Electric Current - AP Physics C Electricity and Magnetism

A conductor carrying a steady current is not in electrostatic equilibrium. Which statement correctly describes the electric field inside this conductor?
Answer: The electric field is non-zero and points in the direction of conventional current. 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.
Which of the following physical quantities is most analogous to electromotive force (emf) in an electrical circuit?
Answer: The force exerted by a pump that circulates water in a closed pipe system. Electromotive force (emf) is the work done per unit charge by a source (like a battery) to move charges and maintain a potential difference, thus sustaining current. In the water-pipe analogy, a pump provides the pressure difference (analogous to potential difference) that does work on the water to make it flow, which is analogous to the role of emf in a circuit.
A Nichrome wire and an aluminum wire have the same dimensions and are subjected to the same potential difference across their ends. The number density of charge carriers in Nichrome is roughly half that in aluminum, and the resistivity of Nichrome is about 60 times that of aluminum. How does the electron drift speed $v_{d,N}$ in Nichrome compare to the drift speed $v_{d,A}$ in aluminum?
Answer: $v_{d,N} \approx \frac{1}{30} v_{d,A}$ 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, $v_{d,N} \approx \frac{1}{30} v_{d,A}$ .
For an ohmic conductor, the relationship between the magnitude of the electric field $E$ inside it and the magnitude of the current density $J$ is given by $E = \rho J$ , where $\rho$ is the resistivity. If a wire of length $L$ and cross-sectional area $A$ is connected to a battery of emf $V$ , which expression represents the current density $J$ ?
Answer: $V / (\rho L)$ For a uniform wire of length $L$ connected to a potential difference $V$ , the electric field inside is uniform with magnitude $E = V/L$ . Substituting this into the equation $E = \rho J$ , we get $V/L = \rho J$ . Solving for the current density $J$ gives $J = V / (\rho L)$ .
A hollow cylindrical conductor has an inner radius of $R_1$ and an outer radius of $R_2$ . A steady current $I$ flows along the conductor, distributed uniformly over its cross-section. What is the magnitude of the current density $J$ within the conductor?
Answer: $I / (\pi (R_2^2 - R_1^2))$ 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 $J = I / A = I / (\pi (R_2^2 - R_1^2))$ .
The conventional current in a horizontal wire is directed from east to west. What is the direction of the drift velocity of the free electrons in the wire?
Answer: West to east, opposite to the conventional current. By convention, the direction of electric current is the direction that positive charge carriers would move. Electrons are negative charge carriers. Therefore, the drift velocity of electrons is in the direction opposite to the conventional current. If the conventional current is from east to west, the electrons drift from west to east.
A potential difference is applied across a cylindrical conductor, resulting in an electric field $E$ and a current density $J$ . If the temperature of the conductor increases, its resistivity $\rho$ increases. Assuming the applied potential difference remains constant, what happens to the current density $J$ ?
Answer: It decreases, because the increased resistivity impedes the flow of charge. The relationship is $E = \rho J$ . The electric field is $E = V/L$ . Since the potential difference $V$ and length $L$ are constant, the electric field $E$ remains constant. From $J = E/\rho$ , if the resistivity $\rho$ increases due to the temperature increase, the current density $J$ must decrease.
The current density vector $\vec{J}$ in a conductor is related to the number density of charge carriers $n$ , their charge $q$ , and their drift velocity $\vec{v}_d$ . Which of the following equations correctly represents this relationship?
Answer: $\vec{J} = n q \vec{v}_d$ Current density $J$ is the current per unit area. Since $I = nqAv_d$ , dividing by area $A$ gives $J = nqv_d$ . As vectors, the direction of $J$ is defined by the direction of the drift velocity of positive charges (or opposite to negative charges). For a charge $q$ (which can be positive or negative), the drift velocity vector $\vec{v}_d$ determines the direction, so $\vec{J} = n q \vec{v}_d$ .
A cylindrical conductor has a non-uniform current density given by $J(r) = J_0 (1 - r/R)$ , where $J_0$ is a constant, $R$ is the radius of the conductor, and $r$ is the radial distance from the center. What is the total current $I$ flowing through the conductor?
Answer: $\frac{1}{3} \pi R^2 J_0$ 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 $I = \int_A J dA = \int_0^R J_0 (1 - r/R) (2\pi r dr) = 2\pi J_0 \int_0^R (r - r^2/R) dr = 2\pi J_0 [\frac{r^2}{2} - \frac{r^3}{3R}]_0^R = 2\pi J_0 (\frac{R^2}{2} - \frac{R^3}{3R}) = 2\pi J_0 (\frac{R^2}{2} - \frac{R^2}{3}) = 2\pi J_0 (\frac{R^2}{6}) = \frac{1}{3} \pi R^2 J_0$ .
In a metallic conductor, a steady current is established. Which of the following best describes the motion of the electrons that constitute the current?
Answer: They undergo random thermal motion, superimposed with a small average velocity opposite to the direction of the electric field. Electrons in a metal have high-speed random thermal motion. The electric field superimposes a small net drift velocity ( $v_d$ ) on this random motion. The drift is opposite to the electric field direction because electrons are negatively charged. Collisions with the lattice prevent them from accelerating uniformly.
The current in a wire varies with time according to the equation $I(t) = 4 - 2t$ , where $I$ is in amperes and $t$ is in seconds. What is the total charge that passes a point in the wire from $t=0$ s to $t=2$ s?
Answer: 4 C Current is the rate of flow of charge, $I = dq/dt$ . To find the total charge $Q$ that passes a point, we must integrate the current with respect to time: $Q = \int_{t_1}^{t_2} I(t) dt$ . For the given interval, $Q = \int_{0}^{2} (4 - 2t) dt = [4t - t^2]_0^2 = (4(2) - 2^2) - (0) = 8 - 4 = 4$ C.
If $10^{20}$ electrons pass a given point in a wire every 2.0 seconds, what is the magnitude of the current in the wire?
Answer: 8.0 A The total charge $Q$ is the number of electrons $N$ times the elementary charge $e$ : $Q = N e = (10^{20})(1.6 \times 10^{-19} \text{ C}) = 16$ C. The current $I$ is the total charge passed per unit time: $I = Q / \Delta t = 16 \text{ C} / 2.0 \text{ s} = 8.0$ A.
A steady current flows through a conical conductor of uniform resistivity. The conductor's circular cross-section has a radius that increases linearly from left to right. How does the drift velocity of the charge carriers vary along the length of the conductor from left to right?
Answer: It decreases because the cross-sectional area increases. For a steady current $I$ , the current is the same at all cross-sections. The current is related to drift velocity $v_d$ by $I = nqAv_d$ . Since $I$ , $n$ , and $q$ are constant, $v_d$ is inversely proportional to the cross-sectional area $A$ . As the radius increases from left to right, the area $A$ increases, and therefore the drift velocity $v_d$ must decrease.
The net charge passing through a cross-section of a conductor is described by the function $q(t) = 5t^3 - 2t + 1$ , where $q$ is in coulombs and $t$ is in seconds. What is the instantaneous electric current in the conductor at $t = 2$ s?
Answer: $58$ A Electric current $I$ is defined as the time rate of change of charge, $I(t) = \frac{dq}{dt}$ . Taking the derivative of the given charge function: $I(t) = \frac{d}{dt}(5t^3 - 2t + 1) = 15t^2 - 2$ . Evaluating this at $t = 2$ s gives $I(2) = 15(2)^2 - 2 = 15(4) - 2 = 60 - 2 = 58$ A.
A steady current of $2.0$ A flows through a copper wire. Copper has approximately $8.5 \times 10^{28}$ free electrons per cubic meter. If the wire has a cross-sectional area of $2.0 \times 10^{-6} \text{ m}^2$ , what is the approximate drift velocity of the electrons?
Answer: $7.4 \times 10^{-5} \text{ m/s}$ The relationship between current $I$ , charge carrier density $n$ , elementary charge $q$ , cross-sectional area $A$ , and drift velocity $v_d$ is $I = nqAv_d$ . Solving for $v_d$ gives $v_d = \frac{I}{nqA}$ . Plugging in the values: $v_d = \frac{2.0 \text{ A}}{(8.5 \times 10^{28} \text{ m}^{-3})(1.6 \times 10^{-19} \text{ C})(2.0 \times 10^{-6} \text{ m}^2)} \approx 7.4 \times 10^{-5} \text{ m/s}$ .
In the absence of an electric field, the free electrons in a metallic conductor move randomly with high speeds due to thermal energy. When an electric field is applied, a net current is produced. Why is the resulting drift speed of the electrons typically very small (e.g., $10^{-4}$ m/s)?
Answer: Electrons frequently collide with the ions of the metallic lattice, which repeatedly resets their net forward motion. While the electric field accelerates the electrons, they constantly collide with the stationary ions in the crystal lattice. Each collision effectively randomizes the electron's velocity, and it must be accelerated again by the field. This start-and-stop motion, averaged over many electrons and collisions, results in a very slow net drift speed in the direction opposite the field.
Two cylindrical wires, A and B, are made of the same material and have the same length. Wire A has radius $r$ , and wire B has radius $2r$ . They carry the same uniform current $I$ . How does the magnitude of the current density $J_A$ in wire A compare to the magnitude of the current density $J_B$ in wire B?
Answer: $J_A = 4J_B$ Current density is defined as $J = I/A$ , where $A$ is the cross-sectional area. The area is given by $A = \pi r^2$ . For wire A, $A_A = \pi r^2$ . For wire B, $A_B = \pi (2r)^2 = 4\pi r^2 = 4A_A$ . Since both wires carry the same current $I$ , we have $J_A = I/A_A$ and $J_B = I/A_B = I/(4A_A)$ . Therefore, $J_A = 4J_B$ .
Based on the text, how does current behave in a series vs. parallel circuit?
Answer: Series: current is the same through each element; Parallel: current divides among branches This question tests AP Physics C: Electricity and Magnetism, specifically understanding electric current behavior in series versus parallel circuit configurations. Electric current follows different rules in series and parallel arrangements, which is fundamental to circuit analysis. In series circuits, components share the same current path, while in parallel circuits, current has multiple paths to follow. Choice B is correct because it accurately states that in series circuits, the same current flows through each element (conservation of charge), while in parallel circuits, the total current divides among the available branches according to their resistances. Choice A is incorrect because it reverses the behaviors, claiming current splits in series circuits, which violates the principle of charge conservation in a single path. To help students: Use water pipe analogies - series is like a single pipe where all water flows through each section, while parallel is like pipe branches where flow divides. Watch for: Students who memorize rules backwards or confuse current behavior with voltage behavior in these configurations.
A particle with charge $q$ and mass $m$ is accelerated from rest through a potential difference $V_0$ . It then enters a uniform wire of length $L$ and cross-sectional area $A$ , becoming one of the charge carriers. This single particle's contribution to current depends primarily on which property derived from its initial acceleration?
Answer: Its average velocity within the wire, the drift velocity. 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.
Wire 1 and Wire 2 are of the same length and cross-sectional area. Wire 1 is made of aluminum (resistivity $\rho_{Al}$ ) and Wire 2 is made of copper (resistivity $\rho_{Cu}$ ), with $\rho_{Al} > \rho_{Cu}$ . If both wires carry the same current, how does the electric field $E_{Al}$ in the aluminum wire compare to the electric field $E_{Cu}$ in the copper wire?
Answer: $E_{Al} > E_{Cu}$ 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 ( $E_{Al} > E_{Cu}$ ).

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