Multi-Step E&M Problems - Physics 2

1. What is the amplitude of the current induced in the small loop, and in what direction does it flow relative to the solenoid current at the moment when $I(t)$ is at its maximum positive value?
   Answer: Amplitude $\frac{\mu_0 n \omega I_0 \pi r^2}{\mathcal{R}}$; at the moment $I$ is maximum, the induced current is zero because the EMF is proportional to $\cos(\omega t)$, which equals zero when $\sin(\omega t) = 1$. Whenever you see a question about electromagnetic induction with a time-varying current, your first instinct should be Faraday's Law: the induced EMF depends on the rate of change of flux, not the flux itself. Here's the full reasoning. The magnetic field inside the solenoid is $B(t) = \mu_0 n I_0 \sin(\omega t)$. Since the small loop (radius $r < R$) lies entirely inside the solenoid, the flux through it uses the loop's own area: $\Phi = \mu_0 n I_0 \sin(\omega t) \cdot \pi r^2$. Applying Faraday's Law, the induced EMF is $\mathcal{E} = -\frac{d\Phi}{dt} = -\mu_0 n I_0 \omega \pi r^2 \cos(\omega t)$. The amplitude is therefore $\frac{\mu_0 n \omega I_0 \pi r^2}{\mathcal{R}}$ after dividing by resistance. Crucially, at the moment $I$ is maximum, $\sin(\omega t) = 1$, which means $\cos(\omega t) = 0$. The EMF — and thus the induced current — is zero at that instant. This confirms D. Choice A is wrong on both counts: it uses $R$ (solenoid radius) instead of $r$, and incorrectly claims the induced current is maximum and co-directional when $I$ peaks. Choice B uses the correct area $\pi r^2$ but makes the same timing error — claiming EMF is maximum when $I$ is maximum, which confuses flux with its derivative. Choice C gets the timing right (induced current is zero when $\sin(\omega t) = 1$) but uses the wrong area $\pi R^2$ instead of $\pi r^2$. Study tip: Always distinguish between the area that captures flux (the loop's own area when fully inside the solenoid) and the solenoid's cross-section. And remember: EMF peaks when current crosses zero, not when it peaks.
2. For the proton to travel through this region in a straight line at constant speed, what condition must hold, and which field (if either) does net work on the proton during this straight-line motion?
   Answer: The condition $v_0 = E_0/B_0$ must hold so that the electric and magnetic forces cancel; neither field does work on the proton because the electric force is perpendicular to the proton's displacement and the magnetic force is always perpendicular to the velocity. When a charged particle moves through crossed electric and magnetic fields, you're dealing with a velocity selector — a classic setup where force balance determines straight-line motion. For a proton moving in $\hat{x}$ with $\mathbf{E} = E_0\hat{y}$ and $\mathbf{B} = B_0\hat{z}$, the electric force is $qE_0\hat{y}$ and the magnetic force is $q\mathbf{v}\times\mathbf{B} = qv_0B_0(\hat{x}\times\hat{z}) = -qv_0B_0\hat{y}$. For these to cancel, you need $qE_0 = qv_0B_0$, giving the selector condition $v_0 = E_0/B_0$. So far, every answer agrees on this part — the real trap is the work question. B is correct because once you identify that the net force is zero and the proton travels in a straight line along $\hat{x}$, neither field does work. The electric force points in $\hat{y}$, which is perpendicular to the displacement $\Delta x\,\hat{x}$, so $W_E = \mathbf{F}_E \cdot \Delta\mathbf{x} = 0$. The magnetic force is always perpendicular to velocity by definition, so $W_B = 0$ always. Net work is zero and kinetic energy is conserved — not because two nonzero works cancel, but because both are independently zero. A is tempting but wrong: it claims the electric field does positive work while the magnetic field does negative work. Magnetic forces never do work — this is a fundamental fact, not a coincidence of cancellation. C incorrectly computes $W_E = qE_0\Delta x$, forgetting that the electric force is in $\hat{y}$ while displacement is in $\hat{x}$ — they're perpendicular, so the dot product is zero. D inverts the selector condition to $B_0/E_0$ and falsely claims the magnetic force has a component along velocity, which violates the cross-product geometry. Your study tip: always ask "what direction is the force, and what direction is the displacement?" Work requires a parallel component — perpendicularity always means zero work, regardless of force magnitude.
3. At this driving frequency $\omega = 2\omega_0$, which of the following correctly characterizes the phase relationship between the current and the voltage source, and the impedance of the circuit?
   Answer: The current lags the voltage because the inductive reactance exceeds the capacitive reactance at $\omega > \omega_0$; the impedance is $Z = \sqrt{R^2 + \left(\frac{3\omega_0 L}{2}\right)^2}$, since $X_L - X_C = 2\omega_0 L - \dfrac{\omega_0 L}{2} = \dfrac{3\omega_0 L}{2}$. When analyzing an RLC series circuit driven off-resonance, your goal is to compare inductive and capacitive reactances at the given frequency, then calculate impedance from their net difference. At resonance, $\omega_0 = 1/\sqrt{LC}$, which gives us a useful identity: $\omega_0^2 LC = 1$, so $\omega_0 L = 1/(\omega_0 C)$. When the driving frequency is $\omega = 2\omega_0$, the reactances become $X_L = \omega L = 2\omega_0 L$ and $X_C = 1/(\omega C) = 1/(2\omega_0 C) = \omega_0 L/2$. Since $X_L > X_C$, the circuit is net inductive — meaning the current lags the voltage. The net reactance is $X_L - X_C = 2\omega_0 L - \frac{\omega_0 L}{2} = \frac{3\omega_0 L}{2}$, giving impedance $Z = \sqrt{R^2 + \left(\frac{3\omega_0 L}{2}\right)^2}$. This confirms answer A is correct. Answer B makes a critical error: it claims the capacitor dominates above resonance, but the opposite is true — inductive reactance grows with frequency while capacitive reactance shrinks. The impedance formula in B happens to be correct, but the phase reasoning is backwards. Answer C correctly identifies the lag but drops the capacitive term entirely, which is only valid at extremely high frequencies. At $\omega = 2\omega_0$, $X_C$ is still significant and cannot be ignored — omitting it gives the wrong impedance. Answer D confuses harmonics with resonance. True resonance only occurs at $\omega_0$; driving at integer multiples does not reproduce the resonance condition. Study tip: Always calculate both reactances explicitly before concluding which dominates — never assume based on frequency alone.
4. After the second resistor is connected in parallel, compared to the original single-load configuration, how do the RMS secondary voltage and the RMS primary current change?
   Answer: The secondary voltage remains $3V_p$ because it is set by the turns ratio regardless of load; the primary current doubles because power delivered to the secondary doubles when the load resistance is halved, and $P = V_p I_p$. Whenever you see a transformer question, anchor yourself to two governing equations: the voltage ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p}$ and the power conservation condition $P_p = P_s$, which gives $V_p I_p = V_s I_s$. For an ideal transformer, the secondary voltage is determined entirely by the turns ratio and the primary voltage — it does not depend on what load is connected. With $N_s = 3N_p$, the secondary voltage is fixed at $V_s = 3V_p$ regardless of whether one or two resistors are connected. When you add a second $R_L$ in parallel, the equivalent load resistance drops from $R_L$ to $R_L/2$. Since $V_s$ is unchanged, the secondary current doubles: $I_s = V_s / R_{eq}$ goes from $3V_p/R_L$ to $6V_p/R_L$. The power delivered to the secondary therefore doubles. By conservation of energy, the primary must supply twice the power, and since $V_p$ is fixed by the AC source, $I_p$ must double. This confirms C is correct. A is wrong because it concludes the primary current is unchanged — that would violate energy conservation, since more power is being delivered to the load. B is wrong on two counts: secondary voltage in an ideal transformer doesn't drop with increased load (that only happens in real transformers with winding resistance), and the $\sqrt{2}$ factor has no physical basis here. D is tempting but contains a key error: secondary current doubles (not by $\sqrt{2}$) because the resistance is halved, not shared equally in some root-mean sense. Study tip: On transformer problems, always ask two questions in order — (1) what sets the voltage? (turns ratio), then (2) what sets the current? (the load and power conservation). Never let the load affect the voltage in an ideal transformer.
5. What is the surface charge density on the outer surface of the shell, and what is the direction of the electric field in the region $r > b$?
   Answer: Outer surface charge density $\sigma_{outer} = \frac{-2Q}{4\pi b^2}$; field points radially inward (toward the center) for all $r > b$, because the net charge enclosed is negative. Whenever you see a conducting shell with charges, your first move should be Gauss's Law combined with the rule that the electric field inside a conductor is zero. That zero-field condition is what forces charge to redistribute on the inner and outer surfaces. Here's the logic chain: The point charge $+Q$ at the center induces $-Q$ on the inner surface (at radius $a$) to cancel the field inside the conducting material. Since the shell carries a net charge of $-3Q$ total, and $-Q$ is already "used up" on the inner surface, the remaining $-3Q - (-Q) = -2Q$ must reside on the outer surface. This gives an outer surface charge density of $\sigma_{outer} = \frac{-2Q}{4\pi b^2}$. Now apply Gauss's Law for $r > b$: the total enclosed charge is $+Q + (-Q) + (-2Q) = -2Q$, which is negative, so the electric field points radially inward. That's answer A. Choice B is tempting but wrong — it ignores the inner surface induction entirely and treats the full $-3Q$ as sitting on the outer surface, violating charge conservation on the shell. Choice C correctly identifies the outer surface charge density but gets the field direction wrong; a net enclosed charge of $-2Q$ absolutely produces an inward-pointing field — the "central charge dominates" reasoning is fabricated and has no physical basis. Choice D reverses the sign of the outer charge; the outer surface holds $-2Q$, not $+2Q$. Study tip: Always track charge in two steps — inner surface induction first ($-Q_{center}$), then subtract from the shell's net charge to find the outer surface charge. This two-step method prevents all the errors seen in B, C, and D.
6. As the loop is pulled away from the wire at constant velocity, which of the following correctly describes the induced EMF and the direction of the induced current in the loop?
   Answer: EMF: $\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi} \cdot \frac{w}{d(d+w)}$; current flows counterclockwise (when viewed from $+\hat{y}$) to oppose the decreasing flux through the loop. Whenever you see a loop moving near a current-carrying wire, your two jobs are: (1) calculate the EMF using Faraday's law, and (2) determine current direction using Lenz's law. Calculating the EMF. The magnetic field from the wire at distance $x$ is $B(x) = \frac{\mu_0 I_1}{2\pi x}$, directed in $-\hat{y}$ inside the loop (by the right-hand rule, since current flows in $+\hat{z}$). The flux through the loop is $\Phi = \int_d^{d+w} \frac{\mu_0 I_1 h}{2\pi x}\,dx = \frac{\mu_0 I_1 h}{2\pi}\ln\!\left(\frac{d+w}{d}\right)$. Taking the time derivative and using $\dot{d} = v$, the induced EMF becomes $\mathcal{E} = \frac{\mu_0 I_1 h v}{2\pi}\cdot\frac{w}{d(d+w)}$, which matches answer C. Applying Lenz's law. As the loop moves away, $\Phi$ (in $-\hat{y}$) decreases. To oppose this decrease, the induced current must create flux in $-\hat{y}$ inside the loop — meaning the current flows counterclockwise when viewed from $+\hat{y}$. This confirms C. Why the others fail. Answer A gets the EMF formula right but incorrectly states the current is clockwise — clockwise would reinforce the decrease, violating Lenz's law. Answers B and D both use the wrong EMF formula $\frac{1}{d}$, which would only be valid if the loop were infinitesimally thin; integrating across the loop's width $w$ is essential and produces the $\frac{w}{d(d+w)}$ factor. D also gets the current direction wrong. Study tip: Always integrate $B$ across the loop's width — never just evaluate it at the near edge. And remember: Lenz's law opposes change in flux, not motion directly.
7. Immediately after the short circuit is applied ($t = 0^+$), what is the current through the inductor, and what is the voltage across the remaining (non-shorted) resistor?
   Answer: Current: $\mathcal{E}/(2R)$; voltage: $\mathcal{E}/2$. The inductor maintains the pre-short-circuit current, and with only one resistor now in the loop, the full inductor current flows through it, giving voltage $I_0 R = \mathcal{E}/2$. When a circuit contains an inductor, your first instinct should be to apply the inductor's fundamental constraint: current through an inductor cannot change instantaneously. This single rule unlocks the entire problem. Before the short circuit, both resistors are in series with the battery, so the steady-state current is $I_0 = \mathcal{E}/(2R)$. At $t = 0^+$, the inductor "remembers" this current — it immediately freezes at $\mathcal{E}/(2R)$ regardless of what the rest of the circuit is doing. That confirms the first part of answer A. Now apply Kirchhoff's Voltage Law around the new loop (battery, inductor, and the one surviving resistor — the shorted resistor contributes zero voltage). With current $\mathcal{E}/(2R)$ flowing through resistance $R$, the voltage across the remaining resistor is simply: $V_R = I_0 \cdot R = \frac{\mathcal{E}}{2R} \cdot R = \frac{\mathcal{E}}{2}$ This confirms A as correct. B is wrong because it violates the inductor's core constraint — current cannot jump instantaneously to a new steady state. C is tempting but contains a subtle error: if the current is $\mathcal{E}/(2R)$ and the resistor is $R$, Ohm's Law gives $\mathcal{E}/2$, not $\mathcal{E}$. Claiming the full EMF appears across the resistor would require a different (larger) current. D is wrong because the inductor constraining current doesn't mean voltage across resistors drops to zero — Ohm's Law still applies to resistors at every instant. Study tip: In inductor switching problems, always find the pre-switch current first, freeze it at $t = 0^+$, then redraw the circuit and apply KVL/Ohm's Law normally. Never let the new topology change your current instantaneously.
8. Immediately after release, what is the net force on the dipole and what is the net torque about the dipole's center?
   Answer: Net force: zero; net torque: $\tau = pE_0$, directed so as to rotate $\mathbf{p}$ toward alignment with $\mathbf{E}$, because a uniform field exerts no net translational force on a dipole but does exert a torque $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$. When you encounter a dipole in an electric field, train yourself to ask two separate questions: Is the field uniform? and What is the angle between p and E? These two facts determine everything. In a uniform electric field, the force on $+q$ is $+qE_0\hat{z}$ and on $-q$ is $-qE_0\hat{z}$. These are equal and opposite, so they cancel perfectly — the net force is always zero for a dipole in a uniform field, regardless of orientation. For torque, you use $\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$. Here $\mathbf{p} = qd\,\hat{x}$ and $\mathbf{E} = E_0\hat{z}$, so $|\boldsymbol{\tau}| = pE_0\sin\theta$. With $\theta = 90°$ between $\hat{x}$ and $\hat{z}$, you get $|\boldsymbol{\tau}| = pE_0$ — maximum torque, directed to rotate p toward alignment with E. That makes B correct. A is wrong because it invents a net translational force. The equal-and-opposite forces on the two charges cancel — they cannot "add up" to push the dipole anywhere. C contains a critical math error: $\sin(90°) = 1$, not zero. The torque is actually maximum when p ⊥ E, not zero. Students sometimes confuse this with potential energy ($U = -\mathbf{p}\cdot\mathbf{E}$), which is zero at 90°. D double-counts the forces (the two $qE_0$ forces oppose each other, not add) while accidentally getting the torque magnitude right for the wrong reasons. Your study tip: memorize the two key results — uniform field means zero net force always, and torque is maximum (not zero) when the dipole is perpendicular to the field.
9. At a radial distance $r$ such that $a < r < b$, what is the magnitude of the magnetic field, and how does the energy stored per unit length in the magnetic field in this region scale with $I$?
   Answer: $B = \frac{\mu_0 I}{2\pi r}$; the energy per unit length scales as $I^2$ because $u = B^2/(2\mu_0)$ and $B \propto I$, so integrating over the annular region gives energy per unit length $\propto I^2 \ln(b/a)$. Whenever you see a question combining Ampère's Law with energy storage, treat them as two separate sub-problems: first find $B$, then apply the correct energy density formula. For the region $a < r < b$, draw an Amperian loop of radius $r$. The only current enclosed is the full inner current $I$ — the outer shell at radius $b$ lies entirely outside your loop and contributes nothing. Ampère's Law gives: $B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$ Now for energy: the magnetic energy density is $u = \frac{B^2}{2\mu_0}$, so integrating over the annular cross-section yields energy per unit length: $\frac{U}{\ell} = \int_a^b \frac{B^2}{2\mu_0}(2\pi r)\,dr = \frac{\mu_0 I^2}{4\pi}\ln\!\left(\frac{b}{a}\right)$ This scales as $I^2$, confirming D is correct. A is wrong on two counts: it misapplies "Gauss's law for magnetism" (which says no magnetic monopoles, not that fields cancel) and incorrectly treats the geometry like the exterior of a coaxial cable, where both currents are enclosed. Outside $r > b$, the fields cancel — not between the conductors. B uses the field formula for inside the solid conductor ($r < a$), where current is distributed over the cross-section. Between the conductors, all of $I$ is enclosed, so $B = \mu_0 I / 2\pi r$, not $\mu_0 I r / 2\pi a^2$. C gets $B$ right but claims energy scales as $I$ rather than $I^2$. Energy density goes as $B^2$, never $B$ alone — this is a classic trap. Study tip: Always ask "what current does my Amperian loop enclose?" separately from "what is the energy density formula?" — these are two independent steps that must both be correct.
10. After the dielectric is inserted, which of the following correctly describes the changes in both the electric field between the plates and the energy stored in the capacitor?
    Answer: The electric field decreases by a factor of $\kappa$ because the bound surface charges on the dielectric partially cancel the free charges, and the stored energy decreases by a factor of $\kappa$ because $U = Q^2/(2C)$ and $C$ increases while $Q$ remains fixed. When a capacitor is disconnected from its battery before a dielectric is inserted, the key constraint is that the charge $Q$ on the plates is fixed — no charge can flow on or off. This single fact determines everything else. Here's the reasoning: the dielectric becomes polarized, creating bound surface charges that oppose the free charges on the plates. By Gauss's law applied to the full field (including polarization), the net field inside decreases: $E = E_0/\kappa$, where $E_0 = \mathcal{E}/d$ was the original field. Meanwhile, inserting the dielectric increases the capacitance to $C' = \kappa C_0$. Since $Q$ is fixed, the stored energy becomes $U' = Q^2/(2C') = Q^2/(2\kappa C_0) = U_0/\kappa$. Both quantities — field and energy — decrease by the same factor $\kappa$. This is exactly what answer A describes, making it correct. Answer B is wrong on both counts. Gauss's law with only free charges gives $D$, not $E$ — the bound charges absolutely affect $E$. Energy does not increase; it decreases. Answer C correctly identifies the field reduction but gets the energy direction backwards. $U = Q^2/(2C)$ decreases as $C$ increases — you're dividing by a larger number, not multiplying. Many students confuse this with the constant-voltage case ($U = \frac{1}{2}C V^2$), where energy would increase. Answer D is a fabricated-sounding statement with no physical basis; energy is not conserved within the capacitor alone here — it's transferred to mechanical work pulling the dielectric in. Study tip: Always identify whether the battery is connected (voltage fixed) or disconnected (charge fixed) before analyzing a dielectric insertion — the two scenarios give opposite energy trends.

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