4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
The figure shows a cross section through the middle part of a long solenoid carrying a steady current, and three points, 1, 2 and 3. B₁, B₂ and B₃ are the magnitudes of the magnetic field at these points. Which ranking is correct?
Answer and reasoning
AB₂ > B₁ > B₃ A student who pictures the field inside as the field near a single wire, strongest close to the windings, picks this. Inside a long solenoid the field is uniform, the same near the windings as on the axis.
BB₁ = B₂ > B₃Correct A long solenoid’s field is uniform inside, so it is the same on the axis (point 1) as near the windings (point 2), and it is negligible outside (point 3).
CB₁ > B₂ > B₃ A student who expects the field to be strongest on the axis picks this. For a long solenoid the field is the same everywhere inside, on the axis or close to the windings.
DB₁ = B₂ = B₃ A student who thinks the field just outside is as strong as the field inside picks this. For a long solenoid the field outside is negligible, so B₃ is much smaller than B₁ and B₂.
Working A long solenoid has a uniform field inside and a negligible field outside. Points 1 (on the axis) and 2 (inside, near the windings) are both inside: B₁ = B₂ = μ₀nI. Point 3 is outside: B₃ ≈ 0. So B₁ = B₂ > B₃.
The figure shows a long, straight wire carrying current I out of the page and three Amperian loops, 1, 2 and 3, in the plane of the page. Each loop is traversed counterclockwise. Γ₁, Γ₂ and Γ₃ are the values of ∮B⃗ · dℓ⃗ around loops 1, 2 and 3. Which ranking is correct?
Answer and reasoning
AΓ₂ > Γ₁ > Γ₃ A student who thinks a longer loop around the same current has a larger ∮B⃗ · dℓ⃗ ranks loop 2 first. Loop 2’s path is longer, but it runs through weaker field; any loop around the wire gives μ₀I.
BΓ₁ = Γ₂ > Γ₃Correct ∮B⃗ · dℓ⃗ equals μ₀ times the enclosed current, whatever the loop’s size, shape or position. Loops 1 and 2 both enclose the wire, so Γ₁ = Γ₂ = μ₀I; loop 3 encloses no current, so Γ₃ = 0.
CΓ₁ > Γ₂ > Γ₃ A student who thinks the stronger field close to the wire gives a larger ∮B⃗ · dℓ⃗ ranks loop 1 first. The integral depends only on the enclosed current, so loops 1 and 2 give the same value.
DΓ₁ = Γ₂ = Γ₃ A student who counts the wire’s current for every loop near it picks this. Loop 3 does not enclose the wire; the wire’s field acts along loop 3, but its contributions to ∮B⃗ · dℓ⃗ cancel, so Γ₃ = 0.
Working Loops 1 and 2 both enclose the wire, so Γ₁ = Γ₂ = μ₀I (counterclockwise traversal with current out of the page gives a positive value). Loop 3 encloses no current, so Γ₃ = 0. Size, shape and position do not matter. Γ₁ = Γ₂ > Γ₃.
The figure shows two long, straight, parallel wires perpendicular to the page, carrying the currents labeled, and a point P. What is the magnitude of the net magnetic field at P? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A1.8 × 10⁻⁵ TCorrect Each wire is 0.10 m from P. The left wire gives μ₀(6.0 A)/(2π × 0.10 m) = 1.2 × 10⁻⁵ T and the right wire gives 6.0 × 10⁻⁶ T. By the right-hand rule, at P both fields point toward the top of the page, so they add: 1.8 × 10⁻⁵ T.
B6.0 × 10⁻⁶ T A student who thinks opposite currents give opposite fields at P picks this: 1.2 × 10⁻⁵ T − 6.0 × 10⁻⁶ T. Applying the right-hand rule to each wire shows that both fields at P point toward the top of the page, so they add.
C1.2 × 10⁻⁵ T A student who finds the field at P from an Amperian circle around the left wire, counting only the current that circle encloses, picks this. The right wire’s field also acts at P; the net field is the vector sum of both wires’ fields.
D1.8 × 10⁻⁴ T A student who lets each field fall off as 1/r², like a point charge’s, picks this: μ₀(9.0 A)/(2π(0.10 m)²). The field of a long, straight wire is μ₀I/(2πr), falling off as 1/r.
Working Each wire is 0.10 m from P. Left wire (6.0 A, out of the page): B₁ = μ₀I/(2πr) = (2.0 × 10⁻⁷ T·m/A)(6.0 A)/(0.10 m) = 1.2 × 10⁻⁵ T; its field lines circle counterclockwise, so at P, to its right, B⃗₁ points toward the top of the page. Right wire (3.0 A, into the page): B₂ = (2.0 × 10⁻⁷)(3.0)/(0.10) = 6.0 × 10⁻⁶ T; its field lines circle clockwise, so at P, to its left, B⃗₂ also points toward the top of the page. B = 1.2 × 10⁻⁵ + 6.0 × 10⁻⁶ = 1.8 × 10⁻⁵ T. Distractors: fields subtracted, 6.0 × 10⁻⁶ T; left wire only (field taken from the current enclosed by a circle around it through P), 1.2 × 10⁻⁵ T; 1/r² fall-off, μ₀(9.0 A)/(2π(0.10 m)²) = 1.8 × 10⁻⁴.
A parallel-plate capacitor is being charged, with a steady current in the wires connected to its plates. Which statement about the region between the plates is correct?
Answer and reasoning
AThere is no magnetic field, because no charges move across the gap between the two plates. A student who thinks only moving charges produce magnetic fields picks this. It is true that no charge crosses the gap, but a changing electric field also produces a magnetic field.
BThe increasing electric field between the plates produces a magnetic field that circles that region.Correct While the capacitor charges, the electric field between the plates increases. By Maxwell’s addition to Ampère’s law, a changing electric field produces a magnetic field, so a magnetic field circles the region between the plates although no charge crosses the gap.
CA magnetic field is present, produced by the electric field and lasting while that field remains. A student who thinks any electric field produces a magnetic field picks this. It is the change in the electric field that produces the magnetic field; once charging stops and the field is steady, this source is gone.
DA magnetic field is present, produced by charges carried across the gap from plate to plate. A student who thinks the current passes through the capacitor picks this. No charge crosses the insulating gap; charge builds up on the plates, and the magnetic field between them is produced by the changing electric field.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
12.4.A.1 Ampère’s law Fix
Ampère’s law
For steady currents, the line integral of the magnetic field around any closed path equals μ₀ times the net current through a surface bounded by that path: ∮B⃗ · dℓ⃗ = μ₀Ienc. The law holds for every closed path; it gives the field itself only when symmetry makes B constant in magnitude and parallel to dℓ⃗ along the parts of the path that contribute.
Line integral ∮B⃗ · dℓ⃗
The sum of B⃗ · dℓ⃗ = B dℓ cos θ over every element of a closed path, where θ is the angle between the field and the path element. Only the component of B⃗ along the path contributes. SI unit: tesla meter (T·m).
Enclosed current, Ienc
The net current through a surface bounded by the Amperian loop, counted with signs: curl the fingers of the right hand in the direction the loop is traversed, and current in the direction of the thumb is positive. Currents outside the loop are not included. SI unit: ampere (A).
Vacuum permeability, μ₀
The constant in Ampère’s law and in the magnetic fields of currents: μ₀ = 4π × 10⁻⁷ T·m/A.
Current density, J
Current per unit cross-sectional area of a conductor. The current through a surface is I = ∫J⃗ · dA⃗; for uniform J perpendicular to the surface, I = JA. When J depends on the distance s from a cylinder’s axis, the current inside radius r is ∫₀r J(s) 2πs ds. SI unit: A/m².
Field of a long, straight wire
Ampère’s law with a circular loop of radius r centered on the wire gives Bwire = μ₀I/(2πr). The field lines are circles around the wire, with direction given by the right-hand rule. Outside a cylindrical wire, the field depends only on the total current, not on the wire’s radius. SI unit: tesla (T).
Ideal (long) solenoid
A solenoid treated as very long compared with its radius: the field inside is uniform and parallel to the axis, and the field outside is negligible. Solenoids are assumed to be of this kind unless otherwise stated.
Turns per unit length, n
The number of turns of a solenoid per unit of its length, n = N/L, where N is the total number of turns and L the solenoid’s length. SI unit: m⁻¹.
Field inside a long solenoid
A rectangular Amperian loop with one side of length ℓ inside the solenoid encloses nℓ turns, so Bℓ = μ₀nℓI and Bsol = μ₀nI. The field depends on the turns per length and the current, not on the solenoid’s radius or on the position inside it. SI unit: tesla (T).
Students often think For a counterclockwise Amperian loop in the plane of the page, currents into the page count as positive in Ienc. In fact Currents out of the page (⊙). Curl the fingers of the right hand in the direction of traversal: the thumb points out of the page, and that is the positive direction for current. A current into the page (⊗) counts as negative.
Students often think Ampère’s law uses the total current of the conductors near the loop, including currents that pass outside the loop and the whole current of a conductor that the loop lies inside. In fact No. Only the net current through the surface bounded by the loop counts. A current outside the loop changes B⃗ at points on the loop, but its contributions to ∮B⃗ · dℓ⃗ add to zero around the closed path. For a loop inside a conductor, only the part of the current that passes inside the loop is enclosed.
12.4.A.2 Amperian loop Fix
Amperian loop
A closed, imaginary path around a current-carrying conductor, used to apply Ampère’s law. It is not a physical wire. It is chosen to match the symmetry of the field—a circle centered on a long wire or a cylinder, a rectangle for a solenoid or a slab—and it must pass through the point where the field is wanted.
Students often think A longer Amperian loop around the same current has a larger ∮B⃗ · dℓ⃗, and the field along it is stronger, because more of the path lies in the field. In fact No. For every loop that encloses a given current, ∮B⃗ · dℓ⃗ = μ₀Ienc, whatever the loop’s size or shape. On a larger circle around a wire the field is weaker in proportion to the extra length: B = μ₀I/(2πr).
12.4.A.3 Superposition of magnetic fields Fix
Superposition of magnetic fields
The net magnetic field at a point is the vector sum of the fields produced there by each current-carrying conductor. Each field is found separately (for example with Ampère’s law for a symmetric conductor) and the vectors are then added.
Students often think Two equal currents in the same direction produce zero magnetic field at every point that is the same distance from both wires, as they do at the midpoint between them. In fact No. The field is zero only at the midpoint between the wires, where the two fields are equal and opposite. At other points equidistant from the wires, the two field vectors are not opposite: their components along the perpendicular bisector cancel, but their components parallel to the line joining the wires add.
Students often think Between two parallel wires with currents in opposite directions, the two magnetic fields point in opposite directions and subtract. In fact No. Between wires with opposite currents, the two fields point in the same direction, so they add. Between wires with currents in the same direction, the fields point in opposite directions and partly cancel.
12.4.A.4 Maxwell’s addition to Ampère’s law Fix
Maxwell’s addition to Ampère’s law
A changing electric field produces a magnetic field, as a current does: ∮B⃗ · dℓ⃗ = μ₀I + μ₀ε₀(dΦE/dt). For example, a magnetic field circles the region between the plates of a capacitor while it charges, although no charge crosses the gap.
Maxwell’s equations
The collection of equations that fully describe electromagnetism. The fourth is Ampère’s law with Maxwell’s addition: magnetic fields are produced by electric currents and by changing electric fields.
Students often think Only moving charges produce magnetic fields, so there is no magnetic field between the plates of a charging capacitor, where no charges move. In fact Yes. A changing electric field produces a magnetic field (Maxwell’s addition to Ampère’s law). While a capacitor charges, the electric field between its plates increases, and a magnetic field circles the region between the plates although no charge crosses the gap.
Students often think Any electric field, steady or changing, produces a magnetic field around it, so a magnetic field remains between a capacitor’s plates as long as they are charged. In fact No. It is a changing electric field that produces a magnetic field; the term Maxwell added is proportional to dΦE/dt, which is zero for a steady field. Once a capacitor’s charge stops changing, this source of magnetic field between its plates is gone.
14 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 14
The figure shows three long, straight wires, A, B and C, perpendicular to the page, with the current in each labeled; ⊙ means out of the page and ⊗ into the page. The dashed curve is an Amperian loop in the plane of the page, traversed counterclockwise as the arrow shows. What is the value of ∮B⃗ · dℓ⃗ around the loop, in the direction of traversal? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A+2.5 × 10⁻⁶ T·m A student who takes currents into the page as positive for a counterclockwise loop picks this: 5.0 A − 3.0 A = +2.0 A. With the fingers curled along the counterclockwise traversal, the thumb points out of the page, so B’s current, into the page, is the negative one.
B−2.5 × 10⁻⁶ T·mCorrect Curling the right hand’s fingers counterclockwise, along the traversal, points the thumb out of the page, so A’s current counts as +3.0 A and B’s as −5.0 A; C lies outside the loop and adds nothing. Ienc = −2.0 A, so ∮B⃗ · dℓ⃗ = μ₀Ienc = (4π × 10⁻⁷ T·m/A)(−2.0 A) = −2.5 × 10⁻⁶ T·m.
C−7.5 × 10⁻⁶ T·m A student who counts every nearby current picks this: μ₀(3.0 A − 5.0 A − 4.0 A). Wire C passes outside the loop; its field changes B⃗ at points on the loop, but its contributions to ∮B⃗ · dℓ⃗ add to zero.
D+1.0 × 10⁻⁵ T·m A student who adds the enclosed currents as magnitudes picks this: μ₀(3.0 A + 5.0 A). A and B carry current in opposite directions through the loop, so their contributions to Ienc have opposite signs.
Working Right-hand rule for the traversal: with the fingers curled counterclockwise, the thumb points out of the page, so current out of the page is positive. The loop encloses A (+3.0 A, out) and B (−5.0 A, in); C is outside. Ienc = 3.0 A − 5.0 A = −2.0 A. ∮B⃗ · dℓ⃗ = μ₀Ienc = (4π × 10⁻⁷ T·m/A)(−2.0 A) = −2.5 × 10⁻⁶ T·m. Distractors: sign rule reversed, μ₀(+2.0 A) = +2.5 × 10⁻⁶ T·m; C included, μ₀(3.0 − 5.0 − 4.0) A = −7.5 × 10⁻⁶ T·m; magnitudes added, μ₀(8.0 A) = +1.0 × 10⁻⁵ T·m.
A long, straight wire carries a steady current I. A circular Amperian loop lies in a plane perpendicular to the wire, but the wire passes outside the loop, not through it. Which statement about the loop is correct?
Answer and reasoning
A∮B⃗ · dℓ⃗ = 0, although B⃗ is not zero at any point along the loopCorrect The loop encloses no current, so by Ampère’s law ∮B⃗ · dℓ⃗ = μ₀Ienc = 0. The wire’s field is not zero at any point of the loop; B⃗ · dℓ⃗ is positive along part of the loop and negative along the rest, and the two parts cancel.
B∮B⃗ · dℓ⃗ = 0, because B⃗ is zero at every point on the loop A student who reads zero enclosed current as zero field picks this. ∮B⃗ · dℓ⃗ = 0 says that the contributions around the loop cancel; the wire produces a field at every point of the loop.
C∮B⃗ · dℓ⃗ = μ₀I, because the wire’s field acts all along the loop A student who counts every nearby current as enclosed picks this. The wire passes outside the loop, so it is not enclosed: its field acts along the loop, but its contributions to ∮B⃗ · dℓ⃗ cancel.
D∮B⃗ · dℓ⃗ ≠ 0, since B⃗ is stronger on the side nearer the wire A student who expects the stronger field near the wire to make the integral nonzero picks this. Where the loop passes nearer the wire the field is stronger, but B⃗ · dℓ⃗ there has the opposite sign to the far part of the loop, and for a loop that does not enclose the wire the two cancel exactly.
Working The wire does not pass through the loop, so Ienc = 0 and ∮B⃗ · dℓ⃗ = 0. The wire’s field, μ₀I/(2πs) at distance s, is nonzero at every point of the loop. With B⃗ · dℓ⃗ = (μ₀I/(2π))dφ, where dφ is the angle a path element subtends at the wire, the angle swept along one part of the loop is swept back along the other, so the contributions cancel although B⃗ is stronger on the nearer part.
The figure shows an edge view of a very large, flat conducting slab that carries a current of uniform density J, with a point P and the relevant distances labeled. What is the magnitude of the magnetic field at P?
Answer and reasoning
A2μ₀Jy A student who lets only one long side of the loop contribute picks this: Bℓ = μ₀J(2yℓ). The field is present on both sides of the central plane, pointing opposite ways, so both long sides contribute along the direction of traversal.
Bμ₀Jd/2 A student who uses the slab’s whole current picks this: 2Bℓ = μ₀Jdℓ. That is the field outside the slab; a loop through P encloses only the current within a distance y of the central plane.
C0 A student who thinks the field inside any conductor is zero picks this. That holds for the electrostatic field in a conductor in equilibrium, not for the magnetic field of a current: a loop through P encloses current, so the field there is not zero.
Dμ₀JyCorrect Take a rectangular Amperian loop with long sides of length ℓ parallel to the slab, at distance y on either side of the central plane. By symmetry B⃗ has the same magnitude on both long sides and is parallel to each in the direction of traversal, and it is perpendicular to the short sides: ∮B⃗ · dℓ⃗ = 2Bℓ. The loop encloses J(2yℓ), so 2Bℓ = μ₀J(2yℓ) and B = μ₀Jy.
Working Symmetry: B⃗ is parallel to the slab’s faces and perpendicular to J⃗, has equal magnitude at equal distances from the central plane, and points in opposite directions on the two sides of it. Rectangular Amperian loop in the plane of the page, with long sides of length ℓ parallel to the slab at distance y on each side of the central plane: B⃗ is parallel to dℓ⃗ on both long sides and perpendicular to the short sides, so ∮B⃗ · dℓ⃗ = 2Bℓ. Ienc = J(2y)ℓ. 2Bℓ = μ₀J(2yℓ), so B = μ₀Jy. Distractors: one long side only, Bℓ = μ₀J(2yℓ) → 2μ₀Jy; whole slab’s current, 2Bℓ = μ₀Jdℓ → μ₀Jd/2; field taken as zero inside a conductor → 0.
A very long, solid cylindrical conductor of radius R carries a steady current along its length. The current density is J = J₀s/R, where s is the distance from the axis and J₀ is a constant. What is the magnitude of the magnetic field at a distance r < R from the axis?
Answer and reasoning
Aμ₀J₀r²/(2R) A student who treats the current density as uniform at its value at the loop, J₀r/R, picks this: Ienc = (J₀r/R)(πr²). The density is smaller nearer the axis, so the enclosed current must be added ring by ring.
Bμ₀J₀r²/(4R) A student who multiplies the area inside the loop, πr², by the mean of the density’s values at the axis and at the loop, J₀r/(2R), gets Ienc = πJ₀r³/(2R) and this field. The outer rings have more area than the inner ones and carry more of the current, so the density must be weighted by ring area: Ienc = ∫₀r J(s)2πs ds.
Cμ₀J₀r²/(3R)Correct A coaxial loop of radius r gives ∮B⃗ · dℓ⃗ = B(2πr). The enclosed current is the sum over thin rings of radius s, each carrying J(s)(2πs ds): Ienc = ∫₀r (J₀s/R)2πs ds = 2πJ₀r³/(3R). Then B = μ₀Ienc/(2πr) = μ₀J₀r²/(3R).
Dμ₀J₀r/(4πR) A student who integrates the current density along the radius, ∫₀r J₀s/R ds = J₀r²/(2R), picks this. That quantity is in A/m, not a current; each ring’s area, 2πs ds, is missing.
Working Coaxial circular Amperian loop of radius r: by symmetry B is tangent to the loop and constant on it, so ∮B⃗ · dℓ⃗ = B(2πr). Ienc = ∫₀r (J₀s/R)(2πs ds) = 2πJ₀r³/(3R). B = μ₀Ienc/(2πr) = μ₀J₀r²/(3R). Distractors (checked with sympy): J taken as uniform at its value at r, Ienc = (J₀r/R)(πr²) → μ₀J₀r²/(2R); mean of J at the axis and at the loop, J₀r/(2R), times πr² → μ₀J₀r²/(4R); ∫J ds = J₀r²/(2R) used as the current → μ₀J₀r/(4πR).
Two long, straight, parallel wires a distance d apart each carry current I in the same direction. Point P is a distance d from each wire, so that P and the two wires form an equilateral triangle in a plane perpendicular to the wires. What is the magnitude of the net magnetic field at P?
Answer and reasoning
A√3μ₀I/(2πd)Correct Each wire gives μ₀I/(2πd) at P, perpendicular to the line from that wire to P. The two field vectors make 60° with each other: their components along the perpendicular bisector of the wires cancel, and their components parallel to the line joining the wires add, each μ₀I/(2πd) × cos 30°. The net field is √3μ₀I/(2πd).
Bμ₀I/(πd) A student who adds the two field magnitudes picks this: 2 × μ₀I/(2πd). The fields at P point in different directions, 60° apart, so only parts of them add.
C2√3μ₀I/(3πd) A student who applies Ampère’s law with a circle through P centered midway between the wires picks this: the circle has radius √3d/2 and encloses 2I, so B(2π · √3d/2) = μ₀(2I) and B = 2√3μ₀I/(3πd). The two currents are not symmetric about that circle’s center, so B is not the same all around it and cannot be taken outside the integral; each wire’s field must be found separately and the vectors added.
D0 A student who extends the midpoint result to every point equidistant from the wires picks this. Only at the midpoint are the two fields opposite; at P they are 60° apart and their components parallel to the line joining the wires add.
Working Each wire’s field at P has magnitude μ₀I/(2πd) and is perpendicular to the line from that wire to P. Place the wires at (−d/2, 0) and (d/2, 0) with P at (0, √3d/2): the fields are along (−√3/2, 1/2) and (−√3/2, −1/2). The components along the perpendicular bisector cancel; the components parallel to the line joining the wires add, each (μ₀I/(2πd))cos 30°. B = 2(μ₀I/(2πd))(√3/2) = √3μ₀I/(2πd). Distractors: magnitudes added, μ₀I/(πd); Ampère’s law applied to a circle through P centered midway between the wires (radius √3d/2, Ienc = 2I), 2√3μ₀I/(3πd); midpoint cancellation extended to P, 0.
A long, straight wire carrying current 2I lies along the axis of a long, thin-walled conducting cylindrical shell of radius R. The shell carries current 5I, spread uniformly around it, in the opposite direction. What is the magnitude of the magnetic field at a distance r > R from the axis?
Answer and reasoning
A(μ₀/(2π))(7I/r) A student who adds the enclosed currents as magnitudes picks this: 2I + 5I. The currents are in opposite directions, so they partly cancel: the net enclosed current is 3I.
B(μ₀/(2π))(5I/r) A student who thinks the conducting shell blocks the wire’s magnetic field picks this, using the shell’s 5I alone. Outside the shell, the field depends on all the current enclosed, the wire’s included.
C(μ₀/(2π))(3I/r)Correct A coaxial circle of radius r > R encloses both the wire and the shell. The currents are opposite, so Ienc = 5I − 2I = 3I. By symmetry B is constant around the circle and tangent to it: B(2πr) = μ₀(3I), which gives (μ₀/(2π))(3I/r).
D(μ₀/(2π))(2I/R) A student who takes the shell itself as the Amperian loop picks this: a circle of radius R around the wire’s 2I. The loop must pass through the point where the field is wanted, at radius r, and there it encloses both currents.
Working Coaxial circular Amperian loop of radius r > R: by symmetry B is tangent to it and constant on it, so B(2πr) = μ₀Ienc. Taking the shell’s current direction as positive, Ienc = 5I − 2I = 3I. B = (μ₀/(2π))(3I/r). Distractors: magnitudes added, 7I; shell taken to block the wire’s field, 5I; loop taken along the shell itself (radius R), enclosing only the wire, (μ₀/(2π))(2I/R).
At a distance r from a long, straight wire carrying current I, the magnetic field has magnitude B₀. The current is tripled. What is the magnitude of the magnetic field at a distance 2r from the wire?
Answer and reasoning
A0.75B₀ A student who lets the field fall off as 1/r², like a point charge’s, picks this: 3 × (1/4). The loop around a long wire has length 2πr, so the field falls off as 1/r.
B3.00B₀ A student who reads Ampère’s law as B = μ₀Ienc, so that the distance does not matter, picks this. μ₀Ienc equals ∮B⃗ · dℓ⃗ = B(2πr), so the field is inversely proportional to r.
C6.00B₀ A student who thinks the field is stronger on a larger Amperian loop, in proportion to its length, picks this: 3 × 2. For every circle around the wire, B(2πr) = μ₀I, so the field on a larger circle is weaker.
D1.50B₀Correct Ampère’s law with a circle of radius r around the wire gives B = μ₀I/(2πr), so B is proportional to I/r. Tripling the current triples the field, and doubling the distance halves it: B = (3/2)B₀ = 1.50B₀.
Working B = μ₀I/(2πr), so B ∝ I/r. B′ = μ₀(3I)/(2π · 2r) = (3/2)B₀ = 1.50B₀. Distractors: 1/r² fall-off, (3/4)B₀ = 0.75B₀; field independent of distance (B = μ₀Ienc), 3.00B₀; field growing with the size of the loop (∝ r), 6.00B₀.
To derive the field inside a long solenoid, a rectangular Amperian loop is drawn with one long side inside the solenoid, parallel to its axis, and the opposite long side outside it; the two short sides cross the windings. Which statement about the contributions of the four sides to ∮B⃗ · dℓ⃗ is correct?
Answer and reasoning
ABoth long sides contribute equally, as the field outside matches the field inside. A student who thinks the field outside a long solenoid is as strong as the field inside picks this. For a long solenoid the field outside is negligible, so the outside side contributes nothing.
BAll four sides contribute, as each lies at least partly in the solenoid’s field. A student who counts every side that lies in a field picks this. B⃗ · dℓ⃗ keeps only the component of B⃗ along the path, and inside the solenoid B⃗ is perpendicular to the short sides, so they add nothing.
COnly the long side inside contributes, as B⃗ is parallel to it there.Correct Inside the solenoid the field is uniform and parallel to the axis, so along the inside long side B⃗ · dℓ⃗ = B dℓ and that side contributes Bℓ. Outside, the field is negligible. On the short sides, B⃗ is perpendicular to dℓ⃗ inside the solenoid and negligible outside, so they contribute nothing.
DOnly the short sides contribute, as B⃗ is perpendicular to them inside. A student who mixes up the dot and cross products picks this. B⃗ · dℓ⃗ = B dℓ cos θ is zero where B⃗ is perpendicular to the path and largest where it is parallel, which is along the inside long side.
The figure shows a cross section of a long solenoid carrying a steady current, with its data labeled, and a rectangular Amperian loop, one long side of which is inside the solenoid. The windings are drawn schematically, not to scale. What is the magnitude of the magnetic field inside the solenoid? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A2.3 × 10⁻² T A student who counts all 600 turns as enclosed by the loop picks this: Bℓ = μ₀(600)I. The loop encloses only the turns along its own length, nℓ = 150 of them.
B2.8 × 10⁻³ T A student who thinks the outside side contributes as much as the inside side picks this: 2Bℓ = μ₀nℓI. For a long solenoid the field outside is negligible, so only the inside side contributes.
C2.3 × 10⁻³ T A student who uses the total number of turns for n picks this: μ₀NI = (4π × 10⁻⁷)(600)(3.0). That product is in T·m, not T; n = N/L is the number of turns per meter.
D5.7 × 10⁻³ TCorrect The turns per length are n = 600/(0.40 m) = 1.5 × 10³ m⁻¹, so the loop encloses nℓ = 150 turns. Only the inside long side contributes: Bℓ = μ₀(nℓ)I, so B = μ₀nI = (4π × 10⁻⁷ T·m/A)(1.5 × 10³ m⁻¹)(3.0 A) = 5.7 × 10⁻³ T.
Working n = N/L = 600/(0.40 m) = 1.5 × 10³ m⁻¹. The loop encloses nℓ = 150 turns, each carrying I, and only its inside long side contributes: Bℓ = μ₀(nℓ)I, so B = μ₀nI = (4π × 10⁻⁷ T·m/A)(1.5 × 10³ m⁻¹)(3.0 A) = 5.7 × 10⁻³ T. Distractors: all 600 turns taken as enclosed, B = μ₀NI/ℓ = 2.3 × 10⁻² T; outside side counted equally, 2Bℓ = μ₀nℓI → 2.8 × 10⁻³ T; n taken as N, μ₀NI = 2.3 × 10⁻³ (in T·m, not T).
A long solenoid with N turns and length L has a field of magnitude B₀ inside it when the current is I. A second long solenoid has the same number of turns N spread over a length 2L and twice the radius, and it carries the same current I. What is the magnitude of the field inside the second solenoid?
Answer and reasoning
A0.50B₀Correct The field inside a long solenoid is B = μ₀nI, where n is the number of turns per unit length. Spreading the same N turns over twice the length halves n, and the radius does not appear, so B = 0.50B₀.
B1.00B₀ A student who takes n as the total number of turns picks this, since N and I are unchanged. n is turns per unit length, which is halved when the same turns are spread over twice the length.
C0.25B₀ A student who pictures the field inside as a wire’s field, weaker farther from the windings, picks this: halving n, then halving again for the doubled radius. Inside a long solenoid the field is uniform and does not depend on the radius.
D2.00B₀ A student who thinks a wider solenoid holds a stronger field picks this: n halved, area quadrupled. B = μ₀nI does not depend on the cross-sectional area.
Working B = μ₀nI with n = N/L. For the second solenoid n′ = N/(2L) = n/2, and the radius does not enter: B′ = B₀/2 = 0.50B₀. Distractors: n taken as N, 1.00B₀; field also taken to fall with the radius, like a wire’s, (1/2)(1/2) = 0.25B₀; field taken to grow with the cross-sectional area, (1/2)(4) = 2.00B₀.
Two long, straight, parallel wires carry equal currents in the same direction. A student claims that the magnetic field is zero at point P, midway between the wires. Which reasoning correctly supports the claim?
Answer and reasoning
AA small Amperian loop drawn around P encloses no current, so Ampère’s law makes the field on it and at P zero. A student who reads zero enclosed current as zero field gives this reasoning. A loop that encloses no current has ∮B⃗ · dℓ⃗ = 0, but the field along it need not be zero; the same argument would ‘prove’ B = 0 at any point not inside a wire.
BThe wires pull on each other with forces of equal size, and these two opposite forces cancel each other out at P. A student who treats the forces between the wires as the field at P gives this reasoning. The forces act on the wires, not at P, and a third-law pair acts on two different objects; the field at P is the vector sum of the two wires’ fields.
CEach wire’s field points straight away from that wire, so at P the two fields point in opposite directions. A student who pictures a wire’s magnetic field as radial, like a line charge’s electric field, gives this reasoning. The claim is right, but the reason is not: a wire’s field circles the wire, and at P the two fields are opposite because of the right-hand rule.
DThe wires’ fields at P are equally strong, and the right-hand rule shows that they point in opposite directions.Correct P is the same distance from both wires, so their fields there have equal magnitudes, μ₀I/(2πr). Applying the right-hand rule to each wire shows that at P the two fields point in opposite directions, so their vector sum is zero.
A long, straight, solid cylindrical wire of radius 2.0 mm carries a steady current of 8.0 A spread uniformly over its cross section. What is the magnitude of the magnetic field inside the wire, 1.0 mm from its axis? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A1.6 × 10⁻³ T A student who uses the wire’s whole current, 8.0 A, picks this. A loop of radius 1.0 mm lies inside the wire and encloses only the current within that radius.
B2.0 × 10⁻⁴ T A student who takes the wire’s surface as the Amperian loop picks this, dividing μ₀(2.0 A) by 2π(2.0 mm). The loop must pass through the point, so its length is 2π(1.0 mm).
C4.0 × 10⁻⁴ TCorrect A coaxial loop of radius 1.0 mm encloses the fraction (1.0/2.0)² = 1/4 of the uniformly spread current, 2.0 A. By symmetry B(2πr) = μ₀Ienc, so B = (4π × 10⁻⁷ T·m/A)(2.0 A)/(2π × 1.0 × 10⁻³ m) = 4.0 × 10⁻⁴ T.
D8.0 × 10⁻⁴ T A student who takes the enclosed current as proportional to r picks this: half the radius, half the current, 4.0 A. With a uniform current density the enclosed current is proportional to the enclosed area, so it is one quarter of 8.0 A.
Working Coaxial circular Amperian loop of radius r = 1.0 mm. Uniform current density: Ienc = I(r²/R²) = (8.0 A)(1.0 mm/2.0 mm)² = 2.0 A. B(2πr) = μ₀Ienc, so B = (4π × 10⁻⁷ T·m/A)(2.0 A)/(2π × 1.0 × 10⁻³ m) = 4.0 × 10⁻⁴ T. Distractors: whole current, μ₀(8.0 A)/(2π × 1.0 × 10⁻³ m) = 1.6 × 10⁻³ T; loop taken as the wire’s surface, μ₀(2.0 A)/(2π × 2.0 × 10⁻³ m) = 2.0 × 10⁻⁴ T; enclosed current taken as proportional to r (4.0 A), 8.0 × 10⁻⁴ T.
A very long, solid cylindrical conductor of radius R has a long cylindrical hole of radius a running parallel to its axis. The hole’s axis is a distance d from the conductor’s axis, with a < d and d + a < R, so the conductor’s axis lies in the metal. The metal carries a steady current of uniform density J along its length; the hole carries no current. What is the magnitude of the magnetic field on the conductor’s axis?
Answer and reasoning
Aμ₀Ja²/(2d)Correct Treat the conductor as a solid cylinder with density J plus a cylinder of density −J filling the hole. On its own axis the solid cylinder’s field is zero. The conductor’s axis lies outside the hole piece, a distance d from its axis, where Ampère’s law gives μ₀(Jπa²)/(2πd) = μ₀Ja²/(2d).
B0 A student who applies B = μ₀Ienc/(2πr) to a small circle about the conductor’s axis, with Ienc = Jπr², picks this: the field would vanish on the axis, as in a solid wire. With the hole off center, the current is not symmetric about that axis, so B is not the same all around such a circle and that step fails.
Cμ₀J(R − a)/2 A student who takes each Amperian loop along a boundary, the conductor’s surface (μ₀JR/2) for the solid cylinder and the hole’s edge (μ₀Ja/2) for the hole, picks this. Each loop must pass through the point on the axis, so its radius is 0 for the solid cylinder and d for the hole piece.
Dμ₀Ja²/(2d²) A student who takes the hole piece’s field outside it to fall off as 1/r², like a point charge’s field, picks this. Ampère’s law with a circular loop of length 2πd gives a 1/d fall-off.
Working Superposition: the conductor equals a solid cylinder of radius R carrying uniform density J plus a cylinder of radius a (filling the hole) carrying density −J. Each piece is symmetric about its own axis, so Ampère’s law applies to each separately. Solid cylinder: a loop of radius s about its axis encloses Jπs², so B = μ₀Js/2, which is zero on its axis (s = 0). Hole cylinder: the conductor’s axis is a distance d > a from the hole’s axis, outside that piece; a loop of radius d about the hole’s axis encloses Jπa², so B(2πd) = μ₀Jπa² and B = μ₀Ja²/(2d). Net field on the axis: μ₀Ja²/(2d). Units: (T·m/A)(A/m²)(m) = T. Distractors (checked with sympy; with J = a = 1, d = 2, R = 4 and μ₀ = 1 the options give 0.25 (key), 0, 1.5, 0.125): B = μ₀Ienc/(2πr) applied to a small circle about the conductor’s axis, Ienc = Jπr² → μ₀Jr/2 → 0 at r = 0; loops taken along the boundaries, radius R for the cylinder (μ₀JR/2) and a for the hole (μ₀Ja/2) → μ₀J(R − a)/2; 1/r² fall-off for the hole piece → μ₀Ja²/(2d²).
A long coaxial cable consists of a solid inner conductor and a coaxial conducting tube of inner radius R and outer radius 2R. The inner conductor carries a steady current I in one direction, and the tube carries a current I in the opposite direction; in each conductor the current is spread uniformly over its cross section. What is the magnitude of the magnetic field inside the tube’s wall, at a distance 3R/2 from the axis?
Answer and reasoning
A17μ₀I/(36πR) A student who adds the enclosed currents as magnitudes picks this: I + 5I/12 = 17I/12. The tube’s current is opposite to the inner conductor’s, so the enclosed part of it must be subtracted.
B7μ₀I/(36πR)Correct A coaxial loop of radius 3R/2 encloses the inner current I and part of the tube’s opposite current. That part is in proportion to area: π((3R/2)² − R²)/(π((2R)² − R²)) = 5/12. So Ienc = 7I/12, and B(2π · 3R/2) = μ₀(7I/12) gives this expression.
Cμ₀I/(6πR) A student who takes the enclosed part of the tube’s current in proportion to the wall thickness inside the loop, (R/2)/R = 1/2, picks this: Ienc = I/2. The current is spread over the wall’s cross-sectional area, and the loop encloses 5/12 of that area.
D0 A student who counts the tube’s whole current, although the loop lies inside the tube’s wall, picks this: I − I = 0. Only the part of the tube’s current inside the loop is enclosed; a loop encloses the tube’s whole current only when its radius exceeds 2R.
Working Coaxial circular Amperian loop of radius r = 3R/2: by symmetry B is tangent to it and constant on it, so B(2πr) = μ₀Ienc. The loop encloses the inner conductor’s current I and the part of the tube’s current inside radius r. The tube’s current density is I/(π((2R)² − R²)) = I/(3πR²), and the wall area inside the loop is π((3R/2)² − R²) = 5πR²/4, so the enclosed part of the tube’s current is 5I/12, in the opposite direction. Ienc = I − 5I/12 = 7I/12. B = μ₀(7I/12)/(2π(3R/2)) = 7μ₀I/(36πR). Distractors (checked with sympy): currents added as magnitudes, I + 5I/12 = 17I/12 → 17μ₀I/(36πR); tube’s enclosed fraction taken as the fraction of the wall thickness, (R/2)/R = 1/2, so Ienc = I/2 → μ₀I/(6πR); tube’s whole current counted, I − I → 0.
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
