12.3 Magnetism and Current-Carrying Wires

1. AVector K
   A student who thinks the field points away from the wire, like an electric field, picks vector K. The field of a long straight wire has no component toward or away from the wire; it is tangent to circles centered on it.
2. BVector L
   A student who curls the fingers the wrong way picks vector L, which is tangent to the circle but clockwise. For a current out of the page, the field circles counterclockwise; at the point to the right of the wire it points toward the top of the page.
3. CVector M Correct
   With the right thumb along the current, out of the page, the fingers curl counterclockwise, so the field lines circle the wire counterclockwise. At the point below the wire, that direction is to the right, tangent to the circle, as vector M shows.
4. DVector N
   A student who thinks the field points along the current picks vector N, out of the page. The field has no component parallel to the wire; it lies in the plane of the page, tangent to circles around the wire.

CED 12.3.A.1.i · Read this in Fix

1. AEach wire is pushed away from the other wire.
   A student who applies 'like repels like' from charges and poles picks this. Currents in the same direction attract; it is antiparallel currents that repel.
2. BEach wire is pulled toward the other wire. Correct
   Each wire is in the other's magnetic field. The field of one wire at the other is perpendicular to it, and the right-hand rule, applied to the second wire's current and the first wire's field, gives a force directed toward the first wire. Parallel currents in the same direction attract.
3. CEach wire is pushed along its own length.
   A student who thinks the magnetic force pushes a wire along its current picks this. The force on each wire is perpendicular to its current, so it pushes the wires toward each other, not along their lengths.
4. DNeither wire exerts a magnetic force on the other.
   A student who thinks magnetic forces act only on magnetic materials picks this. Each current produces a magnetic field, and that field exerts a force on the moving charges in the other wire.

CED 12.3.B.1.ii · Read this in Fix

1. AB is inversely proportional to r².
   A student who expects an inverse-square law, like the electric field of a point charge, picks this. If B were proportional to 1/r², a graph of B against 1/r would curve upward, growing steeper; these points lie on a straight line through the origin.
2. BB is inversely proportional to r. Correct
   The points lie on a straight line through the origin, so B is proportional to 1/r: doubling 1/r, which halves r, doubles B. That is B inversely proportional to r, as B = (μ₀/(2π))(I/r) predicts.
3. CB decreases linearly as r increases.
   A student who sees that the line through the origin shows B inversely proportional to r, but takes "inversely proportional" to mean a linear decrease, picks this. A linear decrease would give a straight line on a graph of B against r, reaching zero at some distance; a straight line through the origin on a graph of B against 1/r means B × r is constant.
4. DB increases in direct proportion to r.
   A student who reads the horizontal axis as r rather than 1/r picks this. Larger values of 1/r mean smaller distances, so the rising line shows B increasing as the distance decreases.

CED 12.3.A.1.ii · Read this in Fix

1. A4.8 × 10⁻⁵ T Correct
   The distance from the central axis is the radius plus the distance from the surface: r = 0.005 m + 0.045 m = 0.050 m. Then B = (μ₀/(2π))(I/r) = (2 × 10⁻⁷ T·m/A)(12 A)/(0.050 m) = 4.8 × 10⁻⁵ T.
2. B5.3 × 10⁻⁵ T
   A student who measures r from the surface of the wire picks this: (2 × 10⁻⁷ T·m/A)(12 A)/(0.045 m). The distance in B = (μ₀/(2π))(I/r) is from the central axis, which is one radius, 0.005 m, farther away: r = 0.050 m.
3. C4.4 × 10⁻⁵ T
   A student who adds the whole diameter to the distance from the surface picks this, using r = 0.055 m. The axis is only one radius, 0.005 m, inside the surface, so r = 0.050 m.
4. D9.6 × 10⁻⁴ T
   A student who uses an inverse-square dependence picks this: (2 × 10⁻⁷ T·m/A)(12 A)/(0.050 m)². The field of a long straight wire is inversely proportional to r, not r².

Working r is measured from the central axis: r = (0.010 m)/2 + 0.045 m = 0.050 m. B = (μ₀/(2π))(I/r) = (4π × 10⁻⁷ T·m/A)(12 A)/(2π × 0.050 m) = 4.8 × 10⁻⁵ T.

CED 12.3.A.1.ii · Read this in Fix

1. AOut of the page at P, into the page at Q
   A student who applies the right-hand rule to the electrons' direction of motion picks this. The rule uses the conventional current, which is opposite to the electrons' drift, so both directions are reversed.
2. BUp the page at P, down the page at Q
   A student who thinks the field points directly away from the wire, like the electric field of a charged line, picks this. The field of a long straight wire is tangent to circles centered on the wire, with no component toward or away from it.
3. CLeftward at P and at Q, along the current
   A student who thinks the field points along the current picks this. The field has no component parallel to the wire; it circles the wire, into the page on one side and out of the page on the other.
4. DInto the page at P and out of the page at Q Correct
   The electrons drift to the right, so the conventional current is to the left. With the right thumb pointing left along the current, the fingers curl into the page above the wire and out of the page below it. The field circles the wire, so it points in opposite directions on opposite sides.

CED 12.3.A.1.iii · Read this in Fix

1. AOut of the page, along the axis Correct
   Curl the fingers of the right hand along the counterclockwise current; the thumb points out of the page. Every part of the loop produces a field at C directed along the loop's axis, out of the page, so the contributions add.
2. BInto the page, along the axis
   A student who curls the fingers the wrong way, or uses the left hand, picks this. For a counterclockwise current in the plane of the page, the right-hand rule gives a field out of the page at the center.
3. CZero, as opposite sides cancel
   A student who reasons that opposite sides carry opposite currents, so their fields cancel, picks this. The currents are opposite, but C lies on opposite sides of them: the top of the loop, with current to the left, and the bottom, with current to the right, both produce fields out of the page at C, so they add.
4. DAround the loop, with the current
   A student who thinks a current's magnetic field points along the current picks this, picturing the field circulating with it. Each part of the loop produces a field perpendicular to that part; at the center all these fields point along the axis, out of the page.

CED 12.3.A.1.iv · Read this in Fix

1. AToward the right side of the page
   A student who applies the right-hand rule with the wrong sense picks this, making each field circle clockwise. For currents out of the page the fields circle counterclockwise, and their sum at P points to the left.
2. BUpward, away from both wires
   A student who thinks each wire's field points directly away from it picks this, since the two radial fields would add to an upward field. The field of a wire is tangent to circles around it, with no component away from the wire.
3. CToward the left side of the page Correct
   Each wire's field at P is tangent to a circle centered on that wire, counterclockwise for a current out of the page. The left wire's field at P points up and to the left; the right wire's points down and to the left. The currents and distances are equal, so the vertical components cancel and the leftward components add.
4. DOut of the page, like the currents
   A student who thinks the field points along the current picks this. The field of a long straight wire has no component parallel to the wire; at P each wire's field lies in the plane of the page.

CED 12.3.A.1.v · Read this in Fix

1. A2.3 × 10⁻⁵ T
   A student who thinks opposite currents produce opposite fields between them picks this, subtracting: 3.0 × 10⁻⁵ T − 0.67 × 10⁻⁵ T. Between antiparallel currents the two fields point the same way, so they add.
2. B8.6 × 10⁻⁴ T
   A student who uses an inverse-square dependence picks this: (2 × 10⁻⁷)(6.0/0.040² + 2.0/0.060²). The field of a long straight wire is inversely proportional to r, not r².
3. C1.6 × 10⁻⁵ T
   A student who uses the 0.10 m separation of the wires as r for both fields picks this: (2 × 10⁻⁷)(6.0 + 2.0)/0.10. Each field must be found at P, using each wire's own distance to P: 0.040 m and 0.060 m.
4. D3.7 × 10⁻⁵ T Correct
   B_X = (2 × 10⁻⁷ T·m/A)(6.0 A)/(0.040 m) = 3.0 × 10⁻⁵ T and B_Y = (2 × 10⁻⁷ T·m/A)(2.0 A)/(0.060 m) = 0.67 × 10⁻⁵ T. By the right-hand rule, at a point between the wires both fields point toward the same side (toward the top of the page if X is on the left), so they add: 3.7 × 10⁻⁵ T.

Working B_X = (μ₀/(2π))(I_X/r_X) = (2 × 10⁻⁷ T·m/A)(6.0 A)/(0.040 m) = 3.0 × 10⁻⁵ T; B_Y = (2 × 10⁻⁷ T·m/A)(2.0 A)/(0.060 m) = 6.7 × 10⁻⁶ T. Take X on the left: X's current is out of the page, so its field circles counterclockwise and at P (to its right) points toward the top of the page. Y's current is into the page, so its field circles clockwise and at P (to its left) also points toward the top of the page. B = 3.0 × 10⁻⁵ T + 6.7 × 10⁻⁶ T = 3.7 × 10⁻⁵ T.

CED 12.3.A.1.v · Read this in Fix

1. AThe magnet attracts copper, as magnets attract all metals.
   A student who thinks magnets attract all metals picks this. The wire stays at rest while the switch is open, so the magnet exerts no noticeable force on the copper itself; a simple attraction also could not reverse when the current reverses.
2. BThe magnetic field exerts a force on the current in the wire. Correct
   The wire swings only when there is a current and only when the magnet is present, and the swing reverses when the current reverses. So the force comes from the magnetic field acting on the moving charges in the wire, and its direction depends on the direction of the current.
3. CThe field exerts a force on the wire only once it moves.
   A student who thinks a wire, like a single charge, must move through the field to have a magnetic force exerted on it picks this. The wire is at rest when the switch closes, yet it starts to swing: the field acts on the charges moving along the wire.
4. DCopper is not magnetic, so the magnet is not the cause.
   A student who thinks magnetic forces act only on magnetic materials picks this. With the magnet removed, the current alone does not make the wire swing, so the magnet is needed; its field exerts a force on the current.

CED 12.3.B.1 · Read this in Fix

1. AThe force on wire 1 is three times that on wire 2.
   A student who thinks the wire with the larger current exerts the larger force picks this. Wire 2's field at wire 1 is three times as strong, but wire 1 carries a third of the current, so the forces are equal, as a Newton's third law pair must be.
2. BWire 2, with the larger current, has three times the force.
   A student who applies F = IℓB with the same field at both wires picks this. Each wire is in the other's field: wire 2 carries 3I in the field of the current I, and wire 1 carries I in the field of 3I, so the forces are equal.
3. CThe forces on the two wires are equal in magnitude. Correct
   The force on a length ℓ of wire 2 is (3I)ℓB₁, where B₁ = μ₀ I/(2πd) is wire 1's field at wire 2; the force on wire 1 is Iℓ B₂, where B₂ = μ₀(3I)/(2πd). Both equal 3μ₀ I²ℓ/(2πd). The two forces are a Newton's third law pair, so they must be equal in magnitude.
4. DNeither wire has a magnetic force exerted on it.
   A student who thinks magnetic forces act only on magnetic materials picks this. Each current produces a magnetic field that exerts a force on the moving charges in the other wire, whatever the wires are made of.

CED 12.3.B.1.i · Read this in Fix

1. A0.30 N Correct
   Only the 0.25 m of wire inside the field has a force exerted on it, and the angle between the current and the field is 30°: F_B = IℓB sin θ = (4.0 A)(0.25 m)(0.60 T)(sin 30°) = 0.30 N.
2. B0.60 N
   A student who ignores the angle between the wire and the field picks this: (4.0 A)(0.25 m)(0.60 T) = 0.60 N. That is the force only if the current crossed the field at 90°; at 30° it is multiplied by sin 30° = 0.50.
3. C0.52 N
   A student who uses cos θ picks this: (4.0 A)(0.25 m)(0.60 T)(cos 30°) = 0.52 N. The force depends on sin θ; it is zero when the current is along the field.
4. D0.48 N
   A student who uses the whole length of the wire picks this: (4.0 A)(0.40 m)(0.60 T)(sin 30°) = 0.48 N. Only the part of the wire inside the field, 0.25 m, has a force exerted on it.

Working F_B = IℓB sin θ with ℓ = 0.25 m (the length inside the field), θ = 30°: F_B = (4.0 A)(0.25 m)(0.60 T)(0.50) = 0.30 N.

CED 12.3.B.1.i · Read this in Fix

1. AToward the bottom edge of the page
   A student who uses the left hand, or reverses the rule, picks this. The right-hand rule for a current to the right in a field into the page gives a force toward the top of the page.
2. BInto the page, along the field
   A student who thinks the magnetic force points along the field picks this. The force is perpendicular to the field, so here it lies in the plane of the page.
3. CRightward, along the current
   A student who thinks the field pushes the wire along its current picks this. The force is perpendicular to the current, so it pushes the wire sideways.
4. DToward the top edge of the page Correct
   With the current to the right and the field into the page, the right-hand rule gives a force toward the top of the page, perpendicular to both the current and the field.

CED 12.3.B.1.ii · Read this in Fix

1. A0.33d
   A student who uses the separation d as the distance for wire 2's field picks this: I/x = 3I/d gives x = d/3. At P the distance from wire 2 is d − x, not d.
2. B0.37d
   A student who treats each wire's field as inversely proportional to the square of the distance picks this: I/x² = 3I/(d − x)² gives x = d/(1 + √3) ≈ 0.37d. The field of a long straight wire is inversely proportional to r, not r².
3. C0.25d Correct
   Between the wires the two fields point in opposite directions, so they cancel where their magnitudes are equal. With P a distance x from wire 1 and d − x from wire 2: I/x = 3I/(d − x), so d − x = 3x and x = d/4 = 0.25d. P is nearer the wire with the smaller current, as it must be.
4. D0.75d
   A student who pairs each current with the distance from the other wire, writing I/(d − x) = 3I/x, picks this. That puts P nearer the wire with the larger current, where its field would be the stronger; each current goes with the distance from its own wire.

Working For parallel currents in the same direction, the right-hand rule gives the two wires' fields in opposite directions at points between the wires, so they can cancel there. With x the distance from wire 1 to P, the distance from wire 2 is d − x. Zero net field: (μ₀/(2π))(I/x) = (μ₀/(2π))(3I/(d − x)), so d − x = 3x and x = d/4 = 0.25d. (Errors: r = d for wire 2 gives I/x = 3I/d, x = d/3 ≈ 0.33d; B ∝ 1/r² gives (d − x)²/x² = 3, x = d/(1 + √3) ≈ 0.37d; pairing each current with the other wire's distance, I/(d − x) = 3I/x, gives x = 3d/4 = 0.75d.)

CED 12.3.A.1.v · Read this in Fix

1. A2 Correct
   F_B = IℓB sin θ, and the current, the length in the field and the field are unchanged, so the force is proportional to sin θ. sin 90° = 1.0 and sin 30° = 0.50, so the force doubles: k = 2.
2. B0
   A student who uses F = IℓB cos θ picks this: cos 90° = 0, so the force would vanish when the wire is perpendicular to the field. The force depends on sin θ and is greatest, not zero, at 90°.
3. C1
   A student who thinks the force is IℓB whatever the orientation of the wire picks this. The force is IℓB sin θ, so turning the wire from 30° to 90° to the field changes it.
4. D3
   A student who takes the force to be proportional to the angle itself picks this: 90°/30° = 3. The force is proportional to sin θ, and sin 90°/sin 30° = 2.

Working F_B = IℓB sin θ. I, ℓ and B do not change, so F_B ∝ sin θ: k = sin 90°/sin 30° = 1.0/0.50 = 2. (Errors: IℓB cos θ gives cos 90°/cos 30° = 0; IℓB whatever the angle gives 1; F ∝ θ gives 90°/30° = 3.)

CED 12.3.B.1.i · Read this in Fix

1. Amg sin θ/(ℓB)
   A student who takes the total tension in the strings as equal to the rod's weight picks this: IℓB = (mg) sin θ. With the strings tilted, the tension must exceed mg: T cos θ = mg gives T = mg/cos θ, so IℓB = T sin θ = mg tan θ.
2. Bmg/(ℓB tan θ)
   A student who exchanges sine and cosine when resolving the tension picks this: T sin θ = mg and T cos θ = IℓB give IℓB = mg/tan θ. The strings are at θ to the vertical, so the vertical component of the tension is T cos θ and the horizontal component is T sin θ, giving IℓB = mg tan θ.
3. Cmg tan θ/(ℓB) Correct
   The current (horizontal) is perpendicular to the field (vertical), so F_B = IℓB, directed horizontally. With T the total string tension, T cos θ = mg and T sin θ = IℓB, so tan θ = IℓB/(mg) and I = mg tan θ/(ℓB).
4. Dmg/(ℓB cos θ)
   A student who uses the strings' angle θ in F_B = IℓB sin θ picks this: IℓB sin θ = mg tan θ gives I = mg/(ℓB cos θ). The angle in F_B = IℓB sin θ is the angle between the current and the field; the rod is horizontal and the field vertical, so that angle is 90° and F_B = IℓB.

Working The current (along the horizontal rod) is perpendicular to the vertical field, so the angle between them is 90° and F_B = IℓB sin 90° = IℓB, directed horizontally and perpendicular to the rod. Forces on the rod: weight mg downward, magnetic force IℓB horizontal, and total string tension T along the strings at θ to the vertical. Equilibrium: vertical, T cos θ = mg; horizontal, T sin θ = IℓB. Dividing, tan θ = IℓB/(mg), so I = mg tan θ/(ℓB). (Errors: taking the tension equal to mg gives IℓB = mg sin θ and I = mg sin θ/(ℓB); exchanging sine and cosine gives T sin θ = mg and T cos θ = IℓB, so I = mg/(ℓB tan θ); using the string angle θ in F_B = IℓB sin θ gives IℓB sin θ = mg tan θ and I = mg/(ℓB cos θ).)

CED 12.3.B.1.i · Read this in Fix

1. A0.79μ_0I/(πd) Correct
   B₁ = μ_0I/(2πd) and, at distance √2d, B₂ = B₁/√2. The two fields are 45° apart; adding their components gives 1.5B₁ and 0.5B₁, so B = B₁√(1.5² + 0.5²) = 1.58μ_0I/(2πd) = 0.79μ_0I/(πd).
2. B0.85μ_0I/(πd)
   A student who adds the magnitudes, B₁ + B₂ = (1 + 1/√2)μ_0I/(2πd), picks this: 0.85μ_0I/(πd). The fields point in directions 45° apart, so they add as vectors, and the net field, 0.79μ_0I/(πd), is smaller than the sum of the magnitudes.
3. C0.61μ_0I/(πd)
   A student who combines the two fields with the Pythagorean theorem, √(B₁² + B₂²), picks this: 0.61μ_0I/(πd). That rule applies only to perpendicular vectors; these fields are 45° apart, so their components must be added, giving 0.79μ_0I/(πd).
4. D0.92μ_0I/(πd)
   A student who uses the separation of the wires, d, as the distance from wire 2 to P picks this: two fields of magnitude μ_0I/(2πd), 45° apart, add to 0.92μ_0I/(πd). P is √2d from wire 2 (the hypotenuse of the triangle), so wire 2's field there is smaller by a factor of √2.

Working Wire 1 is a distance d from P: B₁ = (μ₀/(2π))(I/d). Wire 2 is a distance √2d from P (the hypotenuse): B₂ = (μ₀/(2π))(I/(√2d)) = B₁/√2. Each field is perpendicular to the line from its wire to P, so the angle between the two fields equals the angle between those lines, 45°. Place wire 1 at (0, 0), wire 2 at (d, 0) and P at (0, d). For currents out of the page the field lines circle counterclockwise, so B⃗₁ points in the −x direction and B⃗₂ points at 45° between −x and −y. Components, in units of B₁: x, −1 − (1/√2)cos 45° = −1.5; y, −(1/√2)sin 45° = −0.5. |B| = B₁√(1.5² + 0.5²) = 1.58B₁ = 1.58μ_0I/(2πd) = 0.79μ_0I/(πd). (Errors: adding magnitudes gives (1 + 1/√2)B₁ = 0.85μ_0I/(πd); the Pythagorean theorem gives √(1 + 1/2)B₁ = 0.61μ_0I/(πd); using the separation d as wire 2's distance gives two fields of magnitude B₁, 45° apart, with sum √(2 + √2)B₁ = 0.92μ_0I/(πd).)

CED 12.3.A.1.v · Read this in Fix

1. A1.8 × 10⁻² N
   A student who treats the wire's field as inversely proportional to the square of the distance picks this: B = (2 × 10⁻⁷ T·m/A)(12 A)/(0.040 m)² = 1.5 × 10⁻³ T gives F = 1.8 × 10⁻² N. A long straight wire's field falls off as 1/r: B₁ = 6.0 × 10⁻⁵ T and F = 7.2 × 10⁻⁴ N.
2. B4.8 × 10⁻⁴ N
   A student who calculates the field acting on wire 2 from wire 2's own current, 8.0 A, picks this: B = 4.0 × 10⁻⁵ T and F = (8.0 A)(1.5 m)(4.0 × 10⁻⁵ T) = 4.8 × 10⁻⁴ N. The force on wire 2 is exerted by wire 1's field, so B must be calculated from wire 1's current, 12 A.
3. C4.5 × 10⁻³ N
   A student who leaves out the 2π and writes B = μ_0I/r picks this: B = (4π × 10⁻⁷ T·m/A)(12 A)/(0.040 m) = 3.8 × 10⁻⁴ T and F = 4.5 × 10⁻³ N. The field of a long straight wire is B = (μ₀/(2π))(I/r), which is 6.0 × 10⁻⁵ T here.
4. D7.2 × 10⁻⁴ N Correct
   Wire 1's field at wire 2 is B₁ = (μ₀/(2π))(12 A)/(0.040 m) = 6.0 × 10⁻⁵ T, perpendicular to wire 2. The force on 1.5 m of wire 2 is I₂ℓB₁ = (8.0 A)(1.5 m)(6.0 × 10⁻⁵ T) = 7.2 × 10⁻⁴ N.

Working Field of wire 1 at wire 2: B₁ = (μ₀/(2π))(I₁/d) = (2 × 10⁻⁷ T·m/A)(12 A)/(0.040 m) = 6.0 × 10⁻⁵ T. This field is tangent to a circle centered on wire 1, so it is perpendicular to wire 2, which is parallel to wire 1: F_B = I₂ℓB₁ sin 90° = (8.0 A)(1.5 m)(6.0 × 10⁻⁵ T) = 7.2 × 10⁻⁴ N (the wires repel). (Errors: a field proportional to 1/r² gives B = 1.5 × 10⁻³ T and F = 1.8 × 10⁻² N; using wire 2's own current for the field gives B = 4.0 × 10⁻⁵ T and F = 4.8 × 10⁻⁴ N; leaving out the 2π gives B = μ_0I₁/d = 3.8 × 10⁻⁴ T and F = 4.5 × 10⁻³ N.)

CED 12.3.B.1.i · Read this in Fix