Magnetic Flux - Physics 2

1. What is the magnetic flux through the square loop?
   Answer: $\Phi_B = \tfrac{1}{2}B_0 a^3 = 3.125 \times 10^{-2} \text{ Wb}$, because the loop lies in the $xz$-plane with area normal $\hat{y}$; the relevant field component is $B_y = B_0 x$, and integrating gives $\int_0^a\!\int_0^a B_0 x\,dx\,dz = B_0 a \cdot \tfrac{a^2}{2} = \tfrac{1}{2}B_0 a^3$. When calculating magnetic flux, the formula $\Phi_B = \iint \vec{B} \cdot d\vec{A}$ requires you to identify the loop's area normal vector first, then extract only the field component aligned with that normal. The loop lies in the $xz$-plane, so its area element is $d\vec{A} = dx\,dz\,\hat{y}$. This means only $B_y$ contributes to the flux. On the loop, $y = 0$ everywhere, so $B_y = B_0 x$ — which varies with $x$ but not $y$. You must integrate this over the surface: $\Phi_B = \int_0^a\int_0^a B_0\, x\,dx\,dz = B_0 \int_0^a dz \int_0^a x\,dx = B_0 \cdot a \cdot \frac{a^2}{2} = \frac{1}{2}B_0 a^3$ Plugging in: $\frac{1}{2}(4.0)(0.25)^3 = 3.125 \times 10^{-2}\ \text{Wb}$. That's answer C. Answer A makes a correct setup but botches the integral — it writes $B_0 \int_0^a x\,dx \cdot a = B_0 a^2 \cdot a$, incorrectly claiming $\int_0^a x\,dx = a^2$ instead of the correct $a^2/2$. That's an off-by-two error. Answer B confuses which component matters. Yes, $B_x = 0$ on the loop, but $B_x$ is irrelevant — it's perpendicular to $\hat{y}$. The flux comes from $B_y$, which is nonzero. Answer D uses a shortcut that only works for uniform fields: evaluating $B_y$ at the loop's center and multiplying by area. Since $B_y = B_0 x$ varies across the loop, you must integrate. Study tip: Always set up the dot product $\vec{B} \cdot \hat{n}$ before doing anything else — it immediately tells you which field component to integrate and prevents both the B trap and the A trap.
2. What is the magnetic flux through the square loop due to the solenoid's field?
   Answer: $\Phi_B = \mu_0 n I \cdot \pi R_s^2 = 3.95 \times 10^{-5} \text{ Wb}$, because the solenoid's field is confined to the solenoid's cross-sectional area $\pi R_s^2$, and is zero outside; only that area contributes to the flux through the larger square loop. Whenever you see a magnetic flux problem involving a solenoid and a larger loop, the critical question isn't "what is the field?" — it's "where does the field exist?" A solenoid produces a strong, uniform field inside its bore ($B = \mu_0 n I$) and essentially zero field outside. This confinement is the key insight the question is testing. Because the field only exists within the solenoid's circular cross-section, only that area contributes to the flux integral $\Phi_B = \int \vec{B} \cdot d\vec{A}$. Even though the square loop is much larger, the field is zero everywhere outside the solenoid, so those regions contribute nothing. The correct flux is: $\Phi_B = \mu_0 n I \cdot \pi R_s^2 = (4\pi \times 10^{-7})(1000)(2.0)\cdot \pi(0.05)^2 \approx 3.95 \times 10^{-5} \text{ Wb}$ This confirms B is correct. A is the most tempting distractor — it uses $L^2$ as if the solenoid's uniform field fills the entire square loop. This would only be valid if the field extended throughout the loop's area, which it does not. C inverts the physics entirely, claiming the field is zero inside the solenoid and nonzero outside — the exact opposite of reality. D invents a "geometric dilution" ratio $\pi R_s^2 / L^2$ that has no physical basis. Field lines don't spread or dilute just because the surrounding loop is larger. Study tip: Always ask yourself where the field is nonzero before setting up a flux integral — the effective area is determined by the field's spatial extent, not the loop's physical size.
3. What is the magnetic flux through the curved hemispherical surface alone (not the flat circular cap)?
   Answer: $\Phi_{\text{curved}} = +B_0 \pi R^2 = +2.12 \times 10^{-1} \text{ Wb}$, because only the projection of the curved surface onto the $xy$-plane matters, and that projection has area $\pi R^2$; the outward normals on the curved surface have a net $+z$ component integrating to $\pi R^2$. Whenever you encounter magnetic flux through a curved surface, your first instinct should be to reach for Gauss's Law for magnetism: $\oint \vec{B} \cdot d\vec{A} = 0$. This law states that the total magnetic flux through any closed surface is zero — no magnetic monopoles exist. Here's how to apply that insight cleverly. The hemisphere and its flat circular cap together form a closed surface. Therefore: $\Phi_{\text{curved}} + \Phi_{\text{cap}} = 0$ The flat cap lies in the $xy$-plane. Its outward normal points in the $-\hat{z}$ direction (downward, away from the enclosed volume), so: $\Phi_{\text{cap}} = \vec{B} \cdot (-\hat{z}) \cdot \pi R^2 = -B_0 \pi R^2$ Substituting back: $\Phi_{\text{curved}} = +B_0 \pi R^2 = (3.0)(π)(0.15)^2 \approx +2.12 \times 10^{-1} \text{ Wb}$. This confirms B is correct. Physically, only the projection of the curved surface onto the plane perpendicular to $\vec{B}$ — which is simply $\pi R^2$ — determines the net flux. A is wrong because using the full curved surface area $2\pi R^2$ ignores that outward normals on the dome point radially outward, not uniformly against $\hat{z}$ — most normals have only a partial $z$-component, and integration correctly yields $\pi R^2$, not $2\pi R^2$. The sign is also incorrect. C makes the opposite sign error and incorrectly assumes $\vec{B}$ is parallel to every outward normal on the dome. D confuses left-right symmetry with cancellation — the $z$-components of all outward normals on the dome point upward, so they add constructively, not destructively. Your go-to strategy: when flux through an open curved surface looks complicated, close the surface using Gauss's Law for magnetism and compute flux through the simpler flat piece instead. This shortcut appears frequently on Physics 2 exams.
4. What is the magnetic flux through the circular loop?
   Answer: $\Phi_B = B_0 \pi R^2 = 6.28 \times 10^{-2} \text{ Wb}$, because the $\alpha x$ term is antisymmetric about the $y$-axis and its integral over the symmetric circular area vanishes, leaving only the uniform part. When you encounter magnetic flux through a loop in a non-uniform field, your first instinct should be to set up the integral $\Phi_B = \iint \vec{B} \cdot d\vec{A}$ and carefully examine the symmetry of the integrand before calculating anything. Here, $\vec{B} = B_0(1 + \alpha x)\hat{z}$, so the flux becomes $\Phi_B = \iint B_0(1 + \alpha x)\, dA = B_0\iint dA + B_0\alpha\iint x\, dA$. The first integral is simply $B_0 \pi R^2$. For the second, ask yourself: what is $\iint x\, dA$ over a disk centered at the origin? The function $x$ is antisymmetric about the $y$-axis — for every point $(x, y)$ in the disk with value $+x$, there is a mirror point $(-x, y)$ with value $-x$. These cancel exactly, so $\iint x\, dA = 0$. Therefore $\Phi_B = B_0\pi R^2 = (0.50)(\pi)(0.04) \approx 6.28 \times 10^{-2}\ \text{Wb}$, confirming answer A. Answer B is physically unmotivated — evaluating the field at the rim and multiplying by total area has no mathematical justification; flux requires integration, not point-evaluation. Answer C is the most tempting trap. The average value of $x$ over a circle (or disk centered at the origin) is zero, not $R/2$. The mean value $R/2$ might arise in other geometries, but symmetry kills it here. Answer D invents a dimensional inconsistency — multiplying $\alpha$ (units of m⁻¹) by area (m²) produces meters, not a dimensionless correction, and has no physical basis. Strategy tip: Whenever a field contains an odd function of position (like $x$ or $y$) integrated over a symmetric region centered at the origin, check antisymmetry first — it often eliminates an entire term before you do any real calculation.
5. Which expression correctly gives the magnetic flux through the loop?
   Answer: $\Phi_B = (2.0)(0.10)(0.20)\cos 30° = 3.46 \times 10^{-3} \text{ Wb}$, because the flux is $BA\cos\theta$ where $\theta = 30°$ is the angle between $\vec{B}$ and the loop's normal $\hat{n}$. Magnetic flux questions hinge on one critical geometric relationship: flux is defined as $\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta$, where $\theta$ is always the angle between the magnetic field vector $\vec{B}$ and the area's normal vector $\hat{n}$. Keeping this definition precise will save you from the traps built into this problem. Here, $\hat{n}$ makes a 30° angle with $\vec{B}$, so the calculation is direct: $\Phi_B = (2.0)(0.10)(0.20)\cos 30° = 3.46 \times 10^{-3} \text{ Wb}$, confirming that B is correct. A contains a subtle but important error. It correctly identifies that $\cos 60°$ and $\sin 30°$ are numerically equal — both equal 0.5 — but its reasoning is wrong. The angle in the flux formula is measured from $\hat{n}$, not from the plane of the loop. Since $\hat{n}$ is already given as 30° from $\vec{B}$, no angle conversion is needed. Answer A reaches the wrong number by using $\cos 60°$ where $\cos 30°$ belongs. C uses $\sin 30°$, which would be appropriate if you wanted the component of $\vec{B}$ parallel to the loop's plane — but that component passes through the loop without contributing to flux. Only the component along $\hat{n}$ (i.e., perpendicular to the surface) counts. D ignores the tilt entirely. A uniform field only produces $\Phi_B = BA$ when $\vec{B}$ is perfectly aligned with $\hat{n}$ ($\theta = 0°$). Study tip: Always identify which angle is given — angle to the normal or angle to the plane — before plugging into $BA\cos\theta$. If given the angle to the plane, subtract from 90° first.
6. A conducting sphere of radius $R$ encloses a small magnetic dipole at its center. An external uniform field $\vec{B}_{\text{ext}} = B_0\hat{z}$ is also present. What is the net magnetic flux through the spherical surface, and which student's reasoning is correct?
   Answer: The net flux is $\Phi_B = 0$, and the second student is correct: Gauss's law for magnetism guarantees zero net flux through any closed surface regardless of the sources inside or outside, because magnetic monopoles do not exist; however, $\vec{B}$ is certainly nonzero at points on the surface. Whenever you encounter a question involving magnetic flux through a closed surface, your first instinct should be to recall Gauss's law for magnetism: $\oint \vec{B} \cdot d\vec{A} = 0$ for any closed surface, always. This law reflects the fundamental fact that magnetic monopoles do not exist — every field line that enters a closed surface must also exit it. This makes D correct. The net flux through the sphere is $\Phi_B = 0$, full stop — regardless of whether there's a magnetic dipole inside, a uniform external field, or both. The dipole's field lines that exit the sphere must re-enter it (they form closed loops), and the uniform external field $B_0\hat{z}$ contributes equal inward and outward flux. The second student is right: this zero result tells you nothing about $\vec{B}$ at individual surface points. The field is certainly nonzero on the sphere — it's just that positive and negative flux contributions cancel globally. A is wrong because a uniform field threading a closed sphere contributes zero net flux — flux entering one hemisphere exactly cancels flux exiting the other. The expression $B_0\pi R^2$ would describe flux through a flat circular disk, not a closed sphere. B is wrong because a magnetic dipole (unlike an electric charge) is not a source in the Gauss's law sense. Its closed field lines produce zero net flux, not $\mu_0 m/(2\pi R^2)$. C is wrong because the first student's logic is flawed. Zero net flux does not imply the radial component vanishes pointwise — only that positive and negative radial contributions integrate to zero. Study tip: Always distinguish between a global integral condition ($\Phi_B = 0$) and local field values. On exam questions, watch for answer choices that conflate these two levels — it's one of the most common traps in electromagnetism.
7. What is the magnetic flux through this rectangular cross-sectional surface of the toroid?
   Answer: $\Phi_B = \frac{\mu_0 N I h}{2\pi}\ln\!\left(\frac{b}{a}\right) = 2.77 \times 10^{-4} \text{ Wb}$, obtained by integrating $B(r) = \frac{\mu_0 N I}{2\pi r}$ over the rectangular area using the area element $dA = h\,dr$. When the magnetic field varies across a surface, you cannot simply multiply a single field value by the total area — you must integrate. This question tests exactly that skill in the context of a toroid, where $B(r) = \frac{\mu_0 N I}{2\pi r}$ depends on radial position. To find the flux correctly, you set up $\Phi_B = \int B \, dA$. Since the field varies only with $r$ and the surface has constant height $h$, the area element is $dA = h \, dr$. This gives: $\Phi_B = \int_a^b \frac{\mu_0 N I}{2\pi r} \cdot h \, dr = \frac{\mu_0 N I h}{2\pi} \ln\!\left(\frac{b}{a}\right)$ Plugging in values: $\frac{(4\pi \times 10^{-7})(500)(4.0)(0.02)}{2\pi}\ln\!\left(\frac{0.12}{0.08}\right) \approx 2.77 \times 10^{-4} \text{ Wb}$. That confirms A is correct. B is a classic "worst-case" trap — using the maximum field at $r = a$ and multiplying by the full area overestimates the flux, since the field is weaker at larger radii. C silently drops the $r$ from the denominator, essentially assuming $r = 1$ m, which is dimensionally inconsistent and physically meaningless. D gets the integral right in structure but flips the logarithm argument to $\ln(a/b)$, yielding a negative result. Since $a < b$, this would be negative — but flux through a physical cross-section with a well-defined field direction must be positive here; the correct argument is always $\ln(b/a)$ when integrating outward from $a$ to $b$. Your takeaway: whenever a field is nonuniform, integration is mandatory — and always check that your logarithm argument is greater than 1 when the physical flux should be positive.
8. What is the net magnetic flux through the figure-eight circuit as defined by the right-hand rule applied to the direction of traversal?
   Answer: $\Phi_{\text{net}} = 0$, because the right-hand rule applied to the traversal direction assigns opposite area-normal orientations to the two loops, so their fluxes $+B\pi r^2$ and $-B\pi r^2$ cancel exactly. Whenever you see a figure-eight circuit in a magnetic flux problem, your first move should be to carefully apply the right-hand rule to each loop as defined by the single continuous traversal direction — not just by the field direction. Here's the key insight: as you trace the wire continuously through a figure-eight, your fingers curl in opposite senses around the two loops. For one loop, the right-hand rule gives an area normal pointing in the $+\hat{z}$ direction (same as $\vec{B}$), while for the other, it points in the $-\hat{z}$ direction (opposite to $\vec{B}$). The flux through each loop has the same magnitude $\Phi = B\pi r^2 = (1.5)(\pi)(0.10)^2 \approx 4.71 \times 10^{-2} \text{ Wb}$, but opposite signs. The net flux is therefore $\Phi_{\text{net}} = +B\pi r^2 - B\pi r^2 = 0$, confirming B is correct. A is wrong because it ignores the sign convention entirely. Both loops do have equal-magnitude flux, but a figure-eight traversal forces opposite normals, so you must subtract, not add, those contributions. C is wrong because the crossing point is not an electrical node that isolates the loops — it's simply a point where the wire crosses itself. The entire loop is one continuous circuit, and Faraday's law demands you account for all enclosed area with proper sign. D is wrong because area is not halved by the crossing geometry; each loop fully encloses $\pi r^2$, and the issue is cancellation, not geometric reduction. Study tip: In any multi-loop circuit problem, always assign normal directions using the right-hand rule consistently along the traversal path — sign errors here are the most common trap on flux and EMF problems.
9. What is the magnitude of the net magnetic flux through the rectangular loop?
   Answer: $\Phi_B = \frac{\mu_0 I \ell}{2\pi}\left[\ln\!\left(\frac{s+w}{s}\right) + \ln\!\left(\frac{d-s}{d-s-w}\right)\right] = 1.61 \times 10^{-6} \text{ Wb}$, because both wires produce fields in the same direction inside the rectangle (antiparallel currents, both fields point the same way inside the loop), so the two flux contributions add. When two wires carry currents in opposite directions, the key question is: what direction does each wire's magnetic field point inside the rectangular loop? This is where most students go wrong. Use the right-hand rule for each wire separately. Wire 1 (current upward, say) creates a field that points out of the page between the wires. Wire 2 (current downward) also creates a field that points out of the page between the wires — because the current reversal and the fact that you're now on the opposite side of wire 2 combine to keep the same field direction. Both fields point the same way inside the loop, so the flux contributions add. The flux from each wire is found by integrating $B = \frac{\mu_0 I}{2\pi r}$ over the loop's area. For wire 1 (near edge at distance $s$, far edge at $s+w$): $\Phi_1 = \frac{\mu_0 I \ell}{2\pi}\ln\!\left(\frac{s+w}{s}\right)$ For wire 2 (near edge at distance $d-s-w = 0.15\ \text{m}$, far edge at $d-s = 0.25\ \text{m}$): $\Phi_2 = \frac{\mu_0 I \ell}{2\pi}\ln\!\left(\frac{d-s}{d-s-w}\right)$ Since both point the same direction, $\Phi_{net} = \Phi_1 + \Phi_2 \approx 1.61\times10^{-6}\ \text{Wb}$, confirming D. A ignores wire 2 entirely — never discard a wire just because it's "farther." B subtracts the two contributions, which would only be correct if the fields opposed each other inside the loop — they don't. C assumes perfect symmetry and doubles wire 1's contribution, but the loop is not centered between the wires, so $\Phi_1 \neq \Phi_2$. Study tip: Always apply the right-hand rule to each source independently at a test point inside the region of interest before deciding whether fluxes add or subtract — don't assume antiparallel currents mean opposing fields.
10. What is the magnetic flux through the triangular loop due to the wire?
    Answer: $\Phi_B = \frac{\mu_0 I}{2\pi}\left[0.20\ln 2 - 0.10\right] = 3.86 \times 10^{-8} \text{ Wb}$, obtained by correctly identifying that at position $x$, the vertical extent of the triangle is $(0.20 - x)$, then integrating $B(x)$ times this height from $x = 0.10$ to $x = 0.20$. When a current-carrying wire creates a non-uniform magnetic field over an extended region, you must integrate — but the setup of that integral requires careful geometric analysis of the loop itself. The wire along the $z$-axis produces a field $B(x) = \frac{\mu_0 I}{2\pi x}$ directed in the $\hat{y}$ direction throughout the $xz$-plane, which aligns perfectly with the loop's normal. The triangular loop spans $x = 0.10$ m to $x = 0.20$ m. The key geometric insight is how tall the triangle is at each $x$. The three vertices tell you the right-angle corner sits at $(0.10, 0, 0)$, the base extends to $(0.20, 0, 0)$, and the apex reaches $(0.10, 0, 0.10)$. The slanted hypotenuse connects $(0.20, 0, 0)$ to $(0.10, 0, 0.10)$, giving a vertical height at position $x$ of $(0.20 - x)$. The correct flux integral is therefore: $\Phi_B = \int_{0.10}^{0.20} \frac{\mu_0 I}{2\pi x}(0.20 - x)\, dx = \frac{\mu_0 I}{2\pi}\left[0.20\ln 2 - 0.10\right] \approx 3.86 \times 10^{-8} \text{ Wb}$ This confirms C is correct. A uses a constant height of $0.10$ m everywhere, ignoring that the triangle narrows as $x$ increases — it overcounts flux near $x = 0.20$ where the triangle actually has zero height. B commits the "centroid shortcut" error: evaluating the field at the centroid and multiplying by area only works when $B$ is uniform, not when it varies as $1/x$. D replaces the linearly varying height with a constant average height of $0.05$ m — while the average height is numerically correct ($0.10/2$), multiplying by $\ln 2$ treats it as though it's uniformly distributed, which doesn't account for the correlation between the $1/x$ weighting and the height variation. Your strategy: always sketch the loop and write the height (or width) of your integration strip as an explicit function of position before integrating. Non-uniform fields demand that the geometry and the field variation be handled together.

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