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Unit 13: Electromagnetic Induction AP Physics C: E&M - Calculus-Based Review (13.1 - 13.3) Section 13.1 Magnetic Flux Magnetic flux is defined as the integral of the magnetic field over a surface area: \( \Phi_B = \int \vec{B} \cdot d\vec{A} \). For uniform fields, \( \Phi_B = BA\cos\theta \). Q1: A rectangular loop of dimensions \( a \times b \) is placed at a distance \( d \) from a very long straight wire carrying current \( I \), such that the long side \( b \) is parallel to the wire. What is the magnetic flux through the loop? (A) \( \displaystyle\frac{\mu_0 I b}{2\pi} \ln\left(\displaystyle\frac{d+a}{d}\right) \) (B) \( \displaystyle\frac{\mu_0 I a b}{2\pi d} \) (C) \( \displaystyle\frac{\mu_0 I b}{2\pi d^2} (d+a) \) (D) \( \displaystyle\frac{\mu_0 I}{2\pi} \ln\left(\displaystyle\frac{d}{a}\right) \) Field from wire is \( B = \displaystyle\frac{\mu_0 I}{2\pi r} \). Setup integral: \( \Phi_B = \int_{d}^{d+a} \left(\displaystyle\frac{\mu_0 I}{2\pi r}\right) (b \, dr) \). Integrating gives \( \displaystyle\frac{\mu_0 I b}{2\pi} [\ln r]_{d}^{d+a} = \displaystyle\frac{\mu_0 I b}{2\pi} \ln\left(\displaystyle\frac{d+a}{d}\right) \). Correct Answer: A Q2: A flat loop of area \( A \) rotates in a uniform magnetic field \( B \) with a constant angular velocity \( \omega \). At \( t=0 \), the loop is perpendicular to the field. Which expression describes the flux \( \Phi_B(t) \)? (A) \( BA \sin(\omega t) \) (B) \( BA \cos(\omega t) \) (C) \( BA \omega t \) (D) \( BA \cos^2(\omega t) \) \( \Phi_B = BA\cos\theta \). At \( t=0 \), the angle between the normal and the field is \( 0^\circ \) (since it's perpendicular). Thus \( \theta = \omega t \), giving \( \Phi_B = BA\cos(\omega t) \). Correct Answer: B Q3: A non-uniform magnetic field is given by \( \vec{B} = B_0 \left(\displaystyle\frac{x}{L}\right)^2 \hat{k} \). A square loop of side \( L \) lies in the xy-plane with one corner at the origin. What is the total flux through the loop? (A) \( B_0 L^2 \) (B) \( \displaystyle\frac{1}{2} B_0 L^2 \) (C) \( \displaystyle\frac{1}{3} B_0 L^2 \) (D) \( \displaystyle\frac{1}{4} B_0 L^2 \) \( \Phi_B = \int_{0}^{L} \int_{0}^{L} B_0 \left(\displaystyle\frac{x}{L}\right)^2 dx dy = \displaystyle\frac{B_0 L}{L^2} \int_{0}^{L} x^2 dx = \displaystyle\frac{B_0}{L} \left[\displaystyle\frac{x^3}{3}\right]_0^L = \displaystyle\frac{1}{3} B_0 L^2 \). Correct Answer: C Q4: Gauss's Law for Magnetism states that the net magnetic flux through any closed surface is: (A) Proportional to the enclosed current (B) Equal to \( \mu_0 \times \) enclosed charge (C) Zero (D) Dependent on the shape of the surface Because magnetic monopoles do not exist, magnetic field lines always form closed loops. Therefore, for every line entering a closed surface, one must exit. Net flux is always zero. Correct Answer: C Q5: The SI unit for magnetic flux, the Weber (\( \text{Wb} \)), is dimensionally equivalent to which of the following? (A) \( \text{Tesla} \cdot \text{m}^2 \) (B) \( \text{Volt} \cdot \text{second} \) (C) \( \text{Joule} / \text{Ampere} \) (D) All of the above 1. \( \Phi = BA \Rightarrow \text{T} \cdot \text{m}^2 \). 2. Faraday's Law: \( \varepsilon = d\Phi/dt \Rightarrow \text{Volt} \cdot \text{sec} \). 3. \( U = LI^2 \Rightarrow \text{Henry} \cdot \text{Amp}^2 \); \( \text{Wb} = \text{H} \cdot \text{A} \), and \( \text{H} \cdot \text{A} = \text{J} / \text{A} \). Correct Answer: D Section 13.2 Electromagnetic Induction Faraday's Law states \( \varepsilon = -\displaystyle\frac{d\Phi_B}{dt} \). Lenz's Law dictates that the induced current creates a field opposing the change in flux. Q1: A circular loop of radius \( R \) is in a magnetic field directed into the page. The field magnitude decreases as \( B(t) = B_0 e^{-\alpha t} \). The induced EMF at time \( t \) is: (A) \( \pi R^2 B_0 \alpha e^{-\alpha t} \) (B) \( -\pi R^2 B_0 \alpha e^{-\alpha t} \) (C) \( 2\pi R B_0 e^{-\alpha t} \) (D) \( \pi R^2 B_0 e^{-\alpha t} \) \( \Phi_B = B(t) \pi R^2 = \pi R^2 B_0 e^{-\alpha t} \). \( \varepsilon = -d\Phi_B/dt = -(\pi R^2 B_0)(-\alpha e^{-\alpha t}) = \pi R^2 B_0 \alpha e^{-\alpha t} \). Correct Answer: A Q2: A bar of length \( L \) moves with speed \( v \) on frictionless rails connected by a resistor \( R \), perpendicular to a uniform field \( B \). The magnitude of the induced EMF is: (A) \( BL^2 v \) (B) \( BLv / R \) (C) \( BLv \) (D) \( Bv / L \) Motional EMF \( \varepsilon = \displaystyle\frac{d(BLx)}{dt} = BL \displaystyle\frac{dx}{dt} = BLv \). Correct Answer: C Q3: A copper ring is dropped toward the North pole of a stationary bar magnet. As seen from above, the induced current in the ring is: (A) Clockwise (B) Counter-clockwise (C) Zero (D) Alternating North pole field points Up. Dropping the ring increases Upward flux. Lenz's Law requires an induced Downward field. By RHR, a Downward field is created by a Clockwise current. Correct Answer: A Q4: A metal rod of length \( L \) is rotated with angular velocity \( \omega \) about one end in a uniform field \( B \) (perpendicular to rotation). What is the induced EMF between the ends? (A) \( BL^2 \omega \) (B) \( \displaystyle\frac{1}{2} BL^2 \omega \) (C) \( 2 BL^2 \omega \) (D) \( BL \omega \) \( \varepsilon = \int_0^L v(r) B dr = \int_0^L (\omega r) B dr = B \omega [\displaystyle\frac{r^2}{2}]_0^L = \displaystyle\frac{1}{2} B L^2 \omega \). Correct Answer: B Q5: A square loop is being pulled out of a region of uniform magnetic field at a constant velocity \( v \). If the side length is \( L \), how does the induced EMF \( \varepsilon \) behave until the loop is fully out? (A) It increases linearly (B) It stays constant (C) It decreases exponentially (D) It is zero Flux change rate \( d\Phi/dt = B(dA/dt) = B(L v) \). Since \( B, L, v \) are constant, the EMF is constant until the loop is entirely outside the field. Correct Answer: B Section 13.3 Induced Currents and Magnetic Forces Induced current \( I = \varepsilon / R \). This current experiences a magnetic force \( \vec{F} = I \vec{L} \times \vec{B} \), often acting to oppose the motion (Magnetic Braking). Q1: In a motional EMF setup (rod on rails), what external force \( F_{ext} \) is required to keep the rod moving at constant speed \( v \) through a field \( B \)? (A) \( \displaystyle\frac{B^2 L^2 v}{R} \) (B) \( \displaystyle\frac{BLv}{R} \) (C) \( \displaystyle\frac{B L^2 v^2}{R} \) (D) \( B^2 L v \) \( \varepsilon = BLv \Rightarrow I = BLv/R \). Magnetic force \( F_B = ILB = (BLv/R)LB = \displaystyle\frac{B^2 L^2 v}{R} \). For constant \( v \), \( F_{ext} = F_B \). Correct Answer: A Q2: How does the power dissipated as heat in the resistor compare to the rate of work done by the external force in the previous question? (A) Power dissipated is half the work rate (B) Power dissipated equals the work rate (C) Work rate is twice the power dissipated (D) They are unrelated \( P_{heat} = I^2 R = \left(\displaystyle\frac{BLv}{R}\right)^2 R = \displaystyle\frac{B^2 L^2 v^2}{R} \). \( P_{work} = F_{ext} v = \left(\displaystyle\frac{B^2 L^2 v}{R}\right) v = \displaystyle\frac{B^2 L^2 v^2}{R} \). They are identical. Correct Answer: B Q3: A rectangular loop of resistance \( R \) is pushed into a magnetic field. If the magnetic force opposing it is \( F_B \), how would \( F_B \) change if the resistance of the loop were tripled? (A) \( F_B \) would triple (B) \( F_B \) would stay the same (C) \( F_B \) would be one-third as large (D) \( F_B \) would be nine times smaller \( F_B \propto I \). Since \( I = \varepsilon / R \), if \( R \rightarrow 3R \), then \( I \rightarrow I/3 \). Consequently, the magnetic force \( F_B \) becomes \( F_B / 3 \). Correct Answer: C Q4: Eddy currents are induced in a solid conductor moving through a non-uniform field. These currents lead to: (A) Thermal energy dissipation and a braking force (B) Increase in kinetic energy of the conductor (C) Creation of a permanent magnetic field in the conductor (D) Zero net force due to symmetry Eddy currents circulate to oppose the flux change (Lenz's Law). These currents encounter resistance, generating heat, and interact with the field to create a force opposing motion (braking). Correct Answer: A Q5: A loop of wire is falling under gravity through a region where the magnetic field increases with height. The magnetic force on the loop is directed: (A) Downward (B) Upward (C) Horizontally (D) It depends on the loop's orientation As the loop falls, it moves from a stronger field to a weaker field (flux decreases). Lenz's law creates a force to oppose this change in position, resulting in an Upward magnetic force (Magnetic Braking). Correct Answer: B Induction Formulas Recap Flux\( \Phi_B = \int \vec{B} \cdot d\vec{A} \) Faraday's Law\( \varepsilon = -\displaystyle\frac{d\Phi_B}{dt} \) Motional EMF\( \varepsilon = BLv \) Rotating Rod\( \varepsilon = \displaystyle\frac{1}{2}BL^2\omega \) Magnetic Force\( F = \displaystyle\frac{B^2 L^2 v}{R} \) Braking Power\( P = \displaystyle\frac{B^2 L^2 v^2}{R} \) Unit 13: Inductance & Time-Dependent Circuits AP Physics C: E&M Advanced Calculus-Based Review (13.4 - 13.6) Section 13.4 Inductance Inductance measures an object's opposition to changes in current. Self-induced EMF is given by \( \varepsilon = -L \displaystyle\frac{dI}{dt} \), and the energy stored in a magnetic field is \( U_B = \displaystyle\frac{1}{2}LI^2 \). Q1: A long ideal solenoid has \( N \) turns, length \( \ell \), and cross-sectional area \( A \). If the number of turns is doubled while the length and area remain constant, the self-inductance \( L \) will: (A) Remain the same (B) Double (C) Quadruple (D) Increase by a factor of \( \sqrt{2} \) The inductance of a solenoid is \( L = \mu_0 n^2 A \ell = \displaystyle\frac{\mu_0 N^2 A}{\ell} \). Since \( L \propto N^2 \), doubling the turns results in \( 2^2 = 4 \) times the inductance. Correct Answer: C Q2: An inductor carries a current that is increasing at a constant rate of \( 5.0 \text{ A/s} \). If the induced back EMF is measured to be \( 20 \text{ V} \), what is the inductance of the component? (A) \( 0.25 \text{ H} \) (B) \( 4.0 \text{ H} \) (C) \( 100 \text{ H} \) (D) \( 0.40 \text{ H} \) Using \( |\varepsilon| = L \left| \displaystyle\frac{dI}{dt} \right| \), we have \( 20 = L(5.0) \). Solving for \( L \): \( L = \displaystyle\frac{20}{5.0} = 4.0 \text{ H} \). Correct Answer: B Q3: How much energy is stored in the magnetic field of a \( 200 \text{ mH} \) inductor when the current flowing through it is \( 4.0 \text{ A} \)? (A) \( 0.8 \text{ J} \) (B) \( 1.6 \text{ J} \) (C) \( 3.2 \text{ J} \) (D) \( 6.4 \text{ J} \) \( U_B = \displaystyle\frac{1}{2} L I^2 = \displaystyle\frac{1}{2} (0.200 \text{ H}) (4.0 \text{ A})^2 \). \( U_B = 0.1 \times 16 = 1.6 \text{ J} \). Correct Answer: B Q4: Two coils are placed near each other. A current in Coil 1 changing at \( 2.0 \text{ A/s} \) induces an EMF of \( 10 \text{ V} \) in Coil 2. What is the mutual inductance \( M \) between the coils? (A) \( 0.2 \text{ H} \) (B) \( 5.0 \text{ H} \) (C) \( 20 \text{ H} \) (D) \( 0.5 \text{ H} \) \( \varepsilon_2 = M \left| \displaystyle\frac{dI_1}{dt} \right| \). \( 10 = M(2.0) \Rightarrow M = 5.0 \text{ H} \). Correct Answer: B Q5: The magnetic energy density \( u_B \) inside an ideal solenoid is proportional to which of the following? (A) \( I \) (B) \( I^2 \) (C) \( 1/I \) (D) \( \sqrt{I} \) Energy density is \( u_B = \displaystyle\frac{B^2}{2\mu_0} \). Since \( B = \mu_0 n I \) for a solenoid, \( u_B = \displaystyle\frac{(\mu_0 n I)^2}{2\mu_0} = \displaystyle\frac{\mu_0 n^2 I^2}{2} \). Thus, \( u_B \propto I^2 \). Correct Answer: B Section 13.5 Circuits with Resistors and Inductors (LR Circuits) In an LR circuit, current does not reach its steady-state value instantaneously. The time constant is \( \tau = \displaystyle\frac{L}{R} \). For a charging circuit: \( I(t) = \displaystyle\frac{\varepsilon}{R}(1 - e^{-t/\tau}) \). Q1: A battery \( \varepsilon \), a resistor \( R \), and an inductor \( L \) are connected in series. At the instant the switch is closed (\( t=0 \)), the voltage across the inductor is: (A) \( 0 \) (B) \( \varepsilon \) (C) \( \varepsilon / 2 \) (D) \( I R \) At \( t=0 \), the inductor opposes the change in current perfectly, so \( I = 0 \). According to Kirchhoff's Loop Rule: \( \varepsilon - V_R - V_L = 0 \). Since \( V_R = IR = 0 \), then \( V_L = \varepsilon \). Correct Answer: B Q2: In a series LR circuit, how much time is required for the current to reach approximately \( 63\% \) of its maximum steady-state value? (A) \( L/R \) (B) \( R/L \) (C) \( 0.693 L/R \) (D) \( 5 L/R \) The expression for current is \( I(t) = I_{max}(1 - e^{-t/\tau}) \). At \( t = \tau \), \( I = I_{max}(1 - e^{-1}) \approx 0.632 I_{max} \). The time constant \( \tau = L/R \). Correct Answer: A Q3: An inductor with \( L = 10 \text{ H} \) and a resistor \( R = 5 \text{ \(\Omega\)} \) are connected to a \( 20 \text{ V} \) source. After the switch has been closed for a very long time, what is the steady-state current? (A) \( 0 \text{ A} \) (B) \( 2 \text{ A} \) (C) \( 4 \text{ A} \) (D) \( 100 \text{ A} \) As \( t \rightarrow \infty \), the current becomes constant (\( dI/dt = 0 \)). The inductor acts like an ideal wire with zero resistance. Thus, \( I = \varepsilon/R = 20/5 = 4 \text{ A} \). Correct Answer: C Q4: A charging LR circuit has a time constant \( \tau \). At what time \( t \) is the voltage across the inductor equal to the voltage across the resistor? (A) \( \tau \) (B) \( \tau \ln 2 \) (C) \( \tau / 2 \) (D) \( 2 \tau \) We need \( V_L(t) = V_R(t) \). Since \( V_L + V_R = \varepsilon \), this happens when \( V_L = \varepsilon / 2 \). \( \varepsilon e^{-t/\tau} = \varepsilon / 2 \Rightarrow e^{-t/\tau} = 1/2 \Rightarrow -t/\tau = \ln(1/2) = -\ln 2 \). \( t = \tau \ln 2 \). Correct Answer: B Q5: When the switch is opened in a circuit where a current \( I_0 \) was flowing through an inductor \( L \) and resistor \( R \), the current decays as \( I(t) = I_0 e^{-t/\tau} \). What is the total energy dissipated by the resistor during the entire decay process? (A) \( I_0^2 R \tau \) (B) \( \displaystyle\frac{1}{2} L I_0^2 \) (C) \( L I_0^2 \) (D) \( \varepsilon I_0 \) By conservation of energy, all the energy initially stored in the inductor's magnetic field must be dissipated as heat in the resistor. The initial energy was \( U_B = \displaystyle\frac{1}{2} L I_0^2 \). Correct Answer: B Section 13.6 Circuits with Capacitors and Inductors (LC Circuits) An ideal LC circuit (no resistance) undergoes simple harmonic oscillations of charge and current. The angular frequency is \( \omega = \displaystyle\frac{1}{\sqrt{LC}} \). Q1: In an ideal LC circuit, the total energy of the system oscillates between: (A) The electric field of the capacitor and the kinetic energy of the electrons (B) The electric field of the capacitor and the magnetic field of the inductor (C) Thermal energy and magnetic energy (D) Gravitational potential and electric potential Energy is transferred back and forth between the electric field of the capacitor (\( U_E = \displaystyle\frac{q^2}{2C} \)) and the magnetic field of the inductor (\( U_B = \displaystyle\frac{1}{2}LI^2 \)). Total energy \( U_{tot} = U_E + U_B \) is constant. Correct Answer: B Q2: A capacitor with capacitance \( C \) is charged to a maximum charge \( Q \). It is then connected to an inductor \( L \). What is the maximum current \( I_{max} \) that will flow in the circuit? (A) \( Q \sqrt{LC} \) (B) \( \displaystyle\frac{Q}{\sqrt{LC}} \) (C) \( \displaystyle\frac{Q}{LC} \) (D) \( Q L C \) By energy conservation: \( \displaystyle\frac{Q^2}{2C} = \displaystyle\frac{1}{2} L I_{max}^2 \). \( I_{max}^2 = \displaystyle\frac{Q^2}{LC} \Rightarrow I_{max} = \displaystyle\frac{Q}{\sqrt{LC}} \). This is also \( \omega Q \). Correct Answer: B Q3: If the inductance in an LC circuit is quadrupled (\( 4L \)) and the capacitance is halved (\( C/2 \)), the new period of oscillation \( T' \) compared to the original period \( T \) is: (A) \( T' = T \) (B) \( T' = 2T \) (C) \( T' = T \sqrt{2} \) (D) \( T' = 4T \) \( T = 2\pi\sqrt{LC} \). \( T' = 2\pi\sqrt{(4L)(C/2)} = 2\pi\sqrt{2LC} = \sqrt{2} (2\pi\sqrt{LC}) = \sqrt{2} T \). Correct Answer: C Q4: At what time \( t \) (in terms of the period \( T \)) does the energy in an LC circuit first become equally divided between the inductor and the capacitor? (Assume \( q(0) = Q \)) (A) \( T/4 \) (B) \( T/8 \) (C) \( T/2 \) (D) \( T/16 \) Total energy \( U_{tot} \) is constant. We want \( U_E = U_{tot}/2 \). \( \displaystyle\frac{q^2}{2C} = \displaystyle\frac{1}{2} \left( \displaystyle\frac{Q^2}{2C} \right) \Rightarrow q^2 = \displaystyle\frac{Q^2}{2} \Rightarrow q = \displaystyle\frac{Q}{\sqrt{2}} \). Since \( q(t) = Q \cos(\omega t) \), we have \( \cos(\omega t) = \displaystyle\frac{1}{\sqrt{2}} \Rightarrow \omega t = \pi/4 \). \( \left( \displaystyle\frac{2\pi}{T} \right) t = \pi/4 \Rightarrow t = T/8 \). Correct Answer: B Q5: Which differential equation describes the behavior of charge \( q \) in an ideal LC circuit? (A) \( L \displaystyle\frac{dq}{dt} + \displaystyle\frac{q}{C} = 0 \) (B) \( L \displaystyle\frac{d^2q}{dt^2} + R \displaystyle\frac{dq}{dt} + \displaystyle\frac{q}{C} = 0 \) (C) \( L \displaystyle\frac{d^2q}{dt^2} + \displaystyle\frac{q}{C} = 0 \) (D) \( \displaystyle\frac{d^2q}{dt^2} - \displaystyle\frac{q}{LC} = 0 \) Using Kirchhoff's Loop Rule: \( -L \displaystyle\frac{dI}{dt} - \displaystyle\frac{q}{C} = 0 \). Since \( I = \displaystyle\frac{dq}{dt} \), then \( \displaystyle\frac{dI}{dt} = \displaystyle\frac{d^2q}{dt^2} \). Substituting gives \( L \displaystyle\frac{d^2q}{dt^2} + \displaystyle\frac{q}{C} = 0 \). Correct Answer: C Inductance & Circuits Recap Inductor EMF\( \varepsilon = -L \displaystyle\frac{dI}{dt} \) Magnetic Energy\( U_B = \displaystyle\frac{1}{2} L I^2 \) LR Time Constant\( \tau = \displaystyle\frac{L}{R} \) LR Current (Growth)\( I = I_{max}(1 - e^{-t/\tau}) \) LC Freq (\( \omega \))\( \omega = \displaystyle\frac{1}{\sqrt{LC}} \) LC Energy\( \displaystyle\frac{Q^2}{2C} + \displaystyle\frac{1}{2}LI^2 = Const \) How are the sections on LR and LC circuits looking to you? Would you like to adjust the difficulty of the next set, or perhaps dive deeper into Maxwell's equations and electromagnetic waves?

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