Common E&M Pitfalls - Physics 2

1. A long straight wire carries current I in the +x direction. A student uses the Biot–Savart law to find the magnetic field at a point P located at position $\vec{r} = d\,\hat{y}$ directly above the wire. The student sets up the cross product $d\vec{l} \times \hat{r}$ where $d\vec{l} = dx\,\hat{x}$ and states that $\hat{r}$ points from the source element toward P, i.e., $\hat{r} = +\hat{y}$. The student then evaluates $\hat{x} \times \hat{y} = +\hat{z}$ and concludes the field at P points in the $+\hat{z}$ direction. Which of the following correctly assesses this work?
   Answer: The student's setup and result are both correct. In the Biot–Savart law, $\hat{r}$ points from the source to the field point ($+\hat{y}$ here), the cross product $\hat{x} \times \hat{y} = +\hat{z}$ is correct by the right-hand rule, and wrapping the right hand around the wire confirms the field is in $+\hat{z}$ above the wire. Whenever you apply the Biot–Savart law, the single most important convention to lock in is the direction of $\hat{r}$: it always points from the source element to the field point. With that established, everything else follows from standard vector cross product rules. Here, the wire runs along $\hat{x}$, so $d\vec{l} = dx\,\hat{x}$. Point P sits at $+d\hat{y}$ relative to the source element, so $\hat{r} = +\hat{y}$ — source to field point, exactly as the student stated. The cross product $\hat{x} \times \hat{y}$ in a standard right-handed coordinate system equals $+\hat{z}$ (point fingers along $\hat{x}$, curl toward $\hat{y}$, thumb points in $+\hat{z}$). You can independently verify this with the right-hand rule on the wire: thumb along $+\hat{x}$, fingers curl over the top of the wire in the $+\hat{z}$ direction. Everything checks out — A is correct. Choice B incorrectly claims $\hat{x} \times \hat{y} = -\hat{z}$, which is simply a cross product error. The result is $+\hat{z}$, not $-\hat{z}$. Choice C flips the definition of $\hat{r}$, claiming it points from the field point back to the source. This reverses $\hat{r}$ to $-\hat{y}$ and produces the wrong field direction $-\hat{z}$. Choice D commits both errors simultaneously — wrong $\hat{r}$ direction and a corrected cross product — and accidentally arrives at the right answer through two canceling mistakes. A useful study tip: memorize the Biot–Savart $\hat{r}$ direction with the phrase "source to observer," and always sanity-check your result with the right-hand rule on the wire itself.
2. A uniform magnetic field $\vec{B} = B_0\hat{z}$ exists in a region of space. A rectangular conducting loop lies in the xy-plane. At $t = 0$, the loop begins to rotate about the y-axis with angular velocity $\omega$, so the normal to the loop makes angle $\theta = \omega t$ with the z-axis. A student writes the flux as $\Phi = B_0 A \sin(\omega t)$ and then computes the induced EMF as $\mathcal{E} = -\frac{d\Phi}{dt} = -B_0 A \omega \cos(\omega t)$. A classmate says the sign is wrong and the EMF should be $+B_0 A \omega \cos(\omega t)$. A third student says the flux formula is wrong and should be $\Phi = B_0 A \cos(\omega t)$, giving $\mathcal{E} = +B_0 A \omega \sin(\omega t)$. Which of the following correctly resolves the disagreement?
   Answer: The third student is correct: $\Phi = \vec{B}\cdot\hat{n}\,A = B_0 A\cos(\omega t)$, since the dot product depends on $\cos\theta$, not $\sin\theta$. At $t = 0$ the loop lies in the xy-plane so $\hat{n} = \hat{z}$ and flux is at its maximum — consistent only with the cosine formula. Differentiating gives $\mathcal{E} = B_0 A\omega\sin(\omega t)$. Whenever you see a rotating loop in a magnetic field, your first instinct should be to set up the geometry carefully at $t = 0$ before writing any formula. The magnetic flux is defined as $\Phi = \vec{B} \cdot \hat{n}\, A = B_0 A \cos\theta$, where $\theta$ is the angle between $\vec{B}$ and the loop's normal $\hat{n}$. The third student gets this exactly right, and here's why. At $t = 0$, the loop lies flat in the xy-plane, so $\hat{n} = \hat{z}$, which is perfectly aligned with $\vec{B} = B_0\hat{z}$. That means the flux starts at its maximum value $B_0 A$ — a telltale sign you need a cosine, not a sine. Substituting $\theta = \omega t$ gives $\Phi = B_0 A \cos(\omega t)$, and Faraday's law yields $\mathcal{E} = -\frac{d\Phi}{dt} = B_0 A \omega \sin(\omega t)$. This is answer C. Choice A fails at the very first step: $\vec{B} \cdot \hat{n} = B_0\cos\theta$, never $B_0\sin\theta$. Swapping cosine for sine is a classic geometry error. Choice B compounds this by incorrectly claiming Faraday's law doesn't include the minus sign for mechanically driven loops — the minus sign is universal; it comes from energy conservation and Lenz's law, regardless of what drives the rotation. Choice D tries to rescue the wrong flux formula with a sign trick on $\hat{n}$, but a compensating negative on a fundamentally wrong function still gives you $-\sin(\omega t)$, not $\cos(\omega t)$. Study tip: Always check your flux formula at $t = 0$. If the loop and field are aligned at the start, flux must be maximum — that's your cosine check.
3. An infinite line charge with linear charge density $\lambda > 0$ runs along the z-axis. A student uses Gauss's law with a cylindrical Gaussian surface of radius r and length L to find the electric field, and correctly arrives at $E = \frac{\lambda}{2\pi\epsilon_0 r}$. The student then claims: 'By analogy, for an infinite plane of surface charge density $\sigma$, I can use a cylindrical Gaussian surface of cross-sectional area A, and since the enclosed charge is $\sigma A$, the field is $E = \frac{\sigma}{2\pi\epsilon_0}$, independent of distance.' Which of the following identifies the error in the plane-charge argument?
   Answer: The student incorrectly evaluated the flux through the cylindrical surface. Because the field is perpendicular to the plane, it is parallel to the curved side of the cylinder (zero flux there) and perpendicular to only the two flat end-caps. The total flux is $2EA$, not $2\pi r L \cdot E$, so Gauss's law gives $E = \sigma/(2\epsilon_0)$, not $\sigma/(2\pi\epsilon_0)$. When applying Gauss's law, your choice of Gaussian surface must match the symmetry of the source — and critically, you must correctly evaluate the flux through every face of that surface. This question tests whether you understand how flux is calculated differently for a cylinder enclosing a line charge versus a plane charge. For an infinite plane of charge with density $\sigma$, the electric field points perpendicular to the plane (away from it on both sides). A cylindrical "pillbox" Gaussian surface straddling the plane is actually the right choice — but you must evaluate the flux correctly. The field is parallel to the curved side of the cylinder, contributing zero flux there. Only the two flat end-caps, each with area $A$, have field lines passing through them perpendicularly. The total flux is therefore $\Phi = 2EA$. Setting this equal to $Q_{enc}/\epsilon_0 = \sigma A/\epsilon_0$ gives $E = \sigma/(2\epsilon_0)$ — the correct result. The student's error was implicitly treating the geometry like the line-charge case, using the curved-surface flux $2\pi r L \cdot E$ instead of $2EA$. That's why C is correct. A is wrong on two counts: infinite planes do not have spherical symmetry, and Gauss's law doesn't require spherical surfaces — it requires surfaces matched to the actual symmetry. B repeats this same misconception that planar sources demand spherical surfaces, which is simply false. D is wrong because the full enclosed charge is $\sigma A$ (the Gaussian surface encloses all charge on the portion of the plane inside it); there's no reason to halve it. Study tip: Whenever you use a cylindrical Gaussian surface, always ask which faces contribute flux. If the field is perpendicular to the axis (like for a plane), it's the end-caps — not the curved side.
4. The student reasons: 'The equivalent resistance is $R_{eq} = \frac{R_1 R_2}{R_1+R_2} = 3\,\Omega$. The total current is $I_{tot} = \frac{24}{3} = 8\,\text{A}$. Using the current divider, the current through R₁ is $I_1 = I_{tot}\frac{R_2}{R_1+R_2} = 8 \times \frac{12}{16} = 6\,\text{A}$. Therefore the power in R₁ is $P_1 = I_1^2 R_1 = 36 \times 4 = 144\,\text{W}$.' Which of the following identifies any error in this reasoning?
   Answer: The reasoning contains no errors. The current-divider formula $I_1 = I_{tot}\frac{R_2}{R_1+R_2}$ correctly reflects that the branch with smaller resistance carries more current, yielding $I_1 = 6\,\text{A}$ and $P_1 = 144\,\text{W}$. When analyzing parallel circuits, two key facts anchor everything: every branch shares the same voltage, and the current divider sends more current through the smaller resistor. Keep both in mind as you evaluate the student's work. The student's reasoning is actually flawless. The equivalent resistance $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{48}{16} = 3\,\Omega$ is correct for parallel resistors. The total current $I_{tot} = \frac{24}{3} = 8\,\text{A}$ follows directly. The current divider formula $I_1 = I_{tot}\frac{R_2}{R_1+R_2}$ is also correct — notice it uses the opposite resistor in the numerator, which ensures the smaller branch (R₁ = 4 Ω) receives the larger share of current: $I_1 = 8 \times \frac{12}{16} = 6\,\text{A}$. Finally, $P_1 = I_1^2 R_1 = 36 \times 4 = 144\,\text{W}$. You can verify independently: $P_1 = V^2/R_1 = 576/4 = 144\,\text{W}$. Everything checks out, making B correct. Choice A flips the current divider formula to $\frac{R_1}{R_1+R_2}$, which would send less current to the smaller resistor — the opposite of physical reality. Choice C confuses series and parallel resistance formulas; $R_1 + R_2$ applies to series circuits only. Choice D creates a false rule: both $P = V^2/R$ and $P = I^2R$ are always valid — they're algebraically equivalent given Ohm's Law. Study tip: Always verify your current-divider result by checking that the branch with smaller resistance carries more current, and cross-check power using both $V^2/R$ and $I^2R$ — if they disagree, you've made an error somewhere.
5. The student correctly computes the equivalent capacitance of C₁ and C₂ in series as $C_{12} = \frac{C_1 C_2}{C_1 + C_2}$, then adds C₃ in parallel to get $C_{eq} = C_{12} + C_3$. However, when computing the energy stored only in C₃, the student writes $U_3 = \frac{1}{2}C_3 V_3^2$ and substitutes the voltage across the series branch for $V_3$, arguing that voltage divides across the parallel combination. Which statement best identifies the error and its consequence?
   Answer: The student has confused series and parallel voltage rules: in a parallel combination, every branch shares the same terminal voltage V, so $V_3 = V$, not the voltage across the series branch. Using a reduced voltage underestimates $U_3$. Whenever you see a circuit with parallel branches, the single most important rule to anchor yourself to is this: components in parallel share the same voltage. That one principle is what this question tests. In the described circuit, C₃ is connected directly in parallel with the series combination C₁₂. Both branches connect between the same two nodes, which means both branches see the full applied voltage V. Therefore, $V_3 = V$, and the correct energy stored in C₃ is $U_3 = \frac{1}{2}C_3 V^2$. The student's mistake was treating the parallel combination as if voltage divides across branches — but voltage division is a series-circuit behavior, not a parallel one. By substituting a reduced voltage, the student underestimates $U_3$. This confirms B as the correct answer. A is wrong precisely because it calls the substitution correct. The voltage across C₃ does equal the voltage across C₁₂, but both equal the full applied voltage V — not some divided fraction. The student's error wasn't in comparing the two branches; it was in what value they assigned to that shared voltage. C describes a completely different circuit topology than the one in the passage. The problem explicitly states C₃ is in parallel with the series branch, so charge-division reasoning doesn't apply here. D is wrong because the parallel equivalent-capacitance formula $C_{eq} = C_{12} + C_3$ is exactly right. The reciprocal sum applies to series combinations, not parallel ones — confusing the two is a separate (and common) error. Study tip: Always label your circuit topology first — parallel means same voltage, series means same charge. Getting that straight before plugging into any formula prevents nearly every capacitor-circuit mistake.
6. A student applies Kirchhoff's voltage law (KVL) to a single-loop circuit containing a 12 V battery (positive terminal up) and two resistors $R_1$ and $R_2$ in series. Traversing the loop clockwise — the same direction as the conventional current — the student writes: $+12 - I R_1 - I R_2 = 0$. A classmate objects, saying the battery term should be negative because 'you always lose voltage going through a source.' A second classmate says the resistor drops should be positive because 'current flows through them in the traversal direction, so they add voltage.' Which of the following correctly adjudicates this dispute?
   Answer: The original student is correct. When traversing a battery from − to + terminal (gaining potential), the term is positive; when traversing a resistor in the direction of current flow (losing potential), the term is negative. Both objections are wrong. Whenever you apply Kirchhoff's Voltage Law, the golden rule is simple: the sign of each term depends entirely on what happens to electric potential as you traverse that element, not on some fixed rule about sources vs. resistors. For a battery, ask which terminal you enter first. If you enter the negative terminal and exit the positive terminal, you're walking uphill in potential — that's a gain, so the term is positive ($+12\text{ V}$). For a resistor, ask whether you're traveling in the same direction as conventional current. If yes, you're moving from high to low potential — that's a drop, so the term is negative ($-IR$). The original student's equation $+12 - IR_1 - IR_2 = 0$ follows these rules exactly, confirming that answer A is correct. Answer B is wrong because it applies a blanket rule ("batteries are always negative") that ignores traversal direction. If you traversed the battery from $+$ to $-$, then the term would be negative — the sign is situational, not fixed. Answer C commits the opposite error with resistors: current flowing through a resistor in the traversal direction means the resistor is dissipating energy and lowering potential, not adding it. Writing $+IR$ for those elements is backwards. Answer D combines the correct battery sign with C's resistor mistake, producing an equation that can never be satisfied by a physical circuit ($12 + IR_1 + IR_2 = 0$ would require negative current or resistance). Study tip: Memorize two simple traversal rules — battery $- \to +$ gives $+\varepsilon$; resistor in current's direction gives $-IR$ — and apply them mechanically rather than reasoning from energy concepts, which can mislead you under pressure.
7. A classmate claims the student's method is flawed because 'you can't find current by differentiating the capacitor voltage — current should come from the resistor voltage, not the capacitor voltage.' The student responds that since $I = C(dV_C/dt)$, the method is valid. Which of the following correctly assesses both the method and the result?
   Answer: The student's method is valid and the result is correct. The relation $I = C(dV_C/dt)$ is the fundamental capacitor current–voltage relationship, and differentiating $\mathcal{E}(1-e^{-t/RC})$ correctly yields $I = (\mathcal{E}/R)e^{-t/RC}$, which agrees with $I = (\mathcal{E} - V_C)/R$ at all times. When analyzing RC circuits, remember that there are multiple valid paths to find the current — and they must all agree. The two most natural approaches are Kirchhoff's voltage law ($I = (\mathcal{E} - V_C)/R$) and the capacitor's defining relation ($I = C\,dV_C/dt$). A question like this tests whether you recognize that both are equivalent, not competing. The student's method is sound. Since $V_C(t) = \mathcal{E}(1 - e^{-t/RC})$, differentiating gives $dV_C/dt = \frac{\mathcal{E}}{RC}e^{-t/RC}$. Multiplying by $C$ yields $I = \frac{\mathcal{E}}{R}e^{-t/RC}$. You can verify this independently: $\mathcal{E} - V_C = \mathcal{E}\,e^{-t/RC}$, so $I = (\mathcal{E} - V_C)/R = \frac{\mathcal{E}}{R}e^{-t/RC}$. Both routes give identical results — confirming D is correct. A is wrong because the classmate's objection is based on a false premise. Differentiating $V_C$ is perfectly legitimate via $I = C\,dV_C/dt$, and the units work out correctly: $[C \cdot \mathcal{E}/(RC)] = [F \cdot V / (F \cdot \Omega)] = [\text{A}]$. There is no distortion. B is wrong because $I = C\,dV_C/dt$ applies universally — charging and discharging. There is no sign flip for charging; the convention is consistent as long as you define current as flowing into the positive plate. C is wrong because it incorrectly claims there is a stray factor of $C$ left over. The $C$ in the numerator cancels with the $RC$ in the denominator, leaving $1/R$ — not $1/(RC^2)$. Study tip: Whenever you derive a current two different ways in an RC circuit and they match, you've confirmed your work. Always cross-check $I = C\,dV_C/dt$ against $I = V_R/R$ — if they disagree, something is wrong in your algebra, not in the method.
8. The student claims: 'Inserting the dielectric increases capacitance by a factor of $\kappa$, and since $U = \frac{1}{2}CV^2$, the stored energy increases by a factor of $\kappa$.' A second student counters: 'The energy actually decreases after the dielectric is inserted.' Which of the following best resolves this dispute, and what is the correct factor by which the energy changes?
   Answer: The first student's formula is right but the wrong version was used. With constant charge, the energy is $U = Q^2/(2C)$; as C increases by $\kappa$, the energy decreases by a factor of $\kappa$. The second student is correct, and the energy becomes $U_0/\kappa$. Whenever a capacitor problem involves disconnecting the battery before making a change, your first instinct should be: charge Q is conserved, not voltage. This distinction completely changes which energy formula applies. When the battery is disconnected, the charge Q on the plates cannot change — there's nowhere for it to go. Inserting the dielectric increases capacitance to $\kappa C_0$, but since $Q = C_0 V_0$ is fixed, the new voltage drops to $V_0/\kappa$. Now, rather than using $U = \frac{1}{2}CV^2$ (which requires knowing the new voltage), use the constant-charge form: $U = \frac{Q^2}{2C}$. Since Q is unchanged and C increases by $\kappa$, the new energy is $U = \frac{Q^2}{2\kappa C_0} = \frac{U_0}{\kappa}$. The energy decreases by a factor of $\kappa$, confirming D is correct. (The "lost" energy is absorbed by the dielectric as it's pulled into the field — the electric force does work on the slab.) A is wrong because it claims both capacitance and voltage increase, but voltage actually falls when charge is fixed and capacitance rises. B incorrectly states that voltage stays at $V_0$ after disconnection — voltage changes precisely because Q is constant while C changes. C gets the direction right (energy decreases) but the factor wrong; claiming a $\kappa^2$ decrease conflates this situation with a different scenario. Study tip: Memorize both energy formulas — $U = \frac{1}{2}CV^2$ (use when V is constant, i.e., battery connected) and $U = \frac{Q^2}{2C}$ (use when Q is constant, i.e., battery disconnected). Choosing the right one based on what's held fixed is the entire game in these problems.

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