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(SI units of physical quantities are given in parentheses)

Ampère’s law: The line integral of the magnetic field along a closed path (called an Ampèrian path) is pro- portional to the current enclosed by the path, :

(28.1) In analogy to Gauss’ law in electrostatics, Ampère’s law allows us to determine the magnetic field for cur- rent distributions that exhibit planar, cylindrical, or toroidal symmetry.

Biot-Savart law: Expression giving the magnetic field at a point P due to a small segment of a wire carrying a current I:

, (28.12)

where is a unit vector pointing from the segment to the point at which the magnetic field is evaluated. The Biot-Savart law can be used to calculate the mag- netic field of a current-carrying conductor of arbitrary shape by integration: (28.10) BÆ

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Ampèrian path:Closed path along which the magnet- ic field is integrated in Ampère’s law.

Current loop:A current-carrying conductor in the shape of a loop. The magnetic field pattern of a current loop is similar to that of a magnetic dipole.

Magnetic dipole moment (A m2_{): Vector pointing} from the S-pole to the N-pole for a bar magnet or along the axis of a planar current loop in the direction given by the right-hand thumb when the fingers of that hand are curled along the direction of the current through the loop. In an external magnetic field, the magnetic di- pole moment tends to align in the direction of the ex- ternal magnetic field.

Permeability constant (T m/A): Constant relating the current enclosed by an Ampèrian path and the line integral of the magnetic field along that loop. In vacu- um:

.

Solenoid: Long, tightly wound helical coil of wire. The magnetic field of a current-carrying solenoid is similar to that of a bar magnet.

Toroid: Solenoid bent into a circle. The magnetic field of a toroid is completely contained within the wind- ings of the toroid.

o Æ

Solutions

28.1 (a) See Figure S28.1a. (b) The two particles exert

no magnetic forces on each other because it takes a moving charged particle to detect the magnetic field of another moving charged particle. (c) See Figure S28.1b. The forces exerted on the horizontal wire cause a torque that tends to align that wire with the vertical one. Conversely, the forces exerted on the vertical wire cause a torque in the opposite direction, tending to align that wire with the horizontal one.

28.2 See Figure S28.2.

28.3 Comparing Figures 28.6b and 28.7; you see that

the positively charged ring has a magnetic field similar to the field of a bar magnet, with its north pole up. The negatively charged ring thus has its north pole down, and this north pole is directly above the north pole of

the positive ring. Because two north poles repel each other, the interaction is repulsive.

28.4 No. By definition the electric field points from

positive charge carriers to negative charge carriers (Section 23.2). The electric field along the axis passing through the poles of an electric dipole therefore points from the positive end to the negative end. The electric dipole moment, however, by definition points in the opposite direction (Section 23.8).

28.5 (a) Because the magnetic field remains perpen-

dicular to the horizontal sides of the loop, you know from Eq. 27.4 that the magnitude of the magnetic force exerted on each side is , where l is the length of as each side of the loop. Because none of these quantities changes, the magnitude of the force exerted on the horizontal sides stays the same. (b) See Figure S28.5ab. The lever arm of the force exerted on the horizontal sides become smaller as the loop rotates and so the torque caused by these forces decreases. (c) In the initial position, the vertical sides are parallel to the magnetic field, which means the magnetic force ex- erted on them is zero (Eq. 27.7, with q = 0). After the loop has rotated 90°, the vertical sides are perpendicular to and thus subject to an outward magnetic force. For 0 < q < 90°, the magnitude of the force is again given by Eq. 27.7. As you can verify using the right-hand force rule, the forces exerted on the ver- tical sides are directed outward along the rotation axis and so cause no torque. (d) See Figure S28.5c. Because the top and bottom of the loop are now reversed, the direction of the torque reverses.

28.6 See Figures S28.6, where the circular loop is ap-

proximated by a series of vertical and horizontal seg- ments. The vertical segments experience no force, but

FB冷 I 冷lB sin BÆ FBmax冷 I 冷lB P₂ P₁ P₅ P₃ P₄ (b) (a) –e F→_wB_e F→_eB_w F→₁B₂ F→₁B₂ F→₂B₁ F→₂B₁ v → B→ B→ B→ – I I I Figure S28.1 B₁ B₂ B→ 1 2 (a) B₁ B₂ → B 1 2 (b) Figure S28.2

Solutions 33

the horizontal ones do. All the horizontal segments add up to the same length as two sides of a square strad- dling the circle, and thus Eq. 27.4, , tells you that the magnitude of the magnetic force exerted on the horizontal segments is the same as the force ex- erted on the horizontal side of the square loop. Some

FB

max冷 I 冷lB

horizontal segments are closer to the rotation axis, however, and so the moment arms of the forces acting on these closer segments are smaller than the moment arms of the forces acting on the square loop. Therefore the torque on the circular loop is smaller than that on the square loop.

28.7 With the magnetic field on the left stronger than

that on the right, . Therefore the vector sum of the forces exerted on these two sides is nonze- ro and points upward. The effect is a clockwise rota- tion due to the torque and an upward acceleration due to the upward vector sum of the forces.

28.8 Increasing the current increases the magnitude of

the magnetic field and so the line integral increases.

28.9 (a) If the direction of the current is reversed, the

direction of the magnetic field is reversed, and so the algebraic sign of the line integral is reversed. (b) Be- cause, the value of the line integral of the magnetic field around a closed path does not depend on the position of the wire inside the path, the second wire by itself gives rise to the same line integral as the first wire. Adding the second wire thus doubles the value of the line integral. (c) Reversing the current though the sec- ond wire flips the sign of the line integral. If its value was C, it is –C after the current is reversed. The sum of the two line integrals is thus C + (–C) = 0.

28.10 If the path tilts, you can break it down into arcs,

radial, segments, and axial segments, which are seg- ments parallel to the wire. Because the magnetic field is perpendicular to the axial segments, they don’t con- tribute to the line integral. For the same reason, of course, the radial segments contribute nothing. So the

FB left FBright axis Figure S28.6 wire current o_ut R r r Ampèrian loop 2 Ampèrian loop 1

Figure S28.12 Cross section through a current-carrying wire.

θ B B (a) (b) m m pivot pivot lever arm lever arm current out of page current out of page current out of page current

into page _{into page}current

current into page F→B F→B F→B F→B → θ B (c) m pivot F→B F→B → → Figure S28.5

line integral again is that along all the arcs, which is equivalent to a single complete circular path.

28.11 (a) Reversing the direction of the current

through wire 1 changes the sign of its contribution from negative to positive, so that adding the contribu- tions of wires 1 and 3 yields a positive value. (b) The loop does not encircle wire 2, and so reversing the cur- rent through wire 2 does not affect the answer to Ex- ercise 28.2. (c) Reversing the direction of the loop inverts the sign of all the contributions to the line inte- gral, but because the integral is zero, nothing changes.

28.12 See Figure S28.12. Inside the wire, use Am-

pèrian path 1 of radius r < R. For this loop, ,

but the loop encircles only part of the cross section so encircles only part of the current I through the wire. The area of the cross-section of the wire is pR2_{, and the} cross-sectional area of the Ampèrian path 1 is pr2_{, mak-} ing the fraction of the wire cross section enclosed by the loop (pr2)/(pR2_{) = r}2_/R2_{. Thus, the right side of Am-}

père’s law is . Ampère’s law thus

yields , or

. Outside the wire, use Ampèrian path 2 in Figure S28.12. For this loop,

.

Because the loop encircles all of the current I through the wire, , and so the magnetic field out- side the wire is the same as that for a long thin wire:

.

28.13 (a) Figure S28.13 shows a head-on view of the

magnetic fields of the two sheets separately and their sum. The magnetic fields add up to zero between the sheets. Below and above the sheets, the magnetic field

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