4 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 4
A circular loop of wire lies in the plane of the page and carries a counterclockwise current. Which statement correctly describes the magnetic field at the center of the loop?
Answer and reasoning
AIt points out of the page, since every element's contribution points out of the page.Correct For each element, dℓ⃗ points along the counterclockwise current and r̂ points from the element to the center. dℓ⃗ × r̂ points out of the page for every element, so all the contributions add, and the field at the center points out of the page, with magnitude μ₀I/(2R).
BIt points into the page, since every element's contribution points into the page. A student who takes the cross product in the wrong order, r̂ × dℓ⃗, or uses the left hand picks this. With the right-hand rule applied to dℓ⃗ × r̂, a counterclockwise current gives a field out of the page.
CIt is zero, since elements on opposite sides of the loop carry currents in opposite directions. A student who thinks opposite currents give opposite fields picks this. On opposite sides both dℓ⃗ and r̂ are reversed, so dℓ⃗ × r̂ is the same for every element and the contributions add.
DIt lies in the plane of the page, since each element's field points away from that element. A student who draws the magnetic field of each element pointing away from it, like an electric field, picks this. Each contribution is perpendicular to both dℓ⃗ and r̂, which both lie in the page, so it is perpendicular to the page.
A short straight segment of wire carries a current. Which statement correctly describes the magnetic field the segment produces at points around it?
Answer and reasoning
AIt points radially away from the segment, perpendicular to it. A student who pictures the magnetic field like the electric field of a charged wire picks this. The magnetic field has no component toward or away from the segment; it circles it.
BIt points along the segment, in the direction of the current. A student who thinks the field is carried along with the moving charges picks this. dℓ⃗ × r̂ is perpendicular to dℓ⃗, so the field has no component along the segment.
CIt leaves one side of the segment and enters the opposite side. A student who pictures the segment as a small magnet with a north and a south side picks this. The field lines around a current are closed circles, with no beginning or end on the wire.
DIt curls around the segment's line, in planes perpendicular to it.Correct Each contribution dB⃗ is along dℓ⃗ × r̂, which is perpendicular to the segment and to the line to the point. The field vectors are therefore tangent to circles centered on the segment's line, with no component toward, away from or along the segment.
A wire carrying current I includes a circular arc of radius R that subtends an angle θ (in radians) at its center P. The straight wires leading to and from the arc lie along lines through P. Using the Biot–Savart law, what is the magnitude of the magnetic field at P?
Answer and reasoning
Aμ₀Iθ/(4πR)Correct Every element of the arc is a distance R from P and perpendicular to the line to P, so each contributes dB = μ₀I dℓ/(4πR²), all in the same direction. The arc length is Rθ, so B = μ₀I(Rθ)/(4πR²) = μ₀Iθ/(4πR). The straight leads contribute nothing, because dℓ⃗ × r̂ = 0 for them.
Bμ₀I/(2R) A student who uses the full-loop result for any arc picks this. The arc is only a fraction θ/(2π) of a loop, so its field at P is that fraction of μ₀I/(2R): μ₀Iθ/(4πR).
Cμ₀I/(2πR) A student who applies the long-straight-wire result to a curved wire picks this. The arc's field must be found from the Biot–Savart law for its own shape.
Dμ₀Iθ/(4πR²) A student who integrates over the angle, writing dℓ as dθ, keeps the 1/R² of each element and picks this. The length of an element is R dθ, so one factor of R cancels; the units of this option, T/m, show the error.
Working dB = (μ₀/4π)I dℓ sin 90°/R², all contributions along the same direction. ∫dℓ = Rθ → B = μ₀IRθ/(4πR²) = μ₀Iθ/(4πR). Leads: dℓ⃗ ∥ r̂ → 0. Check: θ = 2π gives μ₀I/(2R).
The diagram shows a rectangular loop carrying current I. Part of the loop is in a region of uniform magnetic field B⃗ directed into the page; the rest of the loop is in a region with no field. What is the direction of the net magnetic force on the loop?
Answer and reasoning
ATo the left, out of the field regionCorrect Only the parts of the loop inside the field feel a force. On the right side the current is up the page, and I dℓ⃗ × B⃗ = (up) × (into page) points to the left. The parts of the top and bottom sides inside the field carry opposite currents over equal lengths, so their forces (down and up) cancel. The net force is to the left.
BTo the right, farther into the field region A student who takes the cross product in the wrong order, B⃗ × dℓ⃗, or uses the left hand gets the opposite direction for the force on the right side. With dℓ⃗ up and B⃗ into the page, dℓ⃗ × B⃗ points to the left.
CNone, since the loop is a closed circuit A student who applies 'the net force on a closed loop is zero' without its condition picks this. That result holds in a uniform field; here the left side is outside the field, so nothing balances the force on the right side.
DInto the page, along the magnetic field A student who thinks the magnetic force acts along the field lines picks this. The force on each element, I dℓ⃗ × B⃗, is perpendicular to B⃗, so it lies in the plane of the page.
Working Right side: dℓ⃗ = +ŷ, B⃗ = −ẑ: ŷ × (−ẑ) = −x̂ → left. Top (in field): −x̂ × (−ẑ) = −ŷ; bottom: x̂ × (−ẑ) = +ŷ; equal lengths cancel. Left side: no field. Net: to the left, magnitude ILB with L the right side's length.
In preparation: 0 of 4 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
12.3.A.1 Biot–Savart law Fix
Biot–Savart law
The law giving the magnetic field dB⃗ produced at a point by a short current element I dℓ⃗: dB⃗ = (μ₀/4π)I(dℓ⃗ × r̂)/r², where r is the distance from the element to the point and r̂ the unit vector from the element toward the point. The field of a whole conductor is the vector sum (integral) of the contributions of all its elements.
Current element, I dℓ⃗
A short length dℓ of a current-carrying wire, taken as a vector in the direction of the conventional current, multiplied by the current I. SI unit: ampere meter (A·m).
Magnitude and direction of dB⃗
|dB⃗| = (μ₀/4π)I dℓ sin θ/r², where θ is the angle between dℓ⃗ and r̂; it is largest for points perpendicular to the element and zero on the element's own line. The direction, dℓ⃗ × r̂, is perpendicular to both the element and the line to the point (right-hand rule).
Students often think The field of a current element depends on the component of the element along the line to the point (dℓ cos θ), so it is largest straight ahead of the element and zero beside it. In fact No. dB⃗ is proportional to dℓ⃗ × r̂, whose magnitude is dℓ sin θ, where θ is the angle between the current element and the line to the point. The field is zero at points on the line of the element and largest at points in the direction perpendicular to it.
Students often think The field of a current element at a given distance has magnitude (μ₀/4π)I dℓ/r² in every direction, as if the cross product were simply the product of the magnitudes. In fact Yes. Its magnitude is (μ₀/4π)I dℓ sin θ/r², so at a fixed distance r it is largest in the direction perpendicular to the element and falls to zero along the element's own line.
12.3.A.2 Field lines around a wire segment Fix
Field lines around a wire segment
The magnetic field vectors produced by a small segment of current-carrying wire are tangent to circles centered on the line of the wire, in planes perpendicular to it. The field has no component toward, away from or parallel to the segment.
Students often think The magnetic field of a current-carrying wire points radially away from (or toward) the wire, like the electric field of a charged wire. In fact No. The magnetic field vectors around a segment are tangent to circles centered on the wire, so the field has no component toward or away from the wire. Magnetic field lines around a wire close on themselves.
Students often think The magnetic field of a current-carrying wire points along the wire, in the direction of the current, as if carried along by the moving charges. In fact No. dB⃗ is perpendicular to dℓ⃗, so the field has no component parallel to the segment that produces it. The field lines circle the wire.
12.3.A.3 Field at the center of a circular loop Fix
Field at the center of a circular loop
B = μ₀I/(2R) for a loop of radius R carrying current I, directed along the loop's axis (right-hand rule: fingers along the current, thumb along the field). Every element contributes in the same direction. SI unit: tesla (T).
Field at the center of a circular arc
An arc of radius R subtending an angle θ (in radians) at its center produces B = μ₀Iθ/(4πR) at the center: a fraction θ/(2π) of the full-loop value. Straight lead wires that lie along lines through the center contribute nothing there, because dℓ⃗ × r̂ = 0 for them.
Field on the axis of a circular loop
At a point on the central axis a distance x from the center of a loop of radius R, the components perpendicular to the axis cancel by symmetry and B = μ₀IR²/(2(R² + x²)3/2), directed along the axis; at x = 0 this gives μ₀I/(2R).
Field on the perpendicular bisector of a straight segment
For a straight segment of length 2a carrying current I, at a distance y from its midpoint on its perpendicular bisector, B = μ₀Ia/(2πy√(a² + y²)). For a very long segment (a much greater than y) this approaches μ₀I/(2πy).
Students often think The field at the center of any circular arc is μ₀I/(2R), the full-loop value, whatever fraction of a circle the arc is. In fact No. μ₀I/(2R) is the field at the center of a complete loop. Every element of an arc contributes the same amount, μ₀I dℓ/(4πR²), so an arc subtending an angle θ gives μ₀Iθ/(4πR): a semicircle gives μ₀I/(4R) and a quarter circle μ₀I/(8R).
Students often think The field of any current-carrying wire, whatever its shape or length, is μ₀I/(2πr) at a distance r from it. In fact No. μ₀I/(2πr) is the field of a very long straight wire. The field of a finite segment, a circular arc or a loop has to be found from the Biot–Savart law for that shape: for example, μ₀I/(2R) at the center of a circular loop and less than μ₀I/(2πy) near a finite segment.
12.3.B.1 Magnetic force on a current-carrying wire Fix
Magnetic force on a current-carrying wire
F⃗B = ∫I(dℓ⃗ × B⃗): each element of the wire feels a force perpendicular to both the element and the field. For a straight wire of length L in a uniform field, |F⃗B| = ILB sin θ, where θ is the angle between the wire and the field. SI unit: newton (N).
Force on a closed loop in a uniform field
In a uniform magnetic field the forces on the elements of any closed current loop add to zero (the loop may still experience a torque). If the field differs over the loop, the net force is generally not zero.
Students often think The force on a rod in a magnetic field is ILB, with B taken at the nearest point, even when the field varies along the rod. In fact No. When the field varies along the rod, the force must be found by integrating, F = ∫I B dℓ, because each element feels the field at its own position. Using the field at the nearest point overestimates the force when the field weakens along the rod.
Students often think The magnetic field of a long straight wire falls off as 1/r², as the field of a point charge or a single current element does. In fact No. The field of a long straight wire is μ₀I/(2πr), inversely proportional to the distance. The 1/r² dependence belongs to a single short current element, not to a long wire made of many elements.
10 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 10
A straight piece of wire 1.0 mm long carries a current of 10 A. Point P is 5.0 cm from the piece of wire, and the line from the wire to P makes an angle of 30° with the direction of the current. Treating the piece of wire as a single current element, what is the magnitude of the magnetic field it produces at P? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A3.5 × 10⁻⁷ T A student who uses the cosine of the angle between the element and the line to P gets 3.5 × 10⁻⁷ T. The Biot–Savart law contains dℓ⃗ × r̂, whose magnitude uses sin θ.
B2.0 × 10⁻⁷ TCorrect Biot–Savart law for one element: dB = (μ₀/4π)I dℓ sin θ/r² = (1.0 × 10⁻⁷ T·m/A)(10 A)(1.0 × 10⁻³ m)(sin 30°)/(0.050 m)² = 2.0 × 10⁻⁷ T.
C4.0 × 10⁻⁷ T A student who leaves out the angle factor, as if P were perpendicular to the element, gets 4.0 × 10⁻⁷ T. At 30° the field is only sin 30° = 0.50 of that value.
D1.0 × 10⁻⁸ T A student who divides by r instead of r², as for a long straight wire, gets 1.0 × 10⁻⁸. The field of a single current element is proportional to 1/r².
Working dB = (μ₀/4π)I dℓ sin θ/r² = 1e-7 × 10 × 1.0e-3 × 0.50/(0.050)² = 2.0 × 10⁻⁷ T. (cos 30°: 3.5 × 10⁻⁷; no angle: 4.0 × 10⁻⁷; 1/r: 1.0 × 10⁻⁸.)
A student claims that the magnetic field produced by a short straight segment of current-carrying wire has no component parallel to the segment at any point. Which reasoning correctly supports the claim?
Answer and reasoning
AThe field points radially away from the segment, and a radial direction is perpendicular to the segment. A student who thinks the magnetic field points away from the wire, like the electric field of a charged wire, picks this. The conclusion is true, but the premise is false: the field circles the segment and has no radial component.
BThe field runs from one side of the segment across to the other, like the field between a magnet's poles. A student who pictures the segment as having a north side and a south side picks this. A current-carrying segment has no poles; its field lines are closed circles around it.
CEach element's field is perpendicular to both the element itself and the line from it to the point.Correct By the Biot–Savart law, every element of the segment contributes dB⃗ ∝ dℓ⃗ × r̂. A cross product is perpendicular to each of its factors, so every contribution, and therefore their sum, is perpendicular to dℓ⃗, which lies along the segment.
DParts of the segment ahead of and behind the point produce parallel components that cancel each other. A student who assumes a missing component must come from cancellation picks this. No element produces a parallel component in the first place, since dℓ⃗ × r̂ is perpendicular to dℓ⃗; even a single element, with nothing to cancel it, has none.
A circular loop of radius R carries current I. Point P is on the loop's central axis, a distance x from the center, so every point of the loop is a distance r = √(R² + x²) from P. Using the Biot–Savart law, what is the magnitude of the magnetic field at P?
Answer and reasoning
Aμ₀IR/(2r²) A student who adds the magnitudes of all the contributions, (μ₀I/(4πr²))(2πR), picks this. The contributions point in different directions; only their axial components, a fraction R/r of each, add.
Bμ₀IR²/(2r³)Correct Each element is perpendicular to the line to P, so it contributes dB = μ₀I dℓ/(4πr²), directed perpendicular to that line. By symmetry the components perpendicular to the axis cancel in pairs; the axial component of each is dB(R/r). Adding: B = (μ₀I/(4πr²))(R/r)(2πR) = μ₀IR²/(2r³).
Cμ₀I/(2r) A student who takes the center result μ₀I/(2R) and replaces R with the distance r picks this. Off the plane of the loop the contributions are tilted and partly cancel; the result falls off as 1/r³ far from the loop, not as 1/r.
Dμ₀I/(2πr) A student who treats the loop as if it were a long straight wire a distance r away picks this. The field of a loop must be found from the Biot–Savart law for its own shape.
Working dB = μ₀I dℓ/(4πr²) (dℓ⃗ ⊥ r̂). Axial component: dB cos α with cos α = R/r. B = ∮(μ₀I/(4πr²))(R/r)dℓ = μ₀IR(2πR)/(4πr³) = μ₀IR²/(2r³) = μ₀IR²/(2(R² + x²)3/2). x = 0 → μ₀I/(2R) ✓.
A straight wire segment 0.30 m long carries a current of 4.0 A. Point P lies on the perpendicular bisector of the segment, 0.20 m from its midpoint. What is the magnitude of the magnetic field at P due to the segment? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A4.0 × 10⁻⁶ T A student who uses the long-straight-wire result μ₀I/(2πy) picks this. The segment is finite, and the parts far from the midpoint contribute less than an infinite wire's would, so the field is smaller.
B3.0 × 10⁻⁶ T A student who treats the whole 0.30 m segment as one current element at its midpoint gets (μ₀/4π)I(0.30 m)/(0.20 m)² = 3.0 × 10⁻⁶ T. The segment is not short compared with 0.20 m; its elements are at different distances and angles, so the law must be integrated.
C1.2 × 10⁻⁶ T A student who integrates from the midpoint to one end only (0 to a) gets half the field, 1.2 × 10⁻⁶. The integral must cover the whole segment, from −a to +a.
D2.4 × 10⁻⁶ TCorrect With x measured from the midpoint, B = (μ₀I/4π)∫ y dx/(x² + y²)3/2 from x = −a to +a = μ₀Ia/(2πy√(a² + y²)). With a = 0.15 m and y = 0.20 m, √(a² + y²) = 0.25 m, so B = (2.0 × 10⁻⁷ T·m/A)(4.0 A)(0.15 m)/((0.20 m)(0.25 m)) = 2.4 × 10⁻⁶ T.
Working a = 0.15 m, y = 0.20 m. B = μ₀Ia/(2πy√(a² + y²)) = 4πe-7 × 4.0 × 0.15/(2π × 0.20 × 0.25) = 2.4 × 10⁻⁶ T. (Long wire: 4.0 × 10⁻⁶; single element: 3.0 × 10⁻⁶; half the wire: 1.2 × 10⁻⁶.)
The diagram shows a closed loop made of two semicircular arcs with a common center P, joined by two straight segments that lie along a line through P. The loop carries a steady current. What is the magnitude of the magnetic field at P? Use μ₀ = 4π × 10⁻⁷ T·m/A.
Answer and reasoning
A1.9 × 10⁻⁵ T A student who adds the magnitudes of the two arcs' fields gets 1.26 × 10⁻⁵ T + 6.3 × 10⁻⁶ T = 1.9 × 10⁻⁵ T. The current circulates in opposite senses around P in the two arcs, so their fields point in opposite directions.
B1.3 × 10⁻⁵ T A student who uses the full-loop result μ₀I/(2R) for each semicircle gets 2.5 × 10⁻⁵ T − 1.26 × 10⁻⁵ T = 1.3 × 10⁻⁵ T. A semicircle gives half the full-loop value, μ₀I/(4R).
C6.3 × 10⁻⁶ TCorrect Each semicircle contributes μ₀I/(4R) at P, and the straight segments contribute nothing, because they point along lines through P. The current goes around the outer arc and the inner arc in opposite senses, so their fields at P are opposite: B = μ₀I/(4R₁) − μ₀I/(4R₂) = 1.26 × 10⁻⁵ T − 6.3 × 10⁻⁶ T = 6.3 × 10⁻⁶ T, in the direction of the inner arc's field.
D4.0 × 10⁻⁶ T A student who uses the long-straight-wire result μ₀I/(2πR) for each arc gets 8.0 × 10⁻⁶ T − 4.0 × 10⁻⁶ T = 4.0 × 10⁻⁶ T. Arcs must be treated with the Biot–Savart law: each element is a distance R from P and perpendicular to the line to P.
Working Inner: μ₀I/(4R₁) = 4πe-7 × 4.0/(4 × 0.10) = 1.257 × 10⁻⁵ T. Outer: μ₀I/(4R₂) = 6.28 × 10⁻⁶ T, opposite sense. Straight parts along radial lines: 0. Net: 6.3 × 10⁻⁶ T.
The diagram shows three wires, (1), (2) and (3), each carrying the same current, with the center P of each circular part marked. The straight lead wires lie along lines through P. B₁, B₂ and B₃ are the magnitudes of the magnetic field at P for wires (1), (2) and (3). Which ranking is correct? Treat the gap in wire (1) and the separation of its two leads as negligible.
Answer and reasoning
AB₁ > B₂ > B₃ A student who thinks the field depends only on how much wire is in the arc ranks by arc length: 2πR, πR/2, πR/4. Each element's contribution also depends on its distance from P: halving the radius doubles the field of a given fraction of a circle.
BB₂ = B₃ > B₁ A student who uses μ₀I/(2r) for every arc, whatever fraction of a circle it is, gets μ₀I/(2R), μ₀I/R and μ₀I/R. A semicircle gives half, and a quarter circle a quarter, of the full-loop value.
CB₂ > B₁ = B₃ A student who takes the field to be proportional to θ/r², applying the 1/r² of each element to the whole arc, gets 2π/R², 4π/R² and 2π/R². The arc length rθ grows with r, so the field of an arc is proportional to θ/r.
DB₁ = B₂ > B₃Correct At the center of an arc of radius r subtending angle θ, B = μ₀Iθ/(4πr); the radial leads contribute nothing. (1): θ = 2π, r = R: μ₀I/(2R). (2): θ = π, r = R/2: μ₀I/(4(R/2)) = μ₀I/(2R). (3): θ = π/2, r = R/2: μ₀I/(8(R/2)) = μ₀I/(4R). So B₁ = B₂ > B₃.
A single circular loop of wire carries current I and produces a magnetic field of magnitude B₀ at its center. The same length of wire is rewound into a flat circular coil of two turns and carries the same current I. What is the magnitude of the magnetic field at the center of the coil?
Answer and reasoning
A2B₀, as there are two turns and the radius does not matter A student who counts the turns but ignores the change in radius picks this. The field at the center is inversely proportional to the radius, and the rewound turns have half the radius, which doubles the field again.
B4B₀, as there are two turns, each with half the radiusCorrect Two turns made from the same length of wire have half the radius: N × 2πR is unchanged. At the center of a coil, B = Nμ₀I/(2R): doubling N doubles the field and halving R doubles it again, so B = 4B₀.
C8B₀, as each of the two turns gives four times the field A student who thinks the field of a turn is proportional to 1/R² gives each half-size turn four times the field. The field at the center of a loop is μ₀I/(2R), inversely proportional to R, because a smaller loop also has less wire.
DB₀, as it is the same wire carrying the same current A student who thinks the field is set by the current and the amount of wire picks this. Winding the wire more tightly brings every element closer to the center, which increases the field.
The diagram shows a long straight wire carrying current I₁ and a straight rod, perpendicular to the wire and in the same plane, carrying current I₂. The long wire produces a magnetic field of magnitude μ₀I₁/(2πr) at a distance r from it. The magnitude of the magnetic force on the rod can be written F = (μ₀I₁I₂/(2π))G. What is G?
Answer and reasoning
Aln(L/a) A student who integrates dr/r from r = a to r = L, taking the rod's length as the position of its far end, picks this. The far end of the rod is at a distance a + L from the wire.
BL/a A student who uses the field at the rod's nearer end, μ₀I₁/(2πa), for the whole rod gets F = (μ₀I₁I₂/(2π))(L/a). The field weakens along the rod, so the force must be integrated; the result, ln(1 + L/a), is less than L/a.
Cln(1 + L/a)Correct The field of the long wire is perpendicular to the plane, and it changes along the rod. An element dr of the rod at distance r feels dF = I₂B dr = (μ₀I₁I₂/(2π))(dr/r), all in the same direction (parallel to the long wire). Integrating from r = a to r = a + L gives F = (μ₀I₁I₂/(2π))ln((a + L)/a), so G = ln(1 + L/a).
DL/(a(a + L)) A student who takes the long wire's field to fall off as 1/r² integrates dr/r² from a to a + L and gets 1/a − 1/(a + L) = L/(a(a + L)). The field given is proportional to 1/r, so the integral is a logarithm; this option also has units of 1/m, while G must have none.
Working dF = I₂ B(r) dr = (μ₀I₁I₂/(2π)) dr/r (I₂ dr⃗ ⊥ B⃗; all dF parallel to the wire). F = (μ₀I₁I₂/(2π))∫ₐa+L dr/r = (μ₀I₁I₂/(2π))ln((a + L)/a) → G = ln(1 + L/a). (Limits a→L: ln(L/a); uniform B at a: L/a; 1/r²: L/(a(a + L)).)
A square loop of wire with sides of length s carries a steady current I. Point P is at the center of the square. Using the Biot–Savart law, what is the magnitude of the magnetic field at P?
Answer and reasoning
A(4/π)μ₀I/s A student who uses the long-straight-wire field μ₀I/(2πr) with r = s/2 for each side gets 4 × μ₀I/(πs). Each side is only s long, so its field at P is smaller than a long wire's: √2μ₀I/(2πs) rather than μ₀I/(πs).
B(2√2/π)μ₀I/sCorrect P lies on the perpendicular bisector of each side, s/2 from it. Integrating along one side from −s/2 to +s/2 gives B₁ = (μ₀I/(4πy))·2a/√(a² + y²) with a = y = s/2, which is √2μ₀I/(2πs). For every side dℓ⃗ × r̂ points the same way at P, so the four contributions add: B = 4B₁ = (2√2/π)μ₀I/s.
C(√2/π)μ₀I/s A student who integrates along each side from its midpoint to one end only (0 to s/2) counts half of each side and gets half the field. The integral must cover the whole side, from −s/2 to +s/2.
D0 A student who thinks opposite sides cancel because their currents flow in opposite directions picks this. For opposite sides both the current direction and the direction from the side to P are reversed, so dℓ⃗ × r̂ is the same for both, and all four sides give fields in the same direction.
Working P is on the perpendicular bisector of each side, y = s/2 from it; each side runs from x = −s/2 to +s/2. One side: B₁ = (μ₀I/4π)∫ y dx/(x² + y²)3/2 from −s/2 to s/2 = (μ₀I/(4πy))·2a/√(a² + y²) with a = y = s/2, so B₁ = (μ₀I/(2πs))·√2 = √2μ₀I/(2πs). At P, dℓ⃗ × r̂ points the same way (perpendicular to the loop) for all four sides, so B = 4B₁ = 2√2μ₀I/(πs) = (2√2/π)μ₀I/s. Distractors: long-wire field at r = s/2 for each side → 4μ₀I/(πs); limits 0 to s/2 → half → (√2/π)μ₀I/s; opposite sides taken to cancel → 0.
A closed loop of wire is made of a circular arc of radius 0.10 m that subtends an angle of 240° at its center P, and the straight wire that joins the two ends of the arc. The loop carries a steady current of 5.0 A. Use μ₀ = 4π × 10⁻⁷ T·m/A. What is the magnitude of the magnetic field at P?
Answer and reasoning
A3.8 × 10⁻⁵ TCorrect The 240° arc gives (μ₀I/(4πR))θ = 2.09 × 10⁻⁵ T. The straight wire subtends 120°, so P lies on its perpendicular bisector, 0.050 m from it, and its half-length is 0.0866 m; integrating the Biot–Savart law along it gives 1.73 × 10⁻⁵ T. P is inside the loop, so the two contributions point the same way and add: 3.8 × 10⁻⁵ T.
B5.6 × 10⁻⁵ T A student who treats the straight wire as a single current element of length 0.173 m at its midpoint picks this: (μ₀/4π)(5.0)(0.173)/(0.050)² = 3.46 × 10⁻⁵ T, added to the arc's 2.09 × 10⁻⁵ T. The ends of the wire are farther from P and at an angle to the line to P, so they contribute less; integrating gives 1.73 × 10⁻⁵ T.
C3.0 × 10⁻⁵ T A student who integrates along the straight wire only from its midpoint to one end picks this, getting half of its 1.73 × 10⁻⁵ T. Both halves of the wire contribute equally at P, so the limits run from one end to the other.
D3.6 × 10⁻⁶ T A student who thinks that the straight wire's current, running the opposite way to the arc's on the other side of P, produces a field opposing the arc's picks this: 2.09 × 10⁻⁵ − 1.73 × 10⁻⁵ T. The current goes around the loop in one sense, so every part of the loop produces a field at P in the same direction, and the contributions add.
Working Arc: every element is a distance R from P and perpendicular to the line to P, so Barc = (μ₀I/(4πR))θ with θ = 240° = 4π/3 rad: Barc = (10⁻⁷)(5.0)(4.19)/0.10 = 2.09 × 10⁻⁵ T. Straight wire: it subtends the remaining 120° at P, so P is on its perpendicular bisector at distance d = R cos 60° = 0.050 m, and its half-length is a = R sin 60° = 0.0866 m. Biot–Savart, integrating along the wire from −a to a: B = (μ₀I/(4πd)) · 2a/√(a² + d²) = (10⁻⁷)(5.0)/(0.050) × 2(0.0866)/0.10 = 1.73 × 10⁻⁵ T. The current goes around the loop in one sense and P is inside the loop, so both contributions point the same way (both out of or both into the plane): B = 2.09 × 10⁻⁵ + 1.73 × 10⁻⁵ = 3.8 × 10⁻⁵ T. Distractors: straight wire as one element I(2a) at its midpoint, (μ₀/4π)I(2a)/d² = 3.46 × 10⁻⁵ → total 5.6 × 10⁻⁵ T; integral from midpoint to one end only, 0.87 × 10⁻⁵ → 3.0 × 10⁻⁵ T; the straight wire's field taken as opposing the arc's, 2.09 × 10⁻⁵ − 1.73 × 10⁻⁵ = 3.6 × 10⁻⁶ T.
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
