12.1 Magnetic Fields

1. AA small steel paper clip Correct
   A magnetic field exerts forces on moving charges, on currents and on magnetic materials. Steel is mostly iron, a ferromagnetic material: the field aligns the clip's domains, and the clip is strongly attracted.
2. BA small charged plastic bead
   A student who thinks a magnetic field exerts a force on a charge at rest picks this. The bead is charged but not moving, and plastic is not a magnetic material, so the field exerts no noticeable magnetic force on it; any attraction toward the magnet would be electric.
3. CAn empty aluminum drink can
   A student who thinks magnets attract all metals picks this. Aluminum is paramagnetic: the field aligns its dipoles only very weakly, so the force on the can is far too small to notice. (Steel food cans, by contrast, are attracted.)
4. DA heavy block of dry wood
   A student who thinks a magnet pulls on an object's mass, as gravity does, picks this. The magnetic force depends on what the object is made of, not on its mass: wood is not a magnetic material, and the force on it is far too small to notice.

CED 12.1.A.1 · Read this in Fix

1. A5.0 mT Correct
   Magnetic fields are vectors, so the net field is the vector sum of the two fields. The diagram shows B₁ = 3.0 mT to the right and B₂ = 4.0 mT up the page, at right angles, so they are the legs of a right triangle whose hypotenuse is the net field: √((3.0 mT)² + (4.0 mT)²) = 5.0 mT.
2. B7.0 mT
   A student who adds field magnitudes as numbers picks this. That is correct only for fields in the same direction. The two fields are at right angles, so the net field is shorter than their sum: 5.0 mT.
3. C2.6 mT
   A student who takes the larger field, 4.0 mT, as the hypotenuse picks this, calculating √((4.0 mT)² − (3.0 mT)²). The two fields being added are both legs of the right triangle; the net field is the hypotenuse, 5.0 mT.
4. D4.0 mT
   A student who thinks only the stronger field acts where two fields overlap picks this. Both magnets contribute at P; the 3.0 mT field changes both the magnitude and the direction of the net field.

Working The fields at P are B₁ = 3.0 mT (to the right) and B₂ = 4.0 mT (up the page), at 90° to each other. |B_net| = √((3.0 mT)² + (4.0 mT)²) = √(25 mT²) = 5.0 mT, directed up and to the right at tan⁻¹(4.0/3.0) = 53° above B₁.

CED 12.1.A.2 · Read this in Fix

1. AA charged plastic ring held at rest on a stand
   A student who thinks charges at rest produce magnetic fields picks this. A charged ring at rest produces an electric field only; a magnetic field needs moving charge.
2. BA charged plastic ring spinning about its central axis Correct
   A magnetic dipole arises when electric charge moves in a circle or rotates. The charge on the spinning ring moves in circles, forming a loop of current, so the ring is a magnetic dipole and produces a magnetic field.
3. CAn uncharged copper ring spinning about its central axis
   A student who thinks rotation alone produces a magnetic field picks this. The copper ring's positive and negative charges move round together, so there is no net circulation of charge and no magnetic field.
4. DAn unmagnetized iron ring, held at rest on a stand
   A student who thinks every piece of iron is a magnet picks this. The atomic dipoles in unmagnetized iron are grouped into domains that point in random directions, so their fields cancel and the ring produces no noticeable field of its own.

CED 12.1.B.1 · Read this in Fix

1. A40°
   A student who thinks the deflection angle is proportional to the magnet's field picks this, doubling 20°. It is tan θ, not θ, that is proportional to B_m; doubling tan 20° gives an angle of 36°.
2. B36° Correct
   A compass needle aligns with the net field. With Earth's field B_E north and the magnet's field B_m east, tan θ = B_m/B_E. Doubling B_m doubles tan θ: tan θ' = 2 tan 20° = 0.73, so θ' = 36°.
3. C43°
   A student who treats Earth's field as the hypotenuse picks this, using sin θ = B_m/B_E. Earth's field and the magnet's field are the two perpendicular legs of the triangle; the needle points along the hypotenuse, so tan θ = B_m/B_E.
4. D10°
   A student who inverts the ratio, tan θ = B_E/B_m, picks this: doubling B_m would then halve tan θ. The side opposite the angle from north is the eastward field, so tan θ = B_m/B_E increases when B_m increases.

Working tan θ = B_m/B_E. Initially tan 20° = 0.364. After doubling B_m: tan θ' = 2 × 0.364 = 0.728, so θ' = tan⁻¹(0.728) = 36°.

CED 12.1.B.2 · Read this in Fix

1. AThe nickel and aluminum rods, which were both attracted
   A student who thinks every material a magnet attracts stays magnetized picks this. Aluminum is paramagnetic: its dipoles align weakly while the field is present but do not remain aligned once it is removed.
2. BAll three rods, since all three of them are metals
   A student who thinks all metals are magnetic picks this. Magnetic behavior depends on composition: aluminum and copper cannot be permanently magnetized.
3. CNone of them, as their dipoles return to random directions
   A student who thinks induced magnetism always disappears picks this. That is true of paramagnetic aluminum, but in ferromagnetic nickel much of the domain alignment can remain, making a permanent magnet.
4. DThe nickel rod alone, as its domains can stay aligned Correct
   Ferromagnetic materials such as iron, nickel and cobalt can be permanently magnetized by an external field, which aligns their magnetic domains; much of the alignment remains after the field is removed. Aluminum is paramagnetic and copper is diamagnetic, and neither keeps any alignment.

CED 12.1.B.3.i · Read this in Fix

1. ANorthward, level with the ground's surface
   A student who thinks Earth's field is horizontal everywhere picks this. That is nearly true only near the equator. The field line through P is tilted into the ground, and a needle free to turn in a vertical plane follows the tilt.
2. BSouthward, rising steeply up out of the ground
   A student who thinks the magnetic pole near geographic north is a north pole picks this, with field lines leaving the Northern Hemisphere. That pole attracts the north end of compass needles, so it is a south magnetic pole, and the field at P points north and into the ground.
3. CStraight down, toward the center of Earth
   A student who thinks Earth's magnetic field points toward its center, like its gravitational field, picks this. Earth's field is approximately a dipole field: at P it points along the tilted field line, northward as well as downward.
4. DNorthward, dipping steeply down into the ground Correct
   The north end of a compass points toward geographic north, so the magnetic pole there is a south pole, and Earth's dipole field lines enter the ground in the Northern Hemisphere. At P the field line is tilted steeply toward the surface, so the needle points northward and down into the ground.

CED 12.1.B.4 · Read this in Fix

1. AHow freely a magnetic field passes through it, so a vacuum has the most
   A student who reads 'permeable' in its everyday sense, as letting something through, and so expects a vacuum, with nothing in the way, to have the greatest permeability, picks this. Permeability measures how strongly a material is magnetized in response to an external field; iron, which is strongly magnetized, has a permeability thousands of times μ₀, so a vacuum does not have the most.
2. BHow strongly it is magnetized in response to an external field Correct
   Magnetic permeability is a measure of the amount of magnetization in a material in response to an external magnetic field: the more strongly the material's dipoles align in a given field, the greater its permeability.
3. CHow much magnetization it keeps after an external field is removed
   A student who confuses permeability with the ability to stay magnetized picks this. Permeability describes the response while the external field is present; whether a material keeps its alignment afterward is a separate property.
4. DHow well it conducts an electric current in a magnetic field
   A student who links magnetic permeability with electrical conduction picks this. Copper conducts extremely well yet has a permeability almost equal to that of a vacuum; permeability concerns the alignment of magnetic dipoles, not the flow of charge.

CED 12.1.C.1 · Read this in Fix

1. A1.3 × 10⁻⁴ T·m/A
   A student who substitutes r = 5.0 cm without converting it to meters picks this. μ₀ and the tesla are SI quantities, so r must be 0.050 m; the result is 100 times too large.
2. B3.2 × 10⁻⁸ T·m/A
   A student who leaves the 2π in the denominator when rearranging picks this, calculating rB/(2πI). Multiplying both sides of B = μI/(2πr) by 2πr and dividing by I gives μ = 2πrB/I.
3. C1.3 × 10⁻⁶ T·m/A Correct
   Solving B = (μ/(2π))(I/r) for μ gives μ = 2πrB/I = 2π(0.050 m)(1.2 × 10⁻⁵ T)/(3.0 A) = 1.3 × 10⁻⁶ T·m/A. This matches μ₀ = 4π × 10⁻⁷ T·m/A = 1.26 × 10⁻⁶ T·m/A to the precision of the data: the permeability of air is very close to that of free space, which is why μ₀ is used for fields in air.
4. D2.0 × 10⁻⁷ T·m/A
   A student who calculates rB/I and stops picks this. That quantity equals μ/(2π); multiplying by 2π gives μ = 1.3 × 10⁻⁶ T·m/A.

Working μ = 2πrB/I = 2π × 0.050 m × 1.2 × 10⁻⁵ T / 3.0 A = 1.26 × 10⁻⁶ T·m/A ≈ 1.3 × 10⁻⁶ T·m/A (2 s.f.), equal to μ₀ = 4π × 10⁻⁷ T·m/A within the precision of the data.

CED 12.1.C.2 · Read this in Fix

1. ABoth are equal to μ₀, since μ₀ is the permeability of all matter
   A student who thinks μ₀ applies to every material picks this. μ₀ is the permeability of free space; the permeability of matter differs from it, enormously for iron and slightly for aluminum.
2. BIron's is far greater than μ₀, and aluminum's is slightly greater than μ₀ Correct
   The permeability of matter differs from that of free space and depends on composition. Ferromagnetic iron becomes strongly magnetized along an external field, so its permeability is far greater than μ₀; paramagnetic aluminum is magnetized only weakly along the field, so its permeability is only slightly greater than μ₀.
3. CIron's is far above μ₀, and aluminum's is zero, as it is not magnetic
   A student who thinks non-ferromagnetic materials do not respond to a magnetic field at all picks this. Aluminum is paramagnetic and responds weakly along the field, so its permeability is slightly greater than μ₀, not zero; even a vacuum has permeability μ₀.
4. DBoth are less than μ₀, since a field passes through a vacuum most easily
   A student who reads permeability as the ease with which a field passes through a material picks this. Permeability measures magnetization in response to a field; iron and aluminum are both magnetized along the field, so both have permeabilities greater than μ₀.

Working Iron is ferromagnetic: its domains align strongly along an external field, so its permeability is far greater than μ₀. Aluminum is paramagnetic: its dipoles align only weakly along the field, so its permeability is only slightly greater than μ₀ (about 1.00002 μ₀).

CED 12.1.C.3 · Read this in Fix

1. AA push away from the north pole, which repels positive charge
   A student who thinks magnetic poles are electric charges, with the north pole positive, picks this. A magnet is electrically neutral; its poles describe the direction of its magnetic dipole, not a charge, so a north pole does not repel a positive charge.
2. BA force along the field line, as an electric field would exert
   A student who expects a magnetic field to act on a charge at rest as an electric field does picks this. The magnetic field exerts a force on a charge only while the charge moves; for a charge at rest there is no magnetic force, whatever the field's strength.
3. CA pull toward the pole, as magnets attract charged objects
   A student who treats magnetism as static electricity picks this. A charged object attracts small neutral objects, but a magnet's field does not attract a charged object at rest; magnetic forces are exerted on moving charges and magnetic materials.
4. DNone, since the sphere is not moving through the field Correct
   A magnetic field exerts a force on moving charges, on currents and on magnetic materials. The sphere is charged but at rest, and its material is not magnetic, so the magnet's field exerts no magnetic force on it. (Any force it does feel near the magnet would be electric in origin.)

CED 12.1.A.1 · Read this in Fix

1. AIt is the field of a bar magnet's north pole, which acts as a source on its own
   A student who thinks a pole can act as a separate source, like a single charge, picks this. The north pole of a bar magnet is always accompanied by its south pole; its field lines curve around and return to the south pole and continue through the magnet, rather than spreading out from one point forever.
2. BIt would need an isolated north pole at O, so no magnetic field has it Correct
   The lines leave O in every direction and none returns, so O would have to be a source with no partner: an isolated north pole. Magnetic fields are produced by dipoles or combinations of dipoles, never by monopoles, and magnetic field lines form closed loops, so this cannot be a magnetic field.
3. CIt is the magnetic field of a small positively charged sphere at rest
   A student who thinks a charge at rest produces a magnetic field as well as an electric field picks this. This pattern is the electric field of a positive point charge. A charge at rest produces no magnetic field; magnetic fields come from dipoles, such as moving charges in loops.
4. DIt could be, if the lines end on south poles far outside the diagram
   A student who thinks magnetic field lines begin on north poles and end on south poles picks this. Magnetic field lines have no ends: they form closed loops. Even if distant south poles existed, point O would still have to be an isolated north pole, and none exists.

CED 12.1.A.1.i · Read this in Fix

1. ABoth bars are magnets, since attraction shows that both of them are
   A student who thinks magnetic attraction proves that both objects are magnets picks this. A magnet also attracts unmagnetized iron. Two magnets would repel in at least one pairing of ends, and no pairing repels here.
2. BEach bar has one pole only: P is a north pole and Q a south pole
   A student who thinks a magnet can have a single pole picks this. Every magnet is a dipole, with a north pole and a south pole; there are no single-pole bars. If P and Q were both magnets, some pairing of their ends would repel.
3. COne of the bars is a magnet and the other is not magnetized Correct
   Attraction shows that at least one bar is a magnet, since two unmagnetized, uncharged bars would not attract. If both were magnets, each would be a dipole with a north pole and a south pole, and bringing like poles together would make them repel. No pairing repels, so exactly one bar is a magnet; the observations do not show which.
4. DThey may both be plain iron, since any two iron bars attract each other
   A student who thinks every piece of iron is a magnet picks this. In unmagnetized iron the domains point in random directions and their fields cancel, so two unmagnetized iron bars do not attract; attraction shows that at least one bar is a magnet.

CED 12.1.A.1.ii · Read this in Fix

1. AGreater at Y, since at X, between the lines, there is no field
   A student who thinks the field exists only along the drawn lines picks this. The lines are only a sample; the field exists at every point, including X, and the crowding of the lines around X shows that it is strong there.
2. BGreater at X, where the field lines are more closely spaced Correct
   On a field-line map the spacing of the lines shows the field's magnitude: the closer together the lines, the stronger the field. The lines are crowded near the magnet's end at X and far apart where they loop out through Y, so the field is greater at X.
3. CThe same at both, as the field strength is set by the magnet alone
   A student who treats the field as one strength belonging to the magnet picks this. The field has a value at each point, and the magnitude of a dipole's field decreases with distance from it: X is much closer to the magnet than Y.
4. DNot comparable, as the lines show only the field's direction
   A student who thinks a field-line map shows direction only picks this. The spacing of the lines also shows the relative magnitude: closely spaced lines mean a strong field.

CED 12.1.A.2 · Read this in Fix

1. AReverse the arrows, so that the lines outside run from S to N
   A student who thinks the external field points from the south pole to the north pole picks this. By definition the external field points away from the north pole, so the arrows in the sketch, from N around to S, are already correct.
2. BMake no change, since field lines begin at N and end at S
   A student who thinks magnetic field lines begin and end on poles, as electric field lines do on charges, picks this. Magnetic field lines have no ends; each one continues through the magnet to form a closed loop.
3. CContinue each line through the magnet, from its S end to its N end Correct
   Magnetic field lines form closed loops. Outside the magnet the sketch correctly shows the field running from the N end around to the S end; to close each loop, the line must continue inside the magnet from the S end back to the N end.
4. DAdd lines inside the magnet pointing from its N end to its S end
   A student who thinks the field inside a magnet points from N to S, as it does outside, picks this. Such lines would meet the outside lines head-on at both ends instead of joining them into loops; inside the magnet the field points from the S end to the N end.

CED 12.1.A.2.i · Read this in Fix

1. AGreater at D, as field lines end at the poles and none is inside
   A student who thinks field lines begin and end at the poles picks this. The lines do not end: each continues through the magnet, where the field is strongest.
2. BZero at both points, since the middle of a magnet is not magnetic
   A student who thinks the middle of a magnet is not magnetic picks this. There is a field both inside the magnet and beside its middle; only the force on a piece of iron is small beside the middle, where the field is weaker.
3. CThe same at both, as the magnet's field has one strength everywhere
   A student who treats a magnet's field as one strength belonging to the magnet picks this. The field has a different magnitude at different points; the crowding of the field lines inside the magnet shows that it is strongest there.
4. DGreater at C, as the closed field lines are crowded inside the magnet Correct
   Magnetic field lines form closed loops, and every loop of a bar magnet passes through the inside of the magnet, from the S end to the N end. Inside, the lines are crowded into the magnet's small cross-section; outside, they spread out, so the field is stronger at C than at D.

CED 12.1.A.2.i · Read this in Fix

1. ATo the right, parallel to the magnet, toward its N end
   A student who thinks the field everywhere points the same way as the magnet, from S to N, picks this. That is the direction inside the magnet. Outside, the field lines return from the N end toward the S end, so beside the middle they point the opposite way.
2. BUp the page, straight away from the magnet
   A student who pictures the field pointing straight out of the magnet, like the field of a point charge, picks this. The field lines of a bar magnet curve from one end to the other; beside the middle they run parallel to the magnet.
3. CTo the left, parallel to the magnet, toward its S end Correct
   Outside a bar magnet the field points away from the north pole and loops around to the south pole. Beside the middle of the magnet, a field line from the N end is passing back toward the S end, so the field at Q points to the left, parallel to the magnet.
4. DIt has none, since the magnet has no field at Q
   A student who thinks the middle of a magnet is not magnetic picks this. The field beside the middle is weaker than near the ends, which is why few filings cling there, but it is not zero; field lines pass by the middle of every magnet.

CED 12.1.A.2.ii · Read this in Fix

1. AThe field aligns the nail's dipoles, so its near end is an opposite pole Correct
   Induced magnetism results from the alignment of dipoles. The field of either pole aligns the nail's domains so that the end nearer the pole has the opposite polarity; opposite poles attract, so the nail is attracted whichever pole is used.
2. BA magnet's poles attract every metal, and the nail is made of a metal, iron
   A student who thinks magnets attract all metals picks this. Many metals, such as aluminum and copper, show no noticeable attraction; the nail is attracted because it is made of iron, a ferromagnetic material whose dipoles the field can align.
3. CThe pole gives the nail's near end an electric charge opposite to its own
   A student who thinks magnetic poles are electric charges picks this. Poles are not charges, and a magnet does not charge the nail; it magnetizes it, by aligning the nail's magnetic dipoles.
4. DThe magnet pulls on the nail's mass, much as Earth pulls on it
   A student who thinks magnetism is a kind of gravity picks this. A magnet does not pull on mass as such: it does not attract a heavier piece of wood or aluminum. It attracts the nail because the nail's iron becomes magnetized.

CED 12.1.B.1.i · Read this in Fix

1. AIt contained no magnetic dipoles until the stroking created them
   A student who thinks an unmagnetized material contains no dipoles picks this. Iron's atoms are magnetic dipoles all along; stroking only turned them, and the domains they form, to point the same way.
2. BThe stroking moved some of the magnet's magnetism into the nail
   A student who thinks of magnetism as a substance that flows from one object to another picks this. Nothing flows into the nail, and the magnet is not used up: its field rearranges the dipoles that the nail already contains.
3. CThe stroking charged its ends oppositely, as rubbing charges a balloon
   A student who treats magnetism as static electricity picks this. Stroking with a magnet does not charge the nail; if it did, the nail would also attract scraps of paper. The nail attracts the steel clip because it is magnetized.
4. DIts domains pointed in random directions until stroking aligned them Correct
   Permanent and induced magnetism both result from the alignment of magnetic dipoles. Before stroking, the nail's domains pointed in random directions and their fields canceled; the magnet's field aligned many of them, and they stayed aligned, so the nail became a magnet.

CED 12.1.B.1.i · Read this in Fix

1. AEnd 2 is north, like end 1, and end 3 is south, like end 4
   A student who thinks a magnet is made of a north half and a south half picks this. Each piece would then have two like poles and no opposite pole. In fact every small part of a magnet is a dipole, so each new end has the polarity opposite to the other end of its piece.
2. BEnds 2 and 3 are not poles, so each piece has only one pole
   A student who thinks a pole can exist on its own picks this. A single north or south pole has never been found; however a magnet is broken, each piece has both a north pole and a south pole.
3. CEnd 2 is now a south pole and end 3 is now a north pole Correct
   No magnetic north pole is ever found in isolation from a south pole: each piece is a dipole. The left piece has its N pole at end 1, so its other end, 2, is a south pole; the right piece has its S pole at end 4, so end 3 is a north pole. Ends 2 and 3 attract, as they did when the magnet was whole.
4. DNeither piece is magnetized, since breaking destroys magnetism
   A student who thinks breaking a magnet destroys its magnetism picks this. Breaking does not turn the dipoles within each piece, so both pieces remain magnets.

Working Each piece is a dipole. Left piece: end 1 is N, so its other end, 2, is S. Right piece: end 4 is S, so its other end, 3, is N. (The alignment of the dipoles inside each piece is unchanged by the break.)

CED 12.1.B.1.ii · Read this in Fix

1. A2.5 N
   A student who thinks magnets always attract picks this, adding a downward pull of 0.50 N to the weight. The facing poles are both north poles; like poles repel, so B pushes A up and the scale reads less than A's weight.
2. B2.0 N
   A student who thinks the two magnetic forces cancel picks this. The force on A and the force on B are exerted on different objects, so they cannot cancel. The upward force on A reduces what the scale must supply.
3. C1.0 N
   A student who adds the two forces of the pair into one 1.0 N push on A picks this, subtracting 2 × 0.50 N from the weight. The force that A exerts on B acts on B; only one magnetic force, 0.50 N upward, acts on A.
4. D1.5 N Correct
   The figure shows the N pole of A facing the N pole of B. Like poles repel, so B pushes A upward with 0.50 N. A is at rest, so the scale's upward force F_s satisfies F_s + 0.50 N = 2.0 N, giving F_s = 1.5 N.

Working Forces on A (at rest): scale force F_s up, magnetic force from B up (like N poles face each other, so they repel) of 0.50 N, weight 2.0 N down. F_s + 0.50 N − 2.0 N = 0, so F_s = 1.5 N; the scale reads 1.5 N.

CED 12.1.B.1.iii · Read this in Fix

1. AThe magnet's field there weakens with distance, so the net field turns north Correct
   The magnitude of a dipole's field decreases with increasing distance from it. The needle points along the net field, the vector sum of Earth's northward field and the magnet's eastward field; as the magnet's field at the compass decreases, the net field turns back toward north.
2. BThe compass is leaving the region the magnet's field fills, which has an edge
   A student who pictures the field as a zone with a definite edge picks this. A magnet's field decreases gradually with distance and never stops at a boundary; the needle turns back toward north as the magnet's contribution to the net field shrinks.
3. CThe magnet's field there keeps its strength but turns to point north
   A student who treats the field as a fixed strength belonging to the magnet picks this. As the magnet moves due east without turning, the direction of its field at the compass stays the same; what changes is its magnitude, which decreases with distance.
4. DThe needle follows only the stronger field, which becomes Earth's field
   A student who thinks only the stronger of two fields acts picks this. The needle aligns with the vector sum of both fields, so it swings back toward north gradually as the magnet's field weakens, not all at once when Earth's field becomes the larger.

CED 12.1.B.1.iv · Read this in Fix

1. AIron is strongly attracted, aluminum weakly attracted, bismuth weakly repelled Correct
   A material's composition determines its magnetic behavior. Iron is ferromagnetic: its domains align strongly with the field, and it is strongly attracted. Aluminum is paramagnetic: its dipoles align weakly with the field, and it is weakly attracted. Bismuth's response is dominated by diamagnetism, a weak alignment opposite to the field, so it is weakly repelled.
2. BIron is strongly attracted, and aluminum and bismuth feel no magnetic force at all
   A student who thinks only ferromagnetic materials interact with a magnetic field picks this. Paramagnetic aluminum and diamagnetic bismuth respond weakly, but with a strong magnet the weak attraction of aluminum and weak repulsion of bismuth can be observed.
3. CAll three are attracted equally strongly, since all three are metals
   A student who thinks magnets attract all metals picks this. Magnetic behavior depends on composition, not on being a metal: of these three metals only iron is strongly attracted.
4. DIron is strongly attracted, aluminum weakly repelled and bismuth weakly attracted
   A student who exchanges paramagnetism and diamagnetism picks this. Paramagnetic aluminum's dipoles align along the field, so it is attracted; bismuth's diamagnetic response is opposite to the field, so it is repelled.

CED 12.1.B.3 · Read this in Fix

1. AGraph 1
   A student who thinks every material that a magnet attracts stays magnetized picks this. Only ferromagnetic materials keep part of their alignment after the field is removed; aluminum's alignment disappears at t₂.
2. BGraph 2
   A student who thinks non-ferromagnetic materials do not interact with a field at all picks this. Aluminum responds weakly but definitely: its dipoles align along the field while it is present.
3. CGraph 4 Correct
   Aluminum is paramagnetic: while the external field is present its dipoles align weakly along the field (M > 0), and after the field is removed they do not remain aligned (M returns to zero). Graph 4 shows exactly this.
4. DGraph 3
   A student who exchanges paramagnetism and diamagnetism picks this. A negative M, alignment opposite to the field, describes a diamagnetic material such as bismuth; paramagnetic aluminum aligns along the field.

Working Paramagnetic aluminum: weak alignment along the field while it is present (M > 0 between t₁ and t₂); no alignment remains after the field is removed (M = 0 after t₂). Only graph 4 has both features.

CED 12.1.B.3.ii · Read this in Fix

1. AMaterials that are ferromagnetic or paramagnetic have no diamagnetism
   A student who thinks each material has one kind of magnetic behavior picks this. Diamagnetism is a property of all materials; in ferromagnetic and paramagnetic materials it is simply outweighed by the stronger alignment along the field.
2. BEvery material has it, though in iron it is hidden by ferromagnetism Correct
   All materials have the property of diamagnetism: their electronic structure produces a usually weak alignment of dipole moments opposite to an external field. In iron this effect is far weaker than the ferromagnetic alignment along the field, so it goes unnoticed.
3. CIt aligns dipoles along the external field, so the material is attracted
   A student who exchanges diamagnetism and paramagnetism picks this. Diamagnetism aligns dipole moments opposite to the external field, so a material in which it dominates is weakly repelled.
4. DNonmetals may have it, but metals are all attracted by magnets
   A student who thinks magnets attract every metal picks this. Some metals, such as bismuth and copper, are weakly repelled by a strong magnet, because diamagnetism dominates their response; and every metal, like every material, has diamagnetism.

CED 12.1.B.3.iii · Read this in Fix

1. AX > Y > Z Correct
   Permeability describes the magnetization produced in response to an external field, so it is compared by the magnetization per unit of external field: the slope of each straight line through the origin. X's slope is 0.5 unit/mT, Y's 0.3 unit/mT and Z's 0.2 unit/mT, so X > Y > Z.
2. BZ > Y > X
   A student who thinks permeability describes how easily a field passes through a material, so that the material that responds least has the most, picks this. Permeability measures the material's magnetization in response to the field: the steeper the line, the greater the permeability.
3. CY > Z > X
   A student who ranks the samples by the greatest magnetization reached picks this. Y and Z reach larger M only because they were placed in stronger fields; per unit of external field, X is magnetized most.
4. DX = Y = Z
   A student who thinks every material has the same permeability, μ₀, picks this. The permeability of matter depends on its composition; the three different slopes show three different permeabilities.

Working Slopes (M per mT): X: 5 units/10 mT = 0.5; Y: 9 units/30 mT = 0.3; Z: 7 units/35 mT = 0.2. Permeability increases with the magnetization produced per unit external field, so X > Y > Z.

CED 12.1.C.1 · Read this in Fix

1. AIt is constant, since M increases whenever B_ext increases
   A student who thinks a material's permeability is a fixed constant picks this. M does increase with B_ext, but not in proportion: the curve bends over, so the magnetization produced per unit of field changes.
2. BIt is greatest at the largest B_ext, where M is the greatest
   A student who equates permeability with the amount of magnetization picks this. At the largest B_ext the magnetization is greatest, but it is smallest in proportion to the field that produces it.
3. CIt is zero at large B_ext, where the graph is almost flat
   A student who reads the flat slope as zero permeability picks this. Where the curve levels off, the iron is still strongly magnetized; its permeability, which compares M with B_ext, is smaller there but far from zero.
4. DIt is not constant: M/B_ext is smaller at large than at small B_ext Correct
   Permeability describes magnetization in response to the external field. At small B_ext, M rises steeply, so M/B_ext is large; at large B_ext, M levels off while B_ext keeps increasing, so M/B_ext is much smaller. The permeability of iron varies with the strength of the external field.

CED 12.1.C.3 · Read this in Fix

1. A2B
   A student who adds the two field magnitudes as numbers picks this. That is correct only when the fields point the same way (θ = 0°). At any other angle part of each field is cancelled by the other, so the net field is less than 2B.
2. B2B cos(θ/2) Correct
   The fields are vectors of equal magnitude, so the net field lies along the bisector of the angle between them, and each field makes θ/2 with it. Each contributes B cos(θ/2) along the bisector, while the perpendicular components, B sin(θ/2) each, cancel. The net field is 2B cos(θ/2): 2B when the fields are parallel and zero when they are opposite.
3. C2B cos θ
   A student who resolves each field along the net field using the full angle θ between the two fields picks this. Each field makes only θ/2 with the net field, which lies midway between them; with θ this expression even gives zero at 90°, where the net field is B√2.
4. DB√2
   A student who combines the fields with the Pythagorean theorem whatever the angle picks this. √(B² + B²) = B√2 is the net field only when the fields are perpendicular (θ = 90°); for other angles the components must be added, giving 2B cos(θ/2).

Working Magnetic fields add as vectors. The two fields have equal magnitude, so their sum lies along the line that bisects the angle between them, and each field makes an angle θ/2 with that line. Along the bisector each contributes B cos(θ/2), giving 2B cos(θ/2); perpendicular to it the components B sin(θ/2) point in opposite directions and cancel. B_net = 2B cos(θ/2). Checks: θ → 0° gives 2B; θ = 90° gives 2B cos 45° = B√2 (the Pythagorean result for perpendicular fields); θ → 180° gives 0. (Errors: adding magnitudes gives 2B; using the full angle θ for each component gives 2B cos θ; using the Pythagorean theorem at any angle gives B√2.)

CED 12.1.A.2 · Read this in Fix

1. A0.58B_E
   A student who takes the whole of Earth's field, B_E, as the northward horizontal field picks this: B_m = B_E tan 30° = 0.58B_E. Earth's field dips 60° below the horizontal, so only its horizontal component, B_E cos 60° = 0.50B_E, lies in the plane in which the needle turns.
2. B0.50B_E
   A student who takes the horizontal component of a field 60° below the horizontal as B_E sin 60° picks this: B_m = (0.866B_E)(tan 30°) = 0.50B_E. The horizontal component is adjacent to the 60° angle, B_E cos 60° = 0.50B_E, so B_m = 0.29B_E.
3. C0.87B_E
   A student who writes tan 30° as the northward field divided by the magnet's field picks this: B_m = (0.50B_E)/(tan 30°) = 0.87B_E. The eastward field is the side opposite the 30° angle measured from north, so tan 30° = B_m/(0.50B_E).
4. D0.29B_E Correct
   The needle follows the horizontal net field. Earth's horizontal component is B_E cos 60° = 0.50B_E, pointing north, and the magnet's field points east, so tan 30° = B_m/(0.50B_E) and B_m = (0.50B_E)(tan 30°) = 0.29B_E.

Working Earth's field has a horizontal component B_E cos 60° = 0.50B_E, pointing north; its vertical component, B_E sin 60°, cannot turn a needle confined to a horizontal plane. The magnet's field B_m is horizontal and points east. The needle lies along the horizontal net field, so tan 30° = B_m/(B_E cos 60°), giving B_m = B_E cos 60° tan 30° = (0.500)(0.577)B_E = 0.29B_E. (Errors: taking the full B_E as the horizontal field gives B_m = B_E tan 30° = 0.58B_E; taking the horizontal component as B_E sin 60° gives (0.866)(0.577)B_E = 0.50B_E; inverting the tangent ratio, tan 30° = B_E cos 60°/B_m, gives B_m = 0.500B_E/0.577 = 0.87B_E.)

CED 12.1.B.2 · Read this in Fix