu_0 = m \cdot \frac{L}{N} \]
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### Part C i. - Vertical axis: Magnetic Field \(B\text{ (mT)}\) [or \(B\text{ (T)}\)] - Horizontal axis: Current \(I\text{ (A)}\)
ii. & iii. - Scale for horizontal axis: \(0\) to \(3.0\text{ A}\), in equal intervals of \(0.5\text{ A}\). - Scale for vertical axis: \(0\) to \(3.0\text{ mT}\), in equal intervals of \(0.5\text{ mT}\). - Plotted points: \((0.50, 0.51)\), \((1.00, 1.02)\), \((1.50, 1.49)\), \((2.00, 2.04)\), \((2.50, 2.52)\). - Best-fit line drawn smoothly and evenly through the plotted points.
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### Part D Selecting two points directly on the best-fit line: - Point 1: \((0.50\text{ A}, 0.50\text{ mT}) = (0.50\text{ A}, 0.50 \times 10^{-3}\text{ T})\) - Point 2: \((2.50\text{ A}, 2.52\text{ mT}) = (2.50\text{ A}, 2.52 \times 10^{-3}\text{ T})\)
Given \(N = 400\) and \(L = 0.50\text{ m}\): \[ \mu_0 = m \left(\frac{L}{N}\right) = (1.01 \times 10^{-3}\text{ T/A}) \left(\frac{0.50\text{ m}}{400}\right) = 1.26 \times 10^{-6}\text{ T}\cdot\text{m/A} \]
评分标准
### Distribution of Points (Total: 10 points)
Part A (2 points) - Point A1: For describing an appropriate experimental setup that includes measuring current with an ammeter connected in series with the solenoid and power supply, and measuring the magnetic field along the central axis inside the solenoid using the sensor. (1 point) - Point A2: For identifying a valid method to reduce experimental uncertainty (e.g., zeroing the sensor to subtract Earth's magnetic field, taking multiple readings at each current setting to calculate an average, or sampling across a wide range of current values). (1 point)
Part B (2 points) - Point B1: For specifying appropriate quantities to graph that yield a linear relationship (e.g., \(B\) vs. \(I\) or \(I\) vs. \(B\)). (1 point) - Point B2: For correctly explaining how \(\mu_0\) is determined from the slope of the linear graph (e.g., relating \(\text{slope} = \mu_0 N / L\) so \(\mu_0 = \text{slope} \cdot L / N\)). (1 point)
Part C (4 points) - Point C1: For correctly identifying the variables plotted on the vertical and horizontal axes consistent with Part B. (1 point) - Point C2: For correctly scaled and labeled axes with units (e.g., \(B\text{ (mT)}\) and \(I\text{ (A)}\)) occupying more than half the grid. (1 point) - Point C3: For accurately plotting the data points from Table 1. (1 point) - Point C4: For drawing a reasonable straight best-fit line that follows the trend of the data points. (1 point)
Part D (2 points) - Point D1: For calculating the slope using two points on the best-fit line (not raw data points unless they lie on the line) and relating that slope to \(\mu_0\). (1 point) - Point D2: For calculating a final value of \(\mu_0\) within the acceptable range of \(1.20 \times 10^{-6}\text{ T}\cdot\text{m/A}\) to \(1.32 \times 10^{-6}\text{ T}\cdot\text{m/A}\) with appropriate units. (1 point)
题目 4 · Free-Response Questions
8 分
Two long, straight, parallel wires, Wire 1 and Wire 2, lie in the \(xy\)-plane and are parallel to the \(x\)-axis. Wire 1 is located at the position \(y = +d\) and carries a constant current \(I_0\) in the \(+x\)-direction. Wire 2 is located at \(y = -d\) and carries an identical constant current \(I_0\) in the \(+x\)-direction.
At a particular instant, Particle 1 (having charge \(+q\)) is located at the origin \((0, 0)\) and is moving with speed \(v_0\) in the \(+x\)-direction. At the same instant, Particle 2 (also having charge \(+q\)) is located at \((0, 2d)\) and is moving with speed \(v_0\) in the \(+x\)-direction. Gravitational and electrostatic interactions between the particles are negligible.
A. Let \(F_1\) be the magnitude of the magnetic force exerted on Particle 1 due to the currents in Wire 1 and Wire 2. Let \(F_2\) be the magnitude of the magnetic force exerted on Particle 2 due to the currents in Wire 1 and Wire 2.
Indicate whether \(F_2\) is greater than, less than, or equal to \(F_1\) by selecting one of the following: - \(F_2 > F_1\) - \(F_2 < F_1\) - \(F_2 = F_1\)
Justify your answer.
B. Derive an expression for the magnitude \(B_{\text{tot}}\) of the net magnetic field at the location of Particle 2 due to the currents in Wire 1 and Wire 2 in terms of \(d\), \(I_0\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
C. Later, the current in Wire 2 is changed so that it carries a current of \(3I_0\) in the \(-x\)-direction, while the current in Wire 1 remains unchanged (\(I_0\) in the \(+x\)-direction). At the instant under consideration, Particle 2 is again at \((0, 2d)\) and moving with speed \(v_0\) in the \(+x\)-direction. Let \(F_{\text{new}}\) be the new magnitude of the magnetic force exerted on Particle 2.
Indicate whether \(F_{\text{new}}\) is greater than, less than, or equal to \(F_2\) by selecting one of the following: - \(F_{\text{new}} > F_2\) - \(F_{\text{new}} < F_2\) - \(F_{\text{new}} = F_2\)
Briefly justify your answer by referencing your derivation in part B.
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解题
Part A: - Selection: \(F_2 > F_1\) - Justification: The magnitude of the magnetic force on a charged particle is given by \(F_B = qvB\sin\theta\). Since both particles have the same charge \(+q\), move with the same speed \(v_0\), and move perpendicular to the magnetic field (which lies along the \(z\)-axis), the magnetic force is directly proportional to the magnitude of the net magnetic field \(B_{\text{net}}\) at the location of each particle. At the origin \((0,0)\), Particle 1 is equidistant (distance \(d\)) from both wires. By the right-hand rule, the magnetic field from Wire 1 points into the page (\(-\hat{k}\)) while the magnetic field from Wire 2 points out of the page (\(+\hat{k}\)). Because the currents and distances are equal, these two fields have equal magnitudes and opposite directions, resulting in \(B_{\text{net}, 1} = 0\), and therefore \(F_1 = 0\). At \((0, 2d)\), Particle 2 is located above both wires. By the right-hand rule, the magnetic fields produced by both Wire 1 and Wire 2 point out of the page (\(+\hat{k}\)). Since both contributions are in the same direction, they add constructively to produce a nonzero net magnetic field \(B_{\text{net}, 2} > 0\). Therefore, \(F_2 > 0 = F_1\), so \(F_2 > F_1\).
Part B: Begin with Ampere's law: \[\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}\] For an Amperian circle of radius \(r\) centered on a long straight wire carrying current \(I\): \[B(2\pi r) = \mu_0 I \implies B(r) = \frac{\mu_0 I}{2\pi r}\] At the position of Particle 2 \((y = 2d)\): - The distance from Wire 1 (at \(y = d\)) is \(r_1 = 2d - d = d\): \[B_1 = \frac{\mu_0 I_0}{2\pi d} \quad (+\hat{k}\text{-direction})\] - The distance from Wire 2 (at \(y = -d\)) is \(r_2 = 2d - (-d) = 3d\): \[B_2 = \frac{\mu_0 I_0}{2\pi (3d)} = \frac{\mu_0 I_0}{6\pi d} \quad (+\hat{k}\text{-direction})\] Since both fields point in the \(+z\)-direction (out of the page): \[B_{\text{tot}} = B_1 + B_2 = \frac{\mu_0 I_0}{2\pi d} + \frac{\mu_0 I_0}{6\pi d} = \frac{3\mu_0 I_0 + \mu_0 I_0}{6\pi d} = \frac{4\mu_0 I_0}{6\pi d} = \frac{2\mu_0 I_0}{3\pi d}\]
Part C: - Selection: \(F_{\text{new}} < F_2\) - Justification: When the current in Wire 2 becomes \(3I_0\) in the \(-x\)-direction, the magnetic field produced by Wire 2 at \(y = 2d\) changes direction to point into the page (\(-\hat{k}\)) by the right-hand rule, and its magnitude becomes: \[B_2' = \frac{\mu_0 (3I_0)}{2\pi(3d)} = \frac{\mu_0 I_0}{2\pi d}\] The field from Wire 1 remains \(B_1 = \frac{\mu_0 I_0}{2\pi d}\) in the \(+\hat{k}\)-direction. Thus, the new net magnetic field at Particle 2 is: \[B_{\text{new}} = B_1 - B_2' = \frac{\mu_0 I_0}{2\pi d} - \frac{\mu_0 I_0}{2\pi d} = 0\] Since the net magnetic field at the location of Particle 2 is now zero, \(F_{\text{new}} = 0\), which is less than \(F_2\).
评分标准
### Part A (3 points) - Point A1: For selecting \(F_2 > F_1\). - Point A2: For correctly relating the magnitude of the magnetic force on each particle to the magnitude of the net magnetic field (e.g., using \(F_B = qvB\) and noting that \(q\) and \(v\) are identical for both particles). - Point A3: For correctly reasoning that the magnetic fields from the two wires cancel out at the position of Particle 1 (giving \(B_{\text{net}, 1} = 0\)) but point in the same direction and reinforce each other at the position of Particle 2 (giving \(B_{\text{net}, 2} > 0\)).
### Part B (3 points) - Point B1: For a multistep derivation that begins with a fundamental physics equation (e.g., \(\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}\) or \(B = \frac{\mu_0 I}{2\pi r}\)). - Point B2: For correct expressions for the magnetic field contributions from each wire at \(y = 2d\) (i.e., \(B_1 = \frac{\mu_0 I_0}{2\pi d}\) and \(B_2 = \frac{\mu_0 I_0}{6\pi d}\)). - Point B3: For correctly adding the two contributions to arrive at the total magnetic field magnitude: \(B_{\text{tot}} = \frac{2\mu_0 I_0}{3\pi d}\).
### Part C (2 points) - Point C1: For selecting \(F_{\text{new}} < F_2\). - Point C2: For a correct justification showing that the new field from Wire 2 opposes the field from Wire 1 and has equal magnitude (\(B_2' = \frac{\mu_0 (3I_0)}{6\pi d} = \frac{\mu_0 I_0}{2\pi d} = B_1\)), resulting in \(B_{\text{new}} = 0\) and hence \(F_{\text{new}} = 0 < F_2\).