Units & Sign Conventions in E&M - Physics 2

1. Which of the following correctly evaluates the student's cross product, sign handling, and final force direction?
   Answer: The cross product is correct ($\hat{x} \times \hat{z} = -\hat{y}$), the sign handling is correct (negative $q$ reverses direction), and the final result $+0.24\hat{y}\text{ N}$ is correct. The student's full calculation contains no errors. When applying the Lorentz force law $\mathbf{F} = q\mathbf{v} \times \mathbf{B}$, you must handle the cross product and the charge's sign as one unified algebraic calculation — not as separate conceptual "reversals." The student's calculation is actually correct. First, verify the cross product: using the right-hand rule and the cyclic identity $\hat{x} \times \hat{y} = \hat{z}$, we get $\hat{x} \times \hat{z} = -\hat{y}$. This is right. Then, plugging everything in: $\mathbf{F} = (-2 \times 10^{-6})(3 \times 10^5)(0.4)(-\hat{y})$ The scalar part gives $(-2 \times 10^{-6})(3 \times 10^5)(0.4) = -0.24$, and multiplying by $-\hat{y}$ yields $+0.24\hat{y} \text{ N}$. The student's arithmetic is clean, and the explanation — that the negative charge reverses the cross-product direction — is a valid conceptual shortcut, not a double-counted sign. Answer B is correct. Answer A claims the student applied the negative sign twice, but that's not what happened. The single factor of $q = -2\times10^{-6}$ is already included in the numerical product; saying "negative charge reverses direction" is just a verbal restatement of that same mathematics, not an additional operation. Answer C incorrectly states $\hat{x} \times \hat{z} = +\hat{y}$, which reverses the cross product's sign — a right-hand rule error. Answer D also gets the cross product wrong (same error as C) and only accidentally recovers the right final answer through compensating mistakes. Study tip: Always treat $q\mathbf{v}\times\mathbf{B}$ as a single algebraic expression. Don't apply the charge sign separately after deciding a direction — let the math handle it all at once to avoid double-counting errors.
2. Which of the following correctly identifies the flaw in the student's dimensional analysis?
   Answer: The student overlooks that $1 \text{ A} = 1 \text{ C/s}$ by definition, so $\text{A/m}^3 = \text{C/(m}^3\cdot\text{s)}$. Both sides carry identical dimensions, and the continuity equation is dimensionally consistent without any conversion factor. Dimensional analysis in electromagnetism often trips students up when they forget the fundamental definitions linking electrical units. When you see a question like this, your first instinct should be to check whether the units involved have hidden relationships — especially between amperes, coulombs, and seconds. The student's arithmetic is actually correct: $[\nabla \cdot \mathbf{J}] = \text{A/m}^3$ and $[\partial\rho/\partial t] = \text{C/(m}^3\cdot\text{s)}$. The mistake is concluding these differ. By definition, the ampere is exactly one coulomb per second: $1\text{ A} = 1\text{ C/s}$. Substituting this into the left side gives $\text{A/m}^3 = \text{C/(s}\cdot\text{m}^3)$, which is identical to the right side. No conversion factor is needed — the continuity equation is perfectly consistent. Answer C correctly identifies this oversight. A is wrong because the divergence operator does divide by length — applying $\nabla\cdot$ to a quantity in A/m² genuinely yields A/m³. The student's divergence step was fine; the error came afterward. B is a fabricated rule. The $\nabla$ operator involves spatial derivatives (dividing by meters), not temporal ones. There is no principle in Maxwell's equations that makes spatial derivatives "carry seconds." This answer invents a correction that produces a dimensionally inconsistent result anyway. D is wrong because $\rho$ is defined as charge per unit volume, C/m³ — not charge per unit volume per unit time. Redefining it to fix a perceived inconsistency that doesn't actually exist is unnecessary and incorrect. Study tip: Memorize the base-unit definitions of common electrical quantities — especially $1\text{ A} = 1\text{ C/s}$ and $1\text{ V} = 1\text{ J/C}$. Many "inconsistencies" in dimensional checks dissolve the moment you expand derived units into their SI base components.
3. Which of the following correctly evaluates the student's unit analysis and physical interpretation?
   Answer: Both the unit analysis and the interpretation are correct. The units of $k q_1 q_2 / r$ reduce to joules, and a negative potential energy means an attractive configuration requires positive external work equal to $|U|$ to separate the charges. When tackling a question like this, you need to evaluate two separate claims independently: dimensional consistency and physical interpretation of sign. For the unit analysis, check that $k q_1 q_2 / r$ yields joules. Since $k$ carries units of $\text{N}\cdot\text{m}^2/\text{C}^2$, multiplying by two charges in coulombs gives $\text{N}\cdot\text{m}^2$, and dividing by meters gives $\text{N}\cdot\text{m} = \text{J}$. This works regardless of whether the charges are expressed in microcoulombs or any other prefix, as long as you convert to base SI units (C) before computing — which the student implicitly did, since she arrived at a numerically correct result of $-1.12 \text{ J}$. For the physical interpretation, a negative $U$ means the charges are in a bound, attractive configuration (opposite signs attract). To pull them apart to infinity — where $U = 0$ by convention — you must increase the potential energy from $-1.12 \text{ J}$ to $0$. That requires an external agent to do positive work of exactly $|U| = 1.12 \text{ J}$. The student's claim is correct, making A the right answer. B is wrong because it reverses the physics: negative $U$ signals attraction, not repulsion. Opposite charges attract, which is precisely why $U$ is negative here. C is wrong in the opposite direction — it confuses "bound system" with needing to remove energy. Separating bound charges requires energy input, not removal. D is wrong on both counts: unit prefixes like $\mu\text{C}$ are perfectly valid as long as you convert properly, and the sign of $U$ absolutely does determine the direction of work for a two-charge system at a known separation. Study tip: Always evaluate unit analysis and physical interpretation as two separate questions — exams frequently make one correct and one wrong to catch students who conflate them.
4. Which of the following most accurately assesses the student's sign convention, flux calculation, and reasoning about Faraday's law?
   Answer: The flux calculation and the zero-EMF result for constant $B$ are both correct. However, the student's final claim is wrong: the sign of the induced EMF depends on whether $B_0$ increases or decreases, not on the sign of $\Phi_B$ alone. The student's statement is only valid for increasing $B_0$. Whenever you see a question involving Faraday's law and sign conventions, your job is to separately evaluate three things: the flux calculation, the EMF result, and any reasoning about hypothetical future scenarios. Here, the student's flux calculation is legitimate. Choosing $d\mathbf{A} = dA(-\hat{z})$ is a perfectly valid convention — the sign of the area vector is a free choice, and the resulting $\Phi_B = -B_0\pi R^2 = -1.57 \times 10^{-3}$ Wb is internally consistent. Since $B_0$ is constant, $d\Phi_B/dt = 0$, so $\mathcal{E} = 0$ is also correct. The problem lies in the student's final claim. The student asserts that a negative flux automatically produces a positive EMF if the field changes — but that's incomplete. If $B_0$ increases, then $\Phi_B$ becomes more negative, giving $d\Phi_B/dt < 0$, so $\mathcal{E} = -d\Phi_B/dt > 0$. But if $B_0$ decreases, $d\Phi_B/dt > 0$, making $\mathcal{E} < 0$. The sign of the induced EMF depends on the rate of change of flux, not on the sign of $\Phi_B$ alone. That's exactly what answer D captures. Answer A is wrong because it claims the positive EMF result holds "regardless of the rate of change" — this ignores the possibility of a decreasing field. Answer B is wrong because it invents a rule that doesn't exist: there is no requirement that the area vector point in $+\hat{z}$; both orientations are valid. Answer C is wrong because magnetic flux is absolutely allowed to be negative — the sign is physically meaningful and convention-dependent. Your takeaway: never conflate the sign of flux with the sign of EMF. Faraday's law responds to how flux changes, so always ask whether the field is increasing or decreasing before drawing conclusions about the induced EMF's direction.
5. Assuming the traversal and sign convention are applied consistently, which statement best evaluates the student's loop equation, her current result, and her terminal-voltage claim?
   Answer: The loop equation is correct, $I = 1 \text{ A}$ is correct, and the terminal-voltage claim is correct: conventional current entering a resistor from one side means that side is at higher potential, consistent with $V = IR = 8 \text{ V}$. Kirchhoff's Voltage Law (KVL) questions test two things simultaneously: your ability to write a correct loop equation and your understanding of what that equation physically means for individual components. Keep both in mind as you evaluate each part of the student's work. The student's loop equation $\mathcal{E} - IR_1 - IR_2 = 0$ is textbook KVL: traversing clockwise, you gain potential across the EMF source and lose potential across each resistor. Solving gives $I = \frac{12}{4+8} = 1 \text{ A}$, which is correct. Her terminal-voltage claim is also correct — in a resistor, conventional current flows from higher potential to lower potential, meaning the side where current enters is at higher potential. With $I = 1 \text{ A}$ through $R_2 = 8\,\Omega$, the voltage across it is $V = IR_2 = 8\text{ V}$, and the entering (left) terminal is indeed at higher potential. Answer A correctly validates all three parts. Answer B reverses the potential rule for resistors — current flows from high to low potential through a resistor, so the entry side is higher, not the exit side. This is a common misconception worth eliminating now. Answer C misapplies the sign convention for EMF sources. When you traverse a battery from − to + (in the direction the source drives current), you gain potential, so the EMF term is positive. There is no sign error here. Answer D introduces a red herring. The problem explicitly states the source is ideal, meaning zero internal resistance. Terminal voltage of $R_2$ is simply $IR_2$, with no ambiguity. Study tip: Always pair KVL with the physical rule — current in a resistor runs from high to low potential. Memorizing this eliminates the trap in B every time.
6. The Biot–Savart law in SI units is $d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I \, d\mathbf{l} \times \hat{r}}{r^2}$. A student claims that the units of $\mu_0$ must be T·m/A (equivalently H/m) so that the expression yields tesla for $d\mathbf{B}$. A second student claims that the factor $\mu_0/(4\pi)$ taken together has units of T·m/A (not $\mu_0$ alone). Which of the following correctly adjudicates this dispute and performs the necessary dimensional analysis?
   Answer: The first student is correct. Since $4\pi$ is dimensionless, it does not contribute units, so $\mu_0$ itself carries T·m/A. Dimensional analysis confirms: $(\text{T}\cdot\text{m/A})(\text{A})(\text{m})/(\text{m}^2) = \text{T}$. Whenever you encounter a dimensional analysis question involving physical constants, the key principle to remember is that pure numbers are dimensionless — they scale a quantity but never contribute or remove units. This question tests whether you understand that $4\pi$ is simply a dimensionless geometric factor. Because $4\pi$ carries no units, $\mu_0$ and $\mu_0/(4\pi)$ must have identical units. To find them, work backward from the Biot–Savart law. You need $dB$ in tesla (T), and the remaining factors contribute $\text{A} \cdot \text{m} / \text{m}^2 = \text{A/m}$. So $\mu_0$ must supply the missing T·m/A to make the product equal T: $\left(\frac{\text{T}\cdot\text{m}}{\text{A}}\right) \cdot \frac{\text{A} \cdot \text{m}}{\text{m}^2} = \text{T} \checkmark$ This confirms Answer A is correct: $\mu_0$ itself has units of T·m/A, and the dimensional check closes perfectly. Answer B is wrong because it introduces a false claim — that $4\pi$ being in the denominator forces you to treat $\mu_0/(4\pi)$ as an inseparable unit-bearing object. Dimensionless numbers never affect units, period. Answer C is wrong because the cross product $d\mathbf{l} \times \hat{r}$ involves a unit vector $\hat{r}$, which is dimensionless. So the cross product contributes only meters (from $d\mathbf{l}$), not an extra factor — the result is indeed tesla, not Wb/m. Answer D is wrong because T·m/A, N/A², and H/m are all equivalent SI units for $\mu_0$. These are not Gaussian-specific forms; the equivalence holds entirely within SI. Study tip: Always check whether a number in a formula is dimensionless before letting it influence your unit analysis — and verify unit equivalences (T·m/A = N/A² = H/m) by expanding each into base SI units: kg·m·s⁻²·A⁻².
7. Which of the following correctly identifies any errors in the student's integral setup, evaluation, or sign reasoning?
   Answer: The setup, evaluation, and reasoning are all correct. The formula $V(a) - V(b) = -\int_b^a E_s \, ds$ is valid because reversing the integration limits introduces a compensating sign flip, yielding $+\frac{\lambda}{2\pi\epsilon_0}\ln(b/a) > 0$, confirming that $a$ (closer to the line) is at higher potential. When working with potential differences, you need to track two things carefully: the definition of the line integral and what happens algebraically when you flip limits. The fundamental relation is $V(a) - V(b) = -\int_b^a \mathbf{E} \cdot d\mathbf{l}$. Notice the limits run from $b$ to $a$ — this is not a typo. The definition states $V(\mathbf{r}) = -\int_{\text{ref}}^{\mathbf{r}} \mathbf{E} \cdot d\mathbf{l}$, so subtracting gives exactly this form. Plugging in the field, $-\int_b^a \frac{\lambda}{2\pi\epsilon_0 s}\,ds = -\frac{\lambda}{2\pi\epsilon_0}[\ln s]_b^a = -\frac{\lambda}{2\pi\epsilon_0}(\ln a - \ln b) = \frac{\lambda}{2\pi\epsilon_0}\ln\!\left(\frac{b}{a}\right)$. Since $b > a$, this is positive — meaning the closer point $a$ is at higher potential, exactly right for a positive line charge. The student's setup, evaluation, and physical conclusion are all correct. B is the answer. Choice A claims there's a sign error in the limits, but the limits $b \to a$ are correct by definition — no error exists to fix. Choice C misapplies the antiderivative evaluation. $[\ln s]_b^a = \ln a - \ln b$, and after applying the leading negative sign you get $+\ln(b/a)$, not $\ln(a/b)$. The student tracked the signs correctly. Choice D incorrectly "corrects" the limits to $a \to b$, which actually introduces an error and produces the wrong sign, falsely suggesting $V(a) < V(b)$. Study tip: Whenever you flip integration limits, you pick up a minus sign. In the potential difference formula, those two negatives (one from the definition, one from flipping limits) cancel — so trust the algebra and always verify the physical sign against intuition.
8. In the context of a parallel-plate capacitor with plate separation $d$, area $A$, and permittivity $\epsilon_0$, the energy stored is $U = Q^2/(2C)$ where $C = \epsilon_0 A/d$. A student argues: 'Since $U = Q^2 d / (2\epsilon_0 A)$, increasing plate separation $d$ at fixed charge $Q$ increases the stored energy. But energy is conserved, so the extra energy must come from the work done by the electric field as the plates attract each other—meaning the electric force between the plates is repulsive, not attractive.' Which of the following correctly identifies the error in the student's reasoning about signs and energy bookkeeping?
   Answer: The student's energy formula and the conclusion that $U$ increases with $d$ are correct, but the sign reasoning is inverted. If an external agent must do positive work to separate the plates against an attractive force, the energy input from the external agent accounts for the increase in $U$. The electric force is attractive, not repulsive, consistent with opposite charges on the plates. When analyzing energy storage in capacitors, the key skill is tracking where energy comes from and what sign the forces have — two separate questions that must not be conflated. The student gets the math right: substituting $C = \epsilon_0 A/d$ into $U = Q^2/(2C)$ gives $U = Q^2 d/(2\epsilon_0 A)$, which genuinely increases as $d$ increases at fixed $Q$. So far, so good. The fatal mistake is in the energy bookkeeping. When you pull the plates apart against an attractive force, you — the external agent — must do positive work. That work is the source of the additional stored energy. The electric force does negative work (it opposes the separation), which is exactly what you expect from an attractive force. The student confused "the system gains energy" with "the electric field must have done positive work," inverting the sign logic entirely. Answer C correctly pinpoints this: the increase in $U$ is funded by the external agent, and the electric force remains attractive, as it must be between opposite charges. Answer A is wrong on two counts: it claims the student's force conclusion is correct (it isn't — the student said repulsive), and its distinction between "field work" and "external agent work" doesn't address the actual error. Answer B is factually incorrect — $U = Q^2/(2C)$ absolutely increases when $C$ decreases, so the premise that $U$ decreases is simply false. Answer D invents a dimensional error that doesn't exist; the units of $Q^2 d/(2\epsilon_0 A)$ do work out to joules correctly. Study tip: Whenever stored energy increases, always ask who supplied that energy — the external agent or the field — before drawing conclusions about force direction. Getting the sign right requires tracking the full energy budget.
9. Maxwell's displacement current density is defined as $\mathbf{J}_d = \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$. A student checks units: $[\epsilon_0][\partial E/\partial t] = (\text{C}^2/(\text{N}\cdot\text{m}^2))(\text{V/(m}\cdot\text{s)})$. She simplifies $\text{V} = \text{N}\cdot\text{m/C}$ and claims the result is $\text{A/m}^2$, which she identifies as the correct SI unit for current density. Which of the following correctly evaluates her unit analysis?
   Answer: Her analysis is correct. Substituting $\text{V} = \text{N}\cdot\text{m/C}$ gives $\frac{\text{C}^2}{\text{N}\cdot\text{m}^2} \cdot \frac{\text{N}\cdot\text{m}}{\text{C}\cdot\text{m}\cdot\text{s}} = \frac{\text{C}}{\text{m}^2 \cdot \text{s}} = \text{A/m}^2$, which is indeed the SI unit of current density. When checking units in electromagnetism, your goal is to systematically substitute known equivalences and track cancellations — treat it like algebraic simplification where units are variables. The student's approach is sound. Starting with $\frac{\text{C}^2}{\text{N}\cdot\text{m}^2} \cdot \frac{\text{V}}{\text{m}\cdot\text{s}}$, she substitutes $\text{V} = \text{N}\cdot\text{m/C}$, giving $\frac{\text{C}^2}{\text{N}\cdot\text{m}^2} \cdot \frac{\text{N}\cdot\text{m}}{\text{C}\cdot\text{m}\cdot\text{s}}$. Now cancel: one factor of N cancels, one factor of m cancels, and one factor of C cancels, leaving $\frac{\text{C}}{\text{m}^2 \cdot \text{s}}$. Since $\text{C/s} = \text{A}$, the result is $\text{A/m}^2$ — exactly the SI unit for volume current density. Answer D is correct. Answer A claims the Newton units don't fully cancel, but if you carefully count the powers — N¹ in the numerator and N¹ in the denominator — they cancel completely. This is simply an arithmetic error in tracking the algebra. Answer B introduces a false claim that using $[\epsilon_0] = \text{C}^2/(\text{N}\cdot\text{m}^2)$ introduces a factor of $4\pi$ error. This is entirely fabricated — $\text{C}^2/(\text{N}\cdot\text{m}^2)$ is a perfectly valid SI expression for $[\epsilon_0]$, equivalent to F/m. The alternative route through F/m also yields $\text{A/m}^2$, confirming both paths agree. Answer C confuses volume current density ($\text{A/m}^2$) with surface (linear) current density ($\text{A/m}$). Displacement current density is indeed a volume quantity, so $\text{A/m}^2$ is correct. Study tip: When verifying units, write every substitution explicitly and tally exponents for each base unit — most unit errors come from losing track of a single power of a variable.

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