Physics 2 Quiz: Magnetic Flux
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Magnetic FluxQuestion 1 of 10

A long solenoid of radius Rs=0.05 mR_s = 0.05 \text{ m} has n=1000 turns/mn = 1000 \text{ turns/m} and carries a current I=2.0 AI = 2.0 \text{ A}. A flat square loop of side L=0.30 mL = 0.30 \text{ m} is placed coaxially with the solenoid so that the solenoid passes through the center of the square loop. Both objects share the same axis. The permeability of free space is μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}.

What is the magnetic flux through the square loop due to the solenoid's field?

ΦB=μ0nI⋅L2=2.26×10−4 Wb\Phi_B = \mu_0 n I \cdot L^2 = 2.26 \times 10^{-4} \text{ Wb}, because the solenoid's uniform field fills the entire area of the square loop, so the full L2L^2 area is used in computing the flux.
ΦB=μ0nI⋅πRs2=3.95×10−5 Wb\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 πRs2\pi R_s^2, and is zero outside; only that area contributes to the flux through the larger square loop.
ΦB=μ0nI⋅(L2−πRs2)=1.87×10−4 Wb\Phi_B = \mu_0 n I \cdot (L^2 - \pi R_s^2) = 1.87 \times 10^{-4} \text{ Wb}, because the field is zero inside the solenoid bore and nonzero in the annular region between the solenoid and the square loop boundary.
ΦB=μ0nI⋅πRs2/L2=4.39×10−7 Wb\Phi_B = \mu_0 n I \cdot \pi R_s^2 / L^2 = 4.39 \times 10^{-7} \text{ Wb}, because the relevant flux is reduced by the ratio of the solenoid cross-section to the loop area, reflecting the geometric dilution of field lines across the larger surface.
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Physics 2 Quiz

Physics 2 Quiz: Magnetic Flux

Practice Magnetic Flux in Physics 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Magnetic Flux, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A long solenoid of radius Rs=0.05 mR_s = 0.05 \text{ m} has n=1000 turns/mn = 1000 \text{ turns/m} and carries a current I=2.0 AI = 2.0 \text{ A}. A flat square loop of side L=0.30 mL = 0.30 \text{ m} is placed coaxially with the solenoid so that the solenoid passes through the center of the square loop. Both objects share the same axis. The permeability of free space is μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}.

What is the magnetic flux through the square loop due to the solenoid's field?

  1. ΦB=μ0nI⋅L2=2.26×10−4 Wb\Phi_B = \mu_0 n I \cdot L^2 = 2.26 \times 10^{-4} \text{ Wb}, because the solenoid's uniform field fills the entire area of the square loop, so the full L2L^2 area is used in computing the flux.
  2. ΦB=μ0nI⋅πRs2=3.95×10−5 Wb\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 πRs2\pi R_s^2, and is zero outside; only that area contributes to the flux through the larger square loop. (correct answer)
  3. ΦB=μ0nI⋅(L2−πRs2)=1.87×10−4 Wb\Phi_B = \mu_0 n I \cdot (L^2 - \pi R_s^2) = 1.87 \times 10^{-4} \text{ Wb}, because the field is zero inside the solenoid bore and nonzero in the annular region between the solenoid and the square loop boundary.
  4. ΦB=μ0nI⋅πRs2/L2=4.39×10−7 Wb\Phi_B = \mu_0 n I \cdot \pi R_s^2 / L^2 = 4.39 \times 10^{-7} \text{ Wb}, because the relevant flux is reduced by the ratio of the solenoid cross-section to the loop area, reflecting the geometric dilution of field lines across the larger surface.
Explanation: 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=μ0nIB = \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 ΦB=∫B⃗⋅dA⃗\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: ΦB=μ0nI⋅πRs2=(4π×10−7)(1000)(2.0)⋅π(0.05)2≈3.95×10−5 Wb\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 L2L^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 πRs2/L2\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.

Question 2

Two infinite parallel wires separated by distance d=0.30 md = 0.30 \text{ m} carry currents I1=10 AI_1 = 10 \text{ A} and I2=10 AI_2 = 10 \text{ A} in opposite directions (antiparallel). A rectangular loop of width w=0.10 mw = 0.10 \text{ m} and height ℓ=0.50 m\ell = 0.50 \text{ m} lies in the plane of the two wires. The near edge of the rectangle is at distance s=0.05 ms = 0.05 \text{ m} from wire 1, so the far edge is at s+w=0.15 ms + w = 0.15 \text{ m} from wire 1 and d−(s+w)=0.15 md - (s+w) = 0.15 \text{ m} from wire 2. The rectangle lies entirely between the two wires. Use μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi\times10^{-7}\text{ T}\cdot\text{m/A}.

What is the magnitude of the net magnetic flux through the rectangular loop?

  1. ΦB=μ0Iℓ2πln⁡ ⁣(s+ws)=1.10×10−6 Wb\Phi_B = \frac{\mu_0 I \ell}{2\pi}\ln\!\left(\frac{s+w}{s}\right) = 1.10 \times 10^{-6} \text{ Wb}, considering only wire 1, since wire 2 is farther from the loop's near edge and its contribution is assumed negligible.
  2. ΦB=μ0Iℓ2π[ln⁡ ⁣(s+ws)−ln⁡ ⁣(d−sd−s−w)]=5.12×10−7 Wb\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] = 5.12 \times 10^{-7} \text{ Wb}, because the two wires carry antiparallel currents, so their field contributions inside the loop point in opposite directions and the net flux is the difference of the two terms.
  3. ΦB=μ0Iℓπln⁡ ⁣(s+ws)=2.20×10−6 Wb\Phi_B = \frac{\mu_0 I \ell}{\pi}\ln\!\left(\frac{s+w}{s}\right) = 2.20 \times 10^{-6} \text{ Wb}, because the two wires are symmetric about the center of the loop, so the total flux is exactly double the contribution from wire 1 alone.
  4. ΦB=μ0Iℓ2π[ln⁡ ⁣(s+ws)+ln⁡ ⁣(d−sd−s−w)]=1.61×10−6 Wb\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. (correct answer)
Explanation: 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=μ0I2πrB = \frac{\mu_0 I}{2\pi r} over the loop's area. For wire 1 (near edge at distance ss, far edge at s+ws+w): Φ1=μ0Iℓ2πln⁡ ⁣(s+ws)\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 md-s-w = 0.15\ \text{m}, far edge at d−s=0.25 md-s = 0.25\ \text{m}): Φ2=μ0Iℓ2πln⁡ ⁣(d−sd−s−w)\Phi_2 = \frac{\mu_0 I \ell}{2\pi}\ln\!\left(\frac{d-s}{d-s-w}\right) Since both point the same direction, Φnet=Φ1+Φ2≈1.61×10−6 Wb\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 Φ1≠Φ2\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.

Question 3

A circular loop of radius R=0.20 mR = 0.20 \text{ m} lies in the xyxy-plane. A non-uniform magnetic field exists in the region given by B⃗=B0(1+αx)z^\vec{B} = B_0(1 + \alpha x)\hat{z}, where B0=0.50 TB_0 = 0.50 \text{ T}, α=2.0 m−1\alpha = 2.0 \text{ m}^{-1}, and xx is the position in meters. The center of the loop is at the origin.

What is the magnetic flux through the circular loop?

  1. ΦB=B0πR2=6.28×10−2 Wb\Phi_B = B_0 \pi R^2 = 6.28 \times 10^{-2} \text{ Wb}, because the αx\alpha x term is antisymmetric about the yy-axis and its integral over the symmetric circular area vanishes, leaving only the uniform part. (correct answer)
  2. ΦB=B0(1+αR)πR2=8.79×10−2 Wb\Phi_B = B_0(1 + \alpha R)\pi R^2 = 8.79 \times 10^{-2} \text{ Wb}, because the effective field is evaluated at the rim of the loop where the field is strongest, weighted by the full area.
  3. ΦB=B0(1+α2R)πR2=7.54×10−2 Wb\Phi_B = B_0\left(1 + \frac{\alpha}{2}R\right)\pi R^2 = 7.54 \times 10^{-2} \text{ Wb}, because the average value of xx over the circular disk equals R/2R/2, which replaces xx in the field expression.
  4. ΦB=B0(1+απR2)πR2=7.54×10−2 Wb\Phi_B = B_0\left(1 + \alpha \pi R^2\right)\pi R^2 = 7.54 \times 10^{-2} \text{ Wb}, because the correction term involves multiplying α\alpha by the area of the loop rather than a length scale.
Explanation: When you encounter magnetic flux through a loop in a non-uniform field, your first instinct should be to set up the integral ΦB=∬B⃗⋅dA⃗\Phi_B = \iint \vec{B} \cdot d\vec{A} and carefully examine the symmetry of the integrand before calculating anything. Here, B⃗=B0(1+αx)z^\vec{B} = B_0(1 + \alpha x)\hat{z}, so the flux becomes ΦB=∬B0(1+αx) dA=B0∬dA+B0α∬x dA\Phi_B = \iint B_0(1 + \alpha x)\, dA = B_0\iint dA + B_0\alpha\iint x\, dA. The first integral is simply B0πR2B_0 \pi R^2. For the second, ask yourself: what is ∬x dA\iint x\, dA over a disk centered at the origin? The function xx is antisymmetric about the yy-axis — for every point (x,y)(x, y) in the disk with value +x+x, there is a mirror point (−x,y)(-x, y) with value −x-x. These cancel exactly, so ∬x dA=0\iint x\, dA = 0. Therefore ΦB=B0πR2=(0.50)(π)(0.04)≈6.28×10−2 Wb\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 xx over a circle (or disk centered at the origin) is zero, not R/2R/2. The mean value R/2R/2 might arise in other geometries, but symmetry kills it here. Answer D invents a dimensional inconsistency — multiplying α\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 xx or yy) integrated over a symmetric region centered at the origin, check antisymmetry first — it often eliminates an entire term before you do any real calculation.

Question 4

A hemispherical surface of radius R=0.15 mR = 0.15 \text{ m} is placed in a uniform magnetic field B⃗=B0z^\vec{B} = B_0\hat{z} with B0=3.0 TB_0 = 3.0 \text{ T}. The rim of the hemisphere lies in the xyxy-plane with the curved surface bulging in the +z+z direction. The outward normal convention is used (normals point away from the enclosed volume).

What is the magnetic flux through the curved hemispherical surface alone (not the flat circular cap)?

  1. Φcurved=−B0⋅2πR2=−4.24×10−1 Wb\Phi_{\text{curved}} = -B_0 \cdot 2\pi R^2 = -4.24 \times 10^{-1} \text{ Wb}, because the outward normal at every point of the dome has a component opposing z^\hat{z}, and the full curved surface area 2πR22\pi R^2 must be used in the flux integral.
  2. Φcurved=+B0πR2=+2.12×10−1 Wb\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 xyxy-plane matters, and that projection has area πR2\pi R^2; the outward normals on the curved surface have a net +z+z component integrating to πR2\pi R^2. (correct answer)
  3. Φcurved=B0⋅2πR2=+4.24×10−1 Wb\Phi_{\text{curved}} = B_0 \cdot 2\pi R^2 = +4.24 \times 10^{-1} \text{ Wb}, because the curved surface area of a hemisphere is 2πR22\pi R^2, and the field is treated as uniformly parallel to the outward normal since the dome faces in the +z+z direction.
  4. Φcurved=0\Phi_{\text{curved}} = 0, because the magnetic field is uniform and the curved surface is symmetric, so equal and opposite flux contributions from the left and right halves of the hemisphere cancel exactly.
Explanation: Whenever you encounter magnetic flux through a curved surface, your first instinct should be to reach for Gauss's Law for magnetism: ∮B⃗⋅dA⃗=0\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: Φcurved+Φcap=0\Phi_{\text{curved}} + \Phi_{\text{cap}} = 0 The flat cap lies in the xyxy-plane. Its outward normal points in the −z^-\hat{z} direction (downward, away from the enclosed volume), so: Φcap=B⃗⋅(−z^)⋅πR2=−B0πR2\Phi_{\text{cap}} = \vec{B} \cdot (-\hat{z}) \cdot \pi R^2 = -B_0 \pi R^2 Substituting back: Φcurved=+B0πR2=(3.0)(π)(0.15)2≈+2.12×10−1 Wb\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 B⃗\vec{B} — which is simply πR2\pi R^2 — determines the net flux. A is wrong because using the full curved surface area 2πR22\pi R^2 ignores that outward normals on the dome point radially outward, not uniformly against z^\hat{z} — most normals have only a partial zz-component, and integration correctly yields πR2\pi R^2, not 2πR22\pi R^2. The sign is also incorrect. C makes the opposite sign error and incorrectly assumes B⃗\vec{B} is parallel to every outward normal on the dome. D confuses left-right symmetry with cancellation — the zz-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.

Question 5

A square loop of side a=0.25 ma = 0.25 \text{ m} is placed with one side along the zz-axis. The loop lies in the xzxz-plane. A magnetic field is given by B⃗=B0y x^+B0x y^\vec{B} = B_0 y\,\hat{x} + B_0 x\,\hat{y}, where B0=4.0 T/mB_0 = 4.0 \text{ T/m} and x,y,zx, y, z are in meters. The loop occupies the region 0≤x≤a0 \leq x \leq a, y=0y = 0, 0≤z≤a0 \leq z \leq a.

What is the magnetic flux through the square loop?

  1. ΦB=B0a3=6.25×10−2 Wb\Phi_B = B_0 a^3 = 6.25 \times 10^{-2} \text{ Wb}, because the area normal is y^\hat{y}, so the relevant component is By=B0xB_y = B_0 x, and integrating over the loop gives B0∫0ax dx⋅a=B0a2⋅a=B0a3B_0 \int_0^a x\,dx \cdot a = B_0 a^2 \cdot a = B_0 a^3.
  2. ΦB=0\Phi_B = 0, because the loop lies entirely in the plane y=0y = 0, where Bx=B0y=0B_x = B_0 y = 0; since the only nonzero component of B⃗\vec{B} on the loop is ByB_y, and y^\hat{y} is parallel to the loop's plane (not the normal), the flux vanishes.
  3. ΦB=12B0a3=3.125×10−2 Wb\Phi_B = \tfrac{1}{2}B_0 a^3 = 3.125 \times 10^{-2} \text{ Wb}, because the loop lies in the xzxz-plane with area normal y^\hat{y}; the relevant field component is By=B0xB_y = B_0 x, and integrating gives ∫0a ⁣∫0aB0x dx dz=B0a⋅a22=12B0a3\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. (correct answer)
  4. ΦB=B0(a2)⋅a=12B0a2=1.25×10−1 Wb\Phi_B = B_0\left(\tfrac{a}{2}\right) \cdot a = \tfrac{1}{2}B_0 a^2 = 1.25 \times 10^{-1} \text{ Wb}, because the field is evaluated at the center of the loop (x=a/2, y=0)(x = a/2,\, y = 0), giving By=B0a/2B_y = B_0 a/2, which is then multiplied by one side length aa rather than the full area.
Explanation: When calculating magnetic flux, the formula ΦB=∬B⃗⋅dA⃗\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 xzxz-plane, so its area element is dA⃗=dx dz y^d\vec{A} = dx\,dz\,\hat{y}. This means only ByB_y contributes to the flux. On the loop, y=0y = 0 everywhere, so By=B0xB_y = B_0 x — which varies with xx but not yy. You must integrate this over the surface: ΦB=∫0a∫0aB0 x dx dz=B0∫0adz∫0ax dx=B0⋅a⋅a22=12B0a3\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: 12(4.0)(0.25)3=3.125×10−2 Wb\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 B0∫0ax dx⋅a=B0a2⋅aB_0 \int_0^a x\,dx \cdot a = B_0 a^2 \cdot a, incorrectly claiming ∫0ax dx=a2\int_0^a x\,dx = a^2 instead of the correct a2/2a^2/2. That's an off-by-two error. Answer B confuses which component matters. Yes, Bx=0B_x = 0 on the loop, but BxB_x is irrelevant — it's perpendicular to y^\hat{y}. The flux comes from ByB_y, which is nonzero. Answer D uses a shortcut that only works for uniform fields: evaluating ByB_y at the loop's center and multiplying by area. Since By=B0xB_y = B_0 x varies across the loop, you must integrate. Study tip: Always set up the dot product B⃗⋅n^\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.

Question 6

A student claims the following: 'If the magnetic flux through a closed surface is zero, then the magnetic field must be zero at every point on that surface.' A second student counters: 'No — Gauss's law for magnetism says the flux through any closed surface is always zero, which means we can deduce nothing about B⃗\vec{B} at individual points from this condition alone.'

A conducting sphere of radius RR encloses a small magnetic dipole at its center. An external uniform field B⃗ext=B0z^\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?

  1. The net flux is ΦB=B0πR2\Phi_B = B_0 \pi R^2, because the external uniform field contributes a net upward flux through the top hemisphere equal to B0πR2B_0\pi R^2, and the dipole field is symmetric so it contributes zero; the first student's claim fails because nonzero flux does not require nonzero field everywhere.
  2. The net flux is ΦB=μ0m/(2πR2)\Phi_B = \mu_0 m / (2\pi R^2), where mm is the dipole moment, because the magnetic dipole inside acts as an effective source whose contribution does not cancel when integrated over the sphere; the second student's reasoning is flawed because it ignores enclosed magnetic sources.
  3. The net flux is ΦB=0\Phi_B = 0, and the first student is correct: zero net flux implies that B⃗\vec{B} must be tangential to the sphere at every surface point, meaning the radial component of the total field vanishes everywhere on the sphere's surface.
  4. The net flux is ΦB=0\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, B⃗\vec{B} is certainly nonzero at points on the surface. (correct answer)
Explanation: Whenever you encounter a question involving magnetic flux through a closed surface, your first instinct should be to recall Gauss's law for magnetism: ∮B⃗⋅dA⃗=0\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 ΦB=0\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 B0z^B_0\hat{z} contributes equal inward and outward flux. The second student is right: this zero result tells you nothing about B⃗\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 B0πR2B_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 μ0m/(2πR2)\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 (ΦB=0\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.

Question 7

A toroidal solenoid has inner radius a=0.08 ma = 0.08 \text{ m}, outer radius b=0.12 mb = 0.12 \text{ m}, and N=500N = 500 total turns carrying current I=4.0 AI = 4.0 \text{ A}. By Ampere's law, the field inside the toroid at radial distance rr from the toroid axis is B(r)=μ0NI2πrB(r) = \frac{\mu_0 N I}{2\pi r}. A flat rectangular surface passes through the interior of the toroid, spanning from r=ar=a to r=br=b radially and having height h=0.02 mh = 0.02 \text{ m} (the cross-sectional height of the toroid). The surface is coplanar with the toroid axis.

What is the magnetic flux through this rectangular cross-sectional surface of the toroid?

  1. ΦB=μ0NIh2πln⁡ ⁣(ba)=2.77×10−4 Wb\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)=μ0NI2πrB(r) = \frac{\mu_0 N I}{2\pi r} over the rectangular area using the area element dA=h drdA = h\,dr. (correct answer)
  2. ΦB=μ0NI2πa⋅h(b−a)=5.00×10−4 Wb\Phi_B = \frac{\mu_0 N I}{2\pi a} \cdot h(b-a) = 5.00 \times 10^{-4} \text{ Wb}, obtained by evaluating the field at the inner radius r=ar = a, where it is strongest, and multiplying by the full rectangular area h(b−a)h(b-a).
  3. ΦB=μ0NI2π⋅h(b−a)=2.00×10−4 Wb\Phi_B = \frac{\mu_0 N I}{2\pi} \cdot h(b-a) = 2.00 \times 10^{-4} \text{ Wb}, obtained by treating the field as uniform with r=1 mr = 1\text{ m} as an implicit unit reference, then multiplying by the cross-sectional area.
  4. ΦB=μ0NIh2πln⁡ ⁣(ab)=−2.77×10−4 Wb\Phi_B = \frac{\mu_0 N I h}{2\pi}\ln\!\left(\frac{a}{b}\right) = -2.77 \times 10^{-4} \text{ Wb}, because the field decreases with increasing rr, so the flux integral must be negative to reflect that the field is stronger near the inner radius.
Explanation: 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)=μ0NI2πrB(r) = \frac{\mu_0 N I}{2\pi r} depends on radial position. To find the flux correctly, you set up ΦB=∫B dA\Phi_B = \int B \, dA. Since the field varies only with rr and the surface has constant height hh, the area element is dA=h drdA = h \, dr. This gives: ΦB=∫abμ0NI2πr⋅h dr=μ0NIh2πln⁡ ⁣(ba)\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: (4π×10−7)(500)(4.0)(0.02)2πln⁡ ⁣(0.120.08)≈2.77×10−4 Wb\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=ar = a and multiplying by the full area overestimates the flux, since the field is weaker at larger radii. C silently drops the rr from the denominator, essentially assuming r=1r = 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)\ln(a/b), yielding a negative result. Since a<ba < 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)\ln(b/a) when integrating outward from aa to bb. 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.

Question 8

A long straight wire carrying current I=5.0 AI = 5.0 \text{ A} runs along the zz-axis. A flat triangular loop with vertices at (0.10, 0, 0)(0.10,\, 0,\, 0), (0.20, 0, 0)(0.20,\, 0,\, 0), and (0.10, 0, 0.10)(0.10,\, 0,\, 0.10) (all coordinates in meters) lies in the xzxz-plane. The area normal is taken as +y^+\hat{y}. Use μ0=4π×10−7 T⋅m/A\mu_0 = 4\pi\times10^{-7}\text{ T}\cdot\text{m/A}.

What is the magnetic flux through the triangular loop due to the wire?

  1. ΦB=μ0I2π(0.10ln⁡2)=6.93×10−8 Wb\Phi_B = \frac{\mu_0 I}{2\pi}\left(0.10\ln 2\right) = 6.93 \times 10^{-8} \text{ Wb}, obtained by integrating B(x)=μ0I2πxB(x) = \frac{\mu_0 I}{2\pi x} times the triangle's height, but incorrectly treating the height as the constant value 0.10 m0.10\text{ m} (the maximum height) rather than the linearly varying limit (0.20−x)(0.20-x).
  2. ΦB=μ0I2π⋅1xˉ⋅A△=4.70×10−8 Wb\Phi_B = \frac{\mu_0 I}{2\pi} \cdot \frac{1}{\bar{x}} \cdot A_{\triangle} = 4.70 \times 10^{-8} \text{ Wb}, obtained by evaluating the field at the triangle's centroid xˉ=(0.10+0.20+0.10)/3≈0.133 m\bar{x} = (0.10+0.20+0.10)/3 \approx 0.133\text{ m} and multiplying by the triangle's area A△=12(0.10)(0.10)=5.0×10−3 m2A_{\triangle} = \frac{1}{2}(0.10)(0.10) = 5.0\times10^{-3}\text{ m}^2.
  3. ΦB=μ0I2π[0.20ln⁡2−0.10]=3.86×10−8 Wb\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 xx, the vertical extent of the triangle is (0.20−x)(0.20 - x), then integrating B(x)B(x) times this height from x=0.10x = 0.10 to x=0.20x = 0.20. (correct answer)
  4. ΦB=μ0I2πln⁡0.200.10⋅(0.05)=3.47×10−8 Wb\Phi_B = \frac{\mu_0 I}{2\pi}\ln\frac{0.20}{0.10}\cdot(0.05) = 3.47 \times 10^{-8} \text{ Wb}, obtained by replacing the triangular height profile with a uniform average height of 0.05 m0.05\text{ m} (half the maximum) and integrating B(x)B(x) over the width 0.10 m0.10\text{ m}.
Explanation: 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 zz-axis produces a field B(x)=μ0I2πxB(x) = \frac{\mu_0 I}{2\pi x} directed in the y^\hat{y} direction throughout the xzxz-plane, which aligns perfectly with the loop's normal. The triangular loop spans x=0.10x = 0.10 m to x=0.20x = 0.20 m. The key geometric insight is how tall the triangle is at each xx. The three vertices tell you the right-angle corner sits at (0.10,0,0)(0.10, 0, 0), the base extends to (0.20,0,0)(0.20, 0, 0), and the apex reaches (0.10,0,0.10)(0.10, 0, 0.10). The slanted hypotenuse connects (0.20,0,0)(0.20, 0, 0) to (0.10,0,0.10)(0.10, 0, 0.10), giving a vertical height at position xx of (0.20−x)(0.20 - x). The correct flux integral is therefore: ΦB=∫0.100.20μ0I2πx(0.20−x) dx=μ0I2π[0.20ln⁡2−0.10]≈3.86×10−8 Wb\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 0.100.10 m everywhere, ignoring that the triangle narrows as xx increases — it overcounts flux near x=0.20x = 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 BB is uniform, not when it varies as 1/x1/x. D replaces the linearly varying height with a constant average height of 0.050.05 m — while the average height is numerically correct (0.10/20.10/2), multiplying by ln⁡2\ln 2 treats it as though it's uniformly distributed, which doesn't account for the correlation between the 1/x1/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.

Question 9

A conducting loop is formed by bending a wire into a figure-eight shape, consisting of two equal circular loops of radius r=0.10 mr = 0.10 \text{ m} lying in the same plane. The two loops share a single crossing point. A uniform magnetic field B⃗=1.5z^ T\vec{B} = 1.5\hat{z} \text{ T} is perpendicular to the plane of the figure-eight. When traversing the wire in one continuous direction, the current flows clockwise around the left loop and counterclockwise around the right loop.

What is the net magnetic flux through the figure-eight circuit as defined by the right-hand rule applied to the direction of traversal?

  1. Φnet=2Bπr2=9.42×10−2 Wb\Phi_{\text{net}} = 2B\pi r^2 = 9.42 \times 10^{-2} \text{ Wb}, because both loops have the same area and the same magnitude of flux; the contributions add since the field is uniform and the geometry is symmetric.
  2. Φnet=0\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 +Bπr2+B\pi r^2 and −Bπr2-B\pi r^2 cancel exactly. (correct answer)
  3. Φnet=Bπr2=4.71×10−2 Wb\Phi_{\text{net}} = B\pi r^2 = 4.71 \times 10^{-2} \text{ Wb}, because only one of the two loops contributes net flux; the crossing point acts as a node that electrically isolates the flux contributions of the two halves.
  4. Φnet=12Bπr2=2.36×10−2 Wb\Phi_{\text{net}} = \frac{1}{2}B\pi r^2 = 2.36 \times 10^{-2} \text{ Wb}, because the crossing point creates a shared boundary, effectively halving the total enclosed area that is counted in computing the circuit's net flux.
Explanation: 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 +z^+\hat{z} direction (same as B⃗\vec{B}), while for the other, it points in the −z^-\hat{z} direction (opposite to B⃗\vec{B}). The flux through each loop has the same magnitude Φ=Bπr2=(1.5)(π)(0.10)2≈4.71×10−2 Wb\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 Φnet=+Bπr2−Bπr2=0\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 πr2\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.

Question 10

A rectangular loop of dimensions a=0.10 ma = 0.10 \text{ m} (along x^\hat{x}) and b=0.20 mb = 0.20 \text{ m} (along y^\hat{y}) is tilted so that its normal vector n^\hat{n} makes an angle of 30°30° with a uniform magnetic field B⃗=2.0z^ T\vec{B} = 2.0\hat{z} \text{ T}. The loop's normal is in the xzxz-plane.

Which expression correctly gives the magnetic flux through the loop?

  1. ΦB=(2.0)(0.10)(0.20)cos⁡60°=2.0×10−3 Wb\Phi_B = (2.0)(0.10)(0.20)\cos 60° = 2.0 \times 10^{-3} \text{ Wb}, because the relevant angle is between B⃗\vec{B} and the plane of the loop (90°−30°=60°90°-30°=60°), not between B⃗\vec{B} and n^\hat{n}.
  2. ΦB=(2.0)(0.10)(0.20)cos⁡30°=3.46×10−3 Wb\Phi_B = (2.0)(0.10)(0.20)\cos 30° = 3.46 \times 10^{-3} \text{ Wb}, because the flux is BAcos⁡θBA\cos\theta where θ=30°\theta = 30° is the angle between B⃗\vec{B} and the loop's normal n^\hat{n}. (correct answer)
  3. ΦB=(2.0)(0.10)(0.20)sin⁡30°=2.0×10−3 Wb\Phi_B = (2.0)(0.10)(0.20)\sin 30° = 2.0 \times 10^{-3} \text{ Wb}, because the flux depends on the component of B⃗\vec{B} lying in the plane of the loop, which is Bsin⁡θB\sin\theta where θ=30°\theta=30°.
  4. ΦB=(2.0)(0.10)(0.20)=4.0×10−3 Wb\Phi_B = (2.0)(0.10)(0.20) = 4.0 \times 10^{-3} \text{ Wb}, because the tilt affects only the EMF induced in the loop and not the instantaneous flux, which always equals BABA for a uniform field.
Explanation: Magnetic flux questions hinge on one critical geometric relationship: flux is defined as ΦB=B⃗⋅A⃗=BAcos⁡θ\Phi_B = \vec{B} \cdot \vec{A} = BA\cos\theta, where θ\theta is always the angle between the magnetic field vector B⃗\vec{B} and the area's normal vector n^\hat{n}. Keeping this definition precise will save you from the traps built into this problem. Here, n^\hat{n} makes a 30° angle with B⃗\vec{B}, so the calculation is direct: ΦB=(2.0)(0.10)(0.20)cos⁡30°=3.46×10−3 Wb\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°\cos 60° and sin⁡30°\sin 30° are numerically equal — both equal 0.5 — but its reasoning is wrong. The angle in the flux formula is measured from n^\hat{n}, not from the plane of the loop. Since n^\hat{n} is already given as 30° from B⃗\vec{B}, no angle conversion is needed. Answer A reaches the wrong number by using cos⁡60°\cos 60° where cos⁡30°\cos 30° belongs. C uses sin⁡30°\sin 30°, which would be appropriate if you wanted the component of B⃗\vec{B} parallel to the loop's plane — but that component passes through the loop without contributing to flux. Only the component along n^\hat{n} (i.e., perpendicular to the surface) counts. D ignores the tilt entirely. A uniform field only produces ΦB=BA\Phi_B = BA when B⃗\vec{B} is perfectly aligned with n^\hat{n} (θ=0°\theta = 0°). Study tip: Always identify which angle is given — angle to the normal or angle to the plane — before plugging into BAcos⁡θBA\cos\theta. If given the angle to the plane, subtract from 90° first.