2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
A closed cylindrical can stands in a uniform magnetic field that is parallel to the can’s axis. Point P is on the curved side of the can. What is the direction of the area vector of a small element of the can’s surface at P?
Answer and reasoning
APerpendicular to the curved side, pointing away from the axisCorrect The area vector of each element is perpendicular to the surface there and, for a closed surface, points outward. At P on the curved side that is radially away from the can’s axis. The field’s direction plays no part; here it lies along the surface, so the flux through that element is zero.
BParallel to the can’s axis, which is the direction of the field A student who thinks the area vector points along the field picks this. The area vector is fixed by the surface: perpendicular to the element at P and outward. The field’s direction does not define it.
CAlong the curved side at P, going around the can’s axis A student who describes a surface by a direction lying along it picks this. The area vector is perpendicular to the surface, not along it.
DIt has none, since the area of the element is a scalar A student who has met area only as a number picks this. For flux, each element of area is a vector: its magnitude is its area and its direction is perpendicular to the surface, outward for a closed surface.
A square loop with sides of 0.20 m lies in the xy-plane, with one side along the y-axis and the opposite side at x = 0.20 m. A magnetic field perpendicular to the plane of the loop has magnitude B = (5.0 T/m²)x² and does not vary with y. What is the magnitude of the magnetic flux through the loop?
Answer and reasoning
A4.0 × 10⁻³ T·m² A student who averages the field at the two edges, 0 and 0.20 T, and multiplies by the area picks this. The field grows as x², so over most of the loop it is below that average; integration gives a mean of one third of 0.20 T.
B8.0 × 10⁻³ T·m² A student who multiplies the area by the largest field, 0.20 T at x = 0.20 m, picks this. The field is weaker over the rest of the loop, so the flux must be integrated.
C1.3 × 10⁻² T·m² A student who integrates B over x alone, ∫₀0.20 5.0x² dx, picks this. That result is in T·m, not T·m²; each strip’s area is (0.20 m)dx.
D2.7 × 10⁻³ T·m²Correct The field depends only on x, so split the loop into strips parallel to the y-axis, each of area (0.20 m)dx. Φ = ∫₀0.20 (5.0x²)(0.20)dx = (5.0)(0.20)(0.20)³/3 = 2.7 × 10⁻³ T·m².
Working Strip parallel to the y-axis, width dx and length 0.20 m: dA = (0.20 m)dx. Φ = ∫₀0.20 (5.0x²)(0.20)dx = (5.0)(0.20)(0.20)³/3 = 2.7 × 10⁻³ T·m². Distractors: mean of the edge values, (0 + 0.20 T)/2 × 0.040 m² = 4.0 × 10⁻³ T·m²; largest value times area, (0.20 T)(0.040 m²) = 8.0 × 10⁻³ T·m²; dA taken as dx, ∫₀0.20 5.0x² dx = 1.3 × 10⁻² (in T·m).
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
13.1.A.1 Magnetic flux, ΦBFix
Magnetic flux, ΦB
A scalar that describes how much magnetic field passes through a surface. For a field B⃗ that is constant over a flat area A⃗, ΦB = B⃗ · A⃗ = BA cos θ, where θ is the angle between B⃗ and the area vector. SI unit: weber (Wb); 1 Wb = 1 T·m².
Angle θ in ΦB = BA cos θ
The angle between the magnetic field and the area vector, which is perpendicular to the surface. The flux is BA when the field is perpendicular to the surface (θ = 0°) and zero when the field lies along the surface (θ = 90°). If the field makes an angle α with the plane of the surface, θ = 90° − α.
Area vector, A⃗
A vector whose magnitude is the area of a flat surface and whose direction is perpendicular to the plane of the surface; for each part of a closed surface it points outward, away from the enclosed volume. SI unit of its magnitude: square meter (m²).
Sign of the magnetic flux
Given by the dot product B⃗ · A⃗: positive when the field has a component along the area vector (θ < 90°), negative when it has a component opposite to it (θ > 90°), and zero when the field lies along the surface.
Students often think The angle θ in ΦB = BA cos θ is measured between the magnetic field and the plane of the loop. In fact No. θ is the angle between B⃗ and the area vector, which is perpendicular to the plane of the loop. If the field makes an angle α with the plane, then θ = 90° − α.
Students often think The flux through a loop in a uniform field is BA whatever the loop’s orientation, since the whole loop is in the field. In fact No. ΦB = BA only when the field is perpendicular to the loop’s plane. In general ΦB = BA cos θ, which is zero when the field lies along the plane.
13.1.A.2 Flux as a surface integral, ΦB = ∫B⃗ · dA⃗ Fix
Flux as a surface integral, ΦB = ∫B⃗ · dA⃗
The total flux through a surface is the sum of B⃗ · dA⃗ over every small element of the surface, each element contributing the field at its own location. It is needed when the field varies over the surface or the surface is curved, and it reduces to ΦB = B⃗ · A⃗ for a uniform field over a flat surface.
Choosing the area element dA
The element is chosen so that the field is constant over it: a strip of length ℓ and width dx (dA = ℓ dx) when B depends on one coordinate x only, and a thin ring of radius r and width dr (dA = 2πr dr) when B depends only on the distance r from a center.
Students often think For a nonuniform field, the flux is BA with B the field’s largest value, such as its value at the edge of the surface nearest the source. In fact No. When B varies over the surface, ΦB = ∫B⃗ · dA⃗: each small element of area contributes the field at its own location. A single value such as the largest gives the wrong total unless the field is uniform.
Students often think The limits of the flux integral are the lengths labeled in the diagram, so for a loop of width b whose near side is a distance a from the wire, x runs from a to b. In fact No. The integration variable x is the distance from the wire, so the limits are the distances of the loop’s near and far sides from the wire: from a to a + b, where a is the gap and b the loop’s width.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
The figure shows a flat loop, seen edge-on, in a uniform magnetic field, with the data labeled. What is the magnitude of the magnetic flux through the loop? Use sin 37° = 0.60 and cos 37° = 0.80.
Answer and reasoning
A2.4 × 10⁻² T·m² A student who uses the angle between the field and the loop’s plane picks this: (0.50 T)(0.060 m²)(cos 37°). θ is measured from the area vector, perpendicular to the plane, so θ = 53°.
B1.8 × 10⁻² T·m²Correct The angle in ΦB = BA cos θ is measured from the area vector, which is perpendicular to the loop, so θ = 90° − 37° = 53° and cos θ = 0.60. ΦB = (0.50 T)(0.060 m²)(0.60) = 1.8 × 10⁻² T·m².
C3.0 × 10⁻² T·m² A student who takes the flux as BA whatever the orientation picks this: (0.50 T)(0.060 m²). The loop is tilted, so only the field component perpendicular to it contributes.
D3.0 × 10⁻¹ T·m² A student who takes the flux to be the field’s component perpendicular to the loop picks this: (0.50 T)(0.60) = 0.30 T. That is a field, in teslas; the flux is that component times the loop’s area.
Working The field makes 37° with the plane of the loop, so the angle between B⃗ and the area vector is θ = 90° − 37° = 53°, and cos 53° = sin 37° = 0.60. ΦB = BA cos θ = (0.50 T)(0.060 m²)(0.60) = 1.8 × 10⁻² T·m². Distractors: angle taken from the plane, (0.50)(0.060)(0.80) = 2.4 × 10⁻² T·m²; orientation ignored, (0.50)(0.060) = 3.0 × 10⁻² T·m²; flux taken as the perpendicular field component, (0.50 T)(0.60) = 0.30, reported as 3.0 × 10⁻¹.
A flat circular loop of radius r lies in a uniform magnetic field with its plane perpendicular to the field, and the magnetic flux through it is Φ₀. It is replaced by a flat circular loop of radius 2r in the same field, with its plane at an angle of 30° to the field. What is the magnitude of the flux through the new loop?
Answer and reasoning
A4.0Φ₀ A student who takes the flux as BA whatever the orientation picks this: four times the area. With the loop tilted, only the component of the field perpendicular to it contributes, which halves the flux here.
B3.5Φ₀ A student who uses the angle between the field and the loop’s plane picks this: 4 cos 30°. The angle in BA cos θ is measured from the area vector, perpendicular to the plane, so θ = 60°.
C2.0Φ₀Correct Doubling the radius multiplies the area by four. The plane makes 30° with the field, so the field makes 60° with the area vector: Φ = B(4A)cos 60° = 2BA = 2.0Φ₀.
D1.0Φ₀ A student who takes the area as proportional to the radius picks this: 2 cos 60°. The area of a circle is πr², so doubling the radius multiplies it by four.
Working Area: π(2r)² = 4πr², four times as large. The plane makes 30° with the field, so the angle between B⃗ and the area vector is 60°. Φ = B(4A)cos 60° = 2Φ₀ = 2.0Φ₀. Distractors: orientation ignored, 4.0Φ₀; angle taken from the plane, 4 cos 30° = 3.5Φ₀; area taken as proportional to radius, 2 cos 60° = 1.0Φ₀.
The figure shows four flat loops, 1 to 4, seen edge-on in a uniform magnetic field B⃗. The loops have equal areas, and the arrow on each loop shows the direction chosen for its area vector. Φ₁, Φ₂, Φ₃ and Φ₄ are the magnetic fluxes through the loops, with their signs. Which ranking is correct?
Answer and reasoning
AΦ₁ = Φ₂ > Φ₄ > Φ₃ A student who treats flux as always positive ranks the magnitudes. Loop 2’s area vector points opposite to the field, so its flux is −BA, the smallest of the four.
BΦ₃ > Φ₄ > Φ₁ = Φ₂ A student who measures θ from each loop’s plane gets BA for loop 3 (plane along the field) and zero for loops 1 and 2. The angle is measured from the area vector, which reverses this: loop 3, lying along the field, has zero flux.
CΦ₁ = Φ₂ = Φ₃ = Φ₄ A student who treats flux as the field itself says the fluxes are equal because the field is the same everywhere. The flux also depends on each loop’s orientation relative to the field, including its sign.
DΦ₁ > Φ₄ > Φ₃ > Φ₂Correct The sign comes from the dot product of B⃗ with each chosen area vector. Loop 1’s area vector is along the field: +BA. Loop 4’s is at 60°: +BA/2. Loop 3’s is perpendicular: 0. Loop 2’s is opposite to the field: −BA.
Working Φ = BA cos θ, with θ measured from the area vector. Loop 1: area vector along B⃗, θ = 0°, Φ₁ = +BA. Loop 2: area vector opposite to B⃗, θ = 180°, Φ₂ = −BA. Loop 3: area vector perpendicular to B⃗, Φ₃ = 0. Loop 4: θ = 60°, Φ₄ = +BA/2. Φ₁ > Φ₄ > Φ₃ > Φ₂.
The figure shows a long, straight wire carrying a steady current I and a rectangular loop in the same plane, with the loop’s dimensions and its distance from the wire labeled. The magnitude of the magnetic flux through the loop can be written as (μ₀Iℓ/(2π)) × F. Which expression is F?
Answer and reasoning
Aln(1 + b/a)Correct The field varies across the loop as μ₀I/(2πx). A strip of length ℓ and width dx at distance x carries flux (μ₀I/(2πx))ℓ dx. Integrating from the near side, x = a, to the far side, x = a + b, gives (μ₀Iℓ/(2π)) ln((a + b)/a), so F = ln((a + b)/a) = ln(1 + b/a).
Bb/a A student who uses the field at the near side, μ₀I/(2πa), for the whole area bℓ picks this. The field falls off across the loop, so the flux must be integrated strip by strip; this overestimates it.
Cln(b/a) A student who integrates from x = a to x = b picks this. x is the distance from the wire, so the far side is at x = a + b; b is the loop’s width, not the far side’s distance.
Db/(a(a + b)) A student who lets the wire’s field fall off as 1/x² picks this: ∫ dx/x² from a to a + b = 1/a − 1/(a + b) = b/(a(a + b)). The field of a long, straight wire is μ₀I/(2πx), so the integral is of dx/x and gives a logarithm; this F would not even give units of flux.
Working At distance x from the wire, B = μ₀I/(2πx), perpendicular to the plane of the loop. A strip of the loop parallel to the wire, of length ℓ and width dx, has dΦ = (μ₀I/(2πx))ℓ dx. Φ = (μ₀Iℓ/(2π))∫aa+b dx/x = (μ₀Iℓ/(2π)) ln((a + b)/a), so F = ln((a + b)/a) = ln(1 + b/a). Distractors (checked with sympy): near-side field over the whole area, (μ₀I/(2πa))(bℓ), F = b/a; limits a to b, F = ln(b/a); field taken as ∝ 1/x², ∫aa+b dx/x² = 1/a − 1/(a + b), F = b/(a(a + b)).
A long, straight wire carrying a steady current passes perpendicularly through the center of a flat circular loop. A student claims that the magnetic flux through the loop is zero. Which reasoning correctly supports the claim?
Answer and reasoning
AThe field enters the surface on one side of the wire and leaves it on the other side, so these two parts cancel. A student who applies the ⊙/⊗ picture of a wire lying in a plane gives this reasoning. Here the wire is perpendicular to the loop, and its field lines lie in the loop’s plane: no field enters or leaves the surface anywhere.
BAt every point of the flat surface inside the loop, B⃗ lies in that surface, so B⃗ · dA⃗ is zero everywhere.Correct The wire’s field lines are circles around the wire in planes perpendicular to it, and the loop’s plane is one of those planes. So at every point of the flat surface the field lies in the surface, perpendicular to dA⃗, and every element contributes zero flux.
CThe current in the wire is steady, and a field that does not change produces no magnetic flux through any loop. A student who thinks flux exists only while a field changes gives this reasoning. A steady field can give any steady flux; the flux here is zero because of the field’s direction.
DThe wire’s field is zero at every point of the flat surface inside the loop, so there is nothing to integrate. A student who equates zero flux with zero field gives this reasoning. The field is not zero on the surface; it is μ₀I/(2πr) at distance r from the wire, but it lies in the surface.
A flat circular loop of radius R lies in a magnetic field perpendicular to its plane. The field’s magnitude depends only on the distance r from the loop’s center: B = kr, where k is a positive constant. What is the magnitude of the magnetic flux through the loop?
Answer and reasoning
AπkR³ A student who multiplies the area πR² by the largest field, kR at the rim, picks this. The field is weaker everywhere inside the rim, so the flux is less than πkR³.
BπkR³/2 A student who averages the field at the center (0) and at the rim (kR) and multiplies by the area picks this. The outer rings have the most area and the largest field, so the mean field over the disk is 2kR/3, not kR/2.
C2πkR³/3Correct The field depends only on r, so split the disk into thin rings of area 2πr dr, each with field kr. Φ = ∫₀R (kr)(2πr dr) = 2πkR³/3.
DkR²/2 A student who integrates the field along the radius alone, ∫₀R kr dr, picks this. That result is in T·m, not T·m²; each ring’s area is 2πr dr.
Working Thin ring of radius r and width dr: dA = 2πr dr. Φ = ∫₀R (kr)(2πr dr) = 2πkR³/3. Distractors (checked with sympy): rim value kR over the whole area, πkR³; mean of the values at the center (0) and the rim (kR) over the area, πkR³/2; dA taken as dr, ∫₀R kr dr = kR²/2 (in T·m). (The numeric check applies only to the key’s leading coefficient, 2.)
A long solenoid of radius a has a uniform magnetic field of magnitude B inside it. A circular loop of radius b > a encircles the solenoid, coaxial with it and far from its ends. What is the magnitude of the magnetic flux through the loop?
Answer and reasoning
AZero, as the field at the loop’s wire is negligible A student who thinks the flux is set by the field at the loop’s wire picks this. The flux depends on the field over the whole surface bounded by the loop, and the solenoid’s field passes through that surface.
Bπa²B, as only the solenoid’s interior has a fieldCorrect The field of a long solenoid is uniform inside it and negligible outside. Over the surface bounded by the loop, only the solenoid’s cross section, of area πa², has a field, and it is perpendicular to the surface there. The flux is πa²B, whatever the loop’s radius.
Cπb²B, as the field spreads out over the whole loop A student who lets the solenoid’s field fill the whole loop picks this. Outside a long solenoid the field is negligible, so only the solenoid’s cross section, πa², contributes.
DB, as the flux is just the field inside the solenoid A student who treats flux as another name for the field picks this. Flux is field times area for a uniform perpendicular field, measured in T·m²; here it is B times the solenoid’s cross-sectional area.
Working The field of a long solenoid is B inside its cross section and negligible outside. Over the flat surface bounded by the loop, only the solenoid’s cross section, of area πa², has a field, and there the field is perpendicular to the surface: Φ = πa²B, whatever the loop’s radius b.
A long, straight, solid cylindrical wire of radius R carries a steady current I spread uniformly over its cross section. A flat rectangular surface of length ℓ lies in a plane that contains the wire’s axis: one of its long sides lies along the axis, and the other is parallel to it, a distance 2R from the axis. What is the magnitude of the magnetic flux through the surface?
Answer and reasoning
A0.11μ₀Iℓ A student who takes the magnetic field inside the metal to be zero, like the electric field inside a conductor, picks this: only the part from R to 2R is counted, (μ₀Iℓ/(2π)) ln 2. A loop inside the wire encloses part of the current, so B = μ₀Is/(2πR²) there.
B0.32μ₀Iℓ A student who multiplies the field’s largest value, μ₀I/(2πR) at the wire’s surface, by the whole area 2Rℓ picks this: μ₀Iℓ/π. The field is weaker everywhere else on the surface, so the flux must be found by integrating.
C0.08μ₀Iℓ A student who averages the field’s values at the two long edges, 0 on the axis and μ₀I/(4πR) at 2R, and multiplies by the area 2Rℓ picks this: μ₀Iℓ/(4π). The field rises and then falls across the surface, so the mean of the edge values is not its average.
D0.19μ₀IℓCorrect Ampère’s law gives B = μ₀Is/(2πR²) inside the wire and μ₀I/(2πs) outside, perpendicular to the surface everywhere. Integrating over strips of area ℓ ds, the part inside the wire gives μ₀Iℓ/(4π) and the part from R to 2R gives (μ₀Iℓ/(2π)) ln 2; the sum is μ₀Iℓ(1 + 2 ln 2)/(4π) = 0.19μ₀Iℓ.
Working Ampère’s law with circular loops coaxial with the wire: inside, B(2πs) = μ₀I(s²/R²), so B = μ₀Is/(2πR²) for s < R; outside, B = μ₀I/(2πs). B⃗ circles the axis, so it is perpendicular to the surface at every point, and a strip of width ds has area ℓ ds. ΦB = ∫₀R (μ₀Is/(2πR²))ℓ ds + ∫R2R (μ₀I/(2πs))ℓ ds = μ₀Iℓ/(4π) + (μ₀Iℓ/(2π)) ln 2 = μ₀Iℓ(1 + 2 ln 2)/(4π) = 0.19μ₀Iℓ. Units: (T·m/A)(A)(m) = T·m² = Wb. Distractors (checked with sympy): field inside the wire taken as zero → (μ₀Iℓ/(2π)) ln 2 = 0.11μ₀Iℓ; largest field, μ₀I/(2πR) at s = R, times the area 2Rℓ → μ₀Iℓ/π = 0.32μ₀Iℓ; mean of the edge values, 0 on the axis and μ₀I/(4πR) at 2R, times 2Rℓ → μ₀Iℓ/(4π) = 0.08μ₀Iℓ.
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
