SOURCES OF MAGNETIC FIELD 28.1 28 G IDENTIFY and SET UP: Use Eq (28.2) to calculate B at each point G G G G G μ qv × rˆ μ0 qv × r r B= = , since rˆ = 4π r 4π r r G G ˆ v = (8.00 × 10 m/s) j and r is the vector from the charge to the point where the field is calculated G EXECUTE: (a) r = (0.500 m) iˆ, r = 0.500 m G G v × r = vrˆj × iˆ = −vrkˆ G μ qv (6.00 × 10−6 C)(8.00 × 106 m/s) ˆ B = − kˆ = −(1 × 10−7 T ⋅ m/A) k 4π r (0.500 m)2 G B = −(1.92 × 10−5 T)kˆ G (b) r = −(0.500 m) ˆj , r = 0.500 m G G G v × r = −vrˆj × ˆj = and B = G (c) r = (0.500 m) kˆ , r = 0.500 m G G v × r = vrˆj × kˆ = vriˆ G (6.00 × 10−6 C)(8.00 × 106 m/s) ˆ B = (1 × 10−7 T ⋅ m/A) i = + (1.92 × 10−5 T)iˆ (0.500 m) G (d) r = −(0.500 m) ˆj + (0.500 m)kˆ , r = (0.500 m)2 + (0.500 m)2 = 0.7071 m G G v × r = v(0.500 m)(− ˆj × ˆj + ˆj × kˆ ) = (4.00 × 106 m /s)iˆ 28.2 G (6.00 × 10−6 C)(4.00 × 106 m /s) ˆ B = (1 × 10−7 T ⋅ m/A) i = +(6.79 × 10−6 T)iˆ (0.7071 m)3 G G G G EVALUATE: At each point B is perpendicular to both v and r B = along the direction of v IDENTIFY: A moving charge creates a magnetic field as well as an electric field μ qv sin φ SET UP: The magnetic field caused by a moving charge is B = , and its electric field is 4π r e since q = e 4π ⑀0 r EXECUTE: Substitute the appropriate numbers into the above equations E= B= μ0 qv sin φ 4π × 10−7 T ⋅ m/A (1.60 × 10−19 C)(2.2 × 106 m/s)sin 90° = = 13 T, out of the page 4π r 4π (5.3 × 10−11 m)2 e (9.00 × 109 N ⋅ m /C2 )(1.60 × 10−19 C) = 5.1 × 1011 N/C, toward the electron 4π ⑀0 r (5.3 × 10−11 m) EVALUATE: There are enormous fields within the atom! E= = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-1 28-2 28.3 Chapter 28 IDENTIFY: A moving charge creates a magnetic field SET UP: The magnetic field due to a moving charge is B = μ0 qv sin φ 4π r EXECUTE: Substituting numbers into the above equation gives μ qv sin φ 4π × 10−7 T ⋅ m/A (1.6 × 10−19 C)(3.0 × 107 m/s)sin 30° (a) B = = 4π r 4π (2.00 × 10−6 m)2 B = 6.00 × 10 –8 T, out of the paper, and it is the same at point B (b) B = (1.00 × 10 –7 T ⋅ m/A)(1.60 × 10 –19 C)(3.00 × 107 m/s)/(2.00 × 10 –6 m) B = 1.20 × 10 –7 T, out of the page (c) B = T since sin(180°) = 28.4 EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the earth’s magnetic field IDENTIFY: Both moving charges produce magnetic fields, and the net field is the vector sum of the two fields SET UP: Both fields point out of the paper, so their magnitudes add, giving B = Balpha + Bel = μ 0v (e sin 40° + 2e sin140°) 4π r EXECUTE: Factoring out an e and putting in the numbers gives B= 4π × 10−7 T ⋅ m/A (1.60 × 10−19 C)(2.50 × 105 m/s) (sin 40° + 2sin140°) 4π (1.75 × 10−9 m)2 B = 2.52 × 10−3 T = 2.52 mT, out of the page 28.5 EVALUATE: At distances very close to the charges, the magnetic field is strong enough to be important G μ qvG × rG IDENTIFY: Apply B = 4π r G SET UP: Since the charge is at the origin, r = xiˆ + yˆj + zkˆ G G G G G EXECUTE: (a) v = vi , r = riˆ; v × r = 0, B = G G G G (b) v = viˆ, r = rˆj; v × r = vrkˆ , r = 0.500 m −7 2 −6 ⎛ μ ⎞ q v (1.0 × 10 N ⋅ s /C )(4.80 × 10 C)(6.80 × 10 m/s) B=⎜ ⎟ = = 1.31 × 10−6 T (0.500 m) ⎝ 4π ⎠ r G q is negative, so B = −(1.31 × 10−6 T)kˆ G G G G (c) v = viˆ, r = (0.500 m)(iˆ + ˆj ); v × r = (0.500 m)vkˆ , r = 0.7071 m (1.0 × 10−7 N ⋅ s /C2 )(4.80 × 10−6 C)(0.500 m)(6.80 × 105 m/s) G G ⎛μ ⎞ B = ⎜ ⎟ q v × r /r = (0.7071 m)3 ⎝ 4π ⎠ G B = 4.62 × 10−7 T B = −(4.62 × 10−7 T) kˆ G G G G (d) v = viˆ, r = rkˆ; v × r = −vrˆj , r = 0.500 m ( ) −7 2 −6 ⎛ μ ⎞ q v (1.0 × 10 N ⋅ s /C )(4.80 × 10 C)(6.80 × 10 m/s) B=⎜ ⎟ = = 1.31 × 10−6 T π (0.500 m) ⎝ ⎠ r G −6 B = (1.31 × 10 T) ˆj G G G EVALUATE: In each case, B is perpendicular to both r and v © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28.6 28-3 G μ qvG × rG IDENTIFY: Apply B = For the magnetic force, apply the results of Example 28.1, except here 4π r the two charges and velocities are different G G v ×r v G G SET UP: In part (a), r = d and r is perpendicular to v in each case, so = For calculating the r d force between the charges, r = 2d μ ⎛ qv q′v′ ⎞ EXECUTE: (a) Btotal = B + B′ = ⎜ + ⎟ 4π ⎝ d d ⎠ B= μ0 ⎛ (8.0 × 10−6 C)(4.5 × 106 m/s) (3.0 × 10−6 C)(9.0 × 106 m/s) ⎞ −4 + ⎜ ⎟⎟ = 4.38 × 10 T 4π ⎜⎝ (0.120 m) (0.120 m) ⎠ G The direction of B is into the page (b) Following Example 28.1 we can find the magnetic force between the charges: μ qq′vv′ (8.00 × 10−6 C)(3.00 × 10−6 C)(4.50 × 106 m/s)(9.00 × 106 m/s) −7 FB = = (10 T ⋅ m/A) 4π r (0.240 m)2 FB = 1.69 × 10−3 N The force on the upper charge points up and the force on the lower charge points 28.7 down The Coulomb force between the charges is qq (8.0 × 10−6 C)(3.0 × 10−6 C) FC = k 22 = (8.99 × 109 N ⋅ m /C2 ) = 3.75 N The force on the upper charge r (0.240 m) points up and the force on the lower charge points down The ratio of the Coulomb force to the magnetic F c2 3.75 N force is C = = = 2.22 × 103 ; the Coulomb force is much larger FB v1v2 1.69 × 10−3 N (c) The magnetic forces are reversed in direction when the direction of only one velocity is reversed but the magnitude of the force is unchanged EVALUATE: When two charges have the same sign and move in opposite directions, the force between them is repulsive When two charges of the same sign move in the same direction, the force between them is attractive G μ qvG × rG G G G IDENTIFY: Apply B = For the magnetic force on q′, use FB = q′v′ × Bq and for the magnetic 4π r G G G force on q use FB = qv × Bq′ G G v ×r v = SET UP: In part (a), r = d and r d μ0qv μ qv′ EXECUTE: (a) q′ = − q; Bq = , into the page; Bq′ = , out of the page 4π d 4π d v qv μ0qv μ (i) v′ = gives B = − 12 = , into the page (ii) v′ = v gives B = 4π d 4π (2d ) ( (iii) v′ = 2v gives B = ) μ0qv , out of the page 4π d G G G μ q 2v′v G (b) The force that q exerts on q′ is given by F = q′v ′ × Bq , so F = Bq is into the page, so the 4π (2d ) force on q′ is toward q The force that q′ exerts on q is toward q′ The force between the two charges is attractive (c) FB = μ0q 2vv′ q2 F , FC = so B = μ0⑀ 0vv′ = μ0⑀ (3.00 × 105 m/s) = 1.00 × 10−6 FC 4π (2d ) 4π ⑀0 (2d ) EVALUATE: When charges of opposite sign move in opposite directions, the force between them is attractive For the values specified in part (c), the magnetic force between the two charges is much smaller in magnitude than the Coulomb force between them © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-4 28.8 Chapter 28 IDENTIFY: Both moving charges create magnetic fields, and the net field is the vector sum of the two The magnetic force on a moving charge is Fmag = qvB sin φ and the electrical force obeys Coulomb’s law SET UP: The magnetic field due to a moving charge is B = μ0 qv sin φ 4π r EXECUTE: (a) Both fields are into the page, so their magnitudes add, giving B = Be + Bp = B= μ0 4π μ0 ⎛ ev ev ⎞ ⎜ + ⎟ sin 90° 4π ⎜ re2 rp2 ⎟ ⎝ ⎠ ⎡ ⎤ 1 (1.60 × 10−19 C)(845,000 m/s) ⎢ + −9 −9 2⎥ (4.00 × 10 m) ⎦ ⎣ (5.00 × 10 m) B = 1.39 × 10 –3 T = 1.39 mT, into the page (b) Using B = B= μ0 qv sin φ 4π r2 , where r = 41 nm and φ = 180° − arctan(5/4) = 128.7°, we get 4π × 10−7 T ⋅ m/A (1.6 × 10−19 C)(845,000 m/s)sin128.7° = 2.58 × 10−4 T, into the page 4π ( 41 × 10−9 m) (c) Fmag = qvB sin 90° = (1.60 × 10−19 C)(845,000 m/s)(2.58 × 10−4 T) = 3.48 × 10−17 N, in the +x-direction Felec = (1/4π ⑀0 )e2 /r = 28.9 28.10 28.11 (9.00 × 109 N ⋅ m /C )(1.60 × 10−19 C) ( 41 × 10−9 m) = 5.62 × 10−12 N, at 51.3° below the + x -axis measured clockwise EVALUATE: The electric force is much stronger than the magnetic force IDENTIFY: A moving charge creates a magnetic field G μ qvG × rG G SET UP: Apply B = r = (0.200 m)iˆ + (−0.300 m) ˆj , and r = 0.3606 m 4π r G G EXECUTE: v × r = [(7.50 × 104 m/s) iˆ + (−4.90 × 104 m/s) ˆj ] × [(0.200 m) iˆ + (−0.300 m) ˆj ], which simplifies to G G v × r = (−2.25 × 104 m /s)kˆ + (9.80 × 103 m /s)kˆ = ( −1.27 × 104 m /s)kˆ G ( −3.00 × 10−6 C)(−1.27 × 104 m /s) ˆ B = (1.00 × 10−7 T ⋅ m/A) k = (9.75 × 10−8 T)kˆ (0.3606 m)3 EVALUATE: We can check the direction of the magnetic field using the right-hand rule, which shows that the field points in the +z-direction IDENTIFY: Apply the Biot-Savart law G G μ qdl × rG SET UP: Apply dB = r = (−0.730 m)2 + (0.390 m) = 0.8267 m 4π r EXECUTE: G G dl × r = [0.500 × 10−3 m] ˆj × [( −0.730 m)iˆ + (0.390 m)kˆ ] = ( +3.65 × 10−4 m )kˆ + (+1.95 × 10−4 m )iˆ G 8.20 A dB = (1.00 × 10−7 T ⋅ m/A) [(3.65 × 10−4 m )kˆ + (1.95 × 10−4 m2 )iˆ] (0.8276 m)3 G dB = (2.83 × 10−10 T)iˆ + (5.28 × 10−10 T)kˆ EVALUATE: The magnetic field lies in the xz-plane IDENTIFY: A current segment creates a magnetic field μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 4π r EXECUTE: Applying the law of Biot and Savart gives 4π × 10−7 T ⋅ m/A (10.0 A)(0.00110 m) sin90° = 4.40 × 10 –7 T, out of the paper (a) dB = 4π (0.0500 m) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-5 (b) The same as above, except r = (5.00 cm) + (14.0 cm)2 and φ = arctan(5/14) = 19.65°, giving dB = 1.67 × 10 –8 T, out of the page (c) dB = since φ = 0° 28.12 EVALUATE: This is a very small field, but it comes from a very small segment of current G G G μ Idl × rˆ μ0 Idl × rG = IDENTIFY: Apply dB = 4π r 4π r μ Idl sin φ , where φ is the angle SET UP: The magnitude of the field due to the current element is dB = 4π r G between r and the current direction EXECUTE: The magnetic field at the given points is: μ Idl sin φ μ0 (200 A)(0.00100 m) dBa = = = 2.00 × 10−6 T 4π 4π r2 (0.100 m) dBb = dBc = dBd = dBe = μ0 Idl sin φ μ0 (200 A)(0.00100 m)sin 45° = = 0.705 × 10−6 T 4π r 4π 2(0.100 m) μ0 Idl sin φ 4π r2 μ0 Idl sin φ 4π r = = μ0 (200 A)(0.00100 m) 4π (0.100 m) μ0 Idl sin(0°) 4π r2 = 2.00 × 10−6 T = μ0 Idl sin φ μ0 (200 A)(0.00100 m) = = 0.545 × 10−6 T 4π r 4π 3(0.100 m) The field vectors at each point are shown in Figure 28.12 G EVALUATE: In each case dB is perpendicular to the current direction Figure 28.12 28.13 IDENTIFY and SET UP: The magnetic field produced by an infinitesimal current element is given by Eq (28.6) G G μ Il × rˆ dB = As in Example 28.2, use this equation for the finite 0.500-mm segment of wire since the 4π r Δl = 0.500-mm length is much smaller than the distances to the field points G G G μ I Δl × rˆ μ0 I Δl × rG B= = 4π r 4π r G I is in the + z -direction, so Δl = (0.500 × 10−3 m)kˆ G EXECUTE: (a) Field point is at x = 2.00 m, y = 0, z = so the vector r from the source point G (at the origin) to the field point is r = (2.00 m) iˆ G G Δl × r = (0.500 × 10−3 m)(2.00 m) kˆ × iˆ = + (1.00 × 10−3 m ) ˆj G (1 × 10−7 T ⋅ m/A)(4.00 A)(1.00 × 10−3 m ) ˆj = (5.00 × 10−11 T) ˆj B= (2.00 m)3 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-6 Chapter 28 G (b) r = (2.00 m) ˆj , r = 2.00 m G G Δl × r = (0.500 × 10−3 m)(2.00 m)kˆ × ˆj = −(1.00 × 10−3 m2 )iˆ G (1 × 10−7 T ⋅ m/A)(4.00 A)(−1.00 × 10−3 m ) B= iˆ = −(5.00 × 10−11 T) iˆ (2.00 m)3 G (c) r = (2.00 m)(iˆ + ˆj ), r = 2(2.00 m) G G Δl × r = (0.500 × 10−3 m)(2.00 m) kˆ × (iˆ + ˆj ) = (1.00 × 10−3 m )( ˆj − iˆ) 28.14 G (1 × 10−7 T ⋅ m/A)(4.00 A)(1.00 × 10−3 m ) ( ˆj − iˆ) = (−1.77 × 10−11 T)( iˆ − ˆj ) B= ⎡ 2(2.00 m) ⎤ ⎣ ⎦ G ˆ (d) r = (2.00 m)k , r = 2.00 m G G G Δl × r = (0.500 × 10−3 m)(2.00 m)kˆ × kˆ = 0; B = G G G EVALUATE: At each point B is perpendicular to both r and Δl B = along the length of the wire IDENTIFY: A current segment creates a magnetic field μ Idl sin φ SET UP: The law of Biot and Savart gives dB = 4π r Both fields are into the page, so their magnitudes add EXECUTE: Applying the law of Biot and Savart for the 12.0-A current gives 4π × 10−7 T ⋅ m/A dB = 4π 28.15 ⎛ 2.50 cm ⎞ (12.0 A)(0.00150 m) ⎜ ⎟ ⎝ 8.00 cm ⎠ = 8.79 × 10 –8 T (0.0800 m)2 The field from the 24.0-A segment is twice this value, so the total field is 2.64 × 10 –7 T, into the page EVALUATE: The rest of each wire also produces field at P We have calculated just the field from the two segments that are indicated in the problem IDENTIFY: A current segment creates a magnetic field μ Idl sin φ Both fields are into the page, so their SET UP: The law of Biot and Savart gives dB = 4π r magnitudes add EXECUTE: Applying the Biot and Savart law, where r = 12 (3.00 cm) + (3.00 cm)2 = 2.121 cm, we have dB = 28.16 28.17 4π × 10−7 T ⋅ m/A (28.0 A)(0.00200 m)sin 45.0° = 1.76 × 10 –5 T, into the paper 4π (0.02121 m) EVALUATE: Even though the two wire segments are at right angles, the magnetic fields they create are in the same direction IDENTIFY: A current segment creates a magnetic field μ Idl sin φ All four fields are of equal magnitude and SET UP: The law of Biot and Savart gives dB = 4π r into the page, so their magnitudes add 4π × 10−7 T ⋅ m/A (15.0 A)(0.00120 m) sin 90° = 2.88 × 10 –6 T, into the page EXECUTE: dB = 4π (0.0500 m) EVALUATE: A small current element causes a small magnetic field IDENTIFY: We can model the lightning bolt and the household current as very long current-carrying wires μ I SET UP: The magnetic field produced by a long wire is B = 2π r EXECUTE: Substituting the numerical values gives (4π × 10−7 T ⋅ m/A)(20,000 A) = × 10 –4 T (a) B = 2π (5.0 m) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-7 (4π × 10−7 T ⋅ m/A)(10 A) = 4.0 × 10 –5 T 2π (0.050 m) EVALUATE: The field from the lightning bolt is about 20 times as strong as the field from the household current IDENTIFY: The long current-carrying wire produces a magnetic field μ I SET UP: The magnetic field due to a long wire is B = 2π r (b) B = 28.18 EXECUTE: First find the current: I = (3.50 × 1018 el/s)(1.60 × 10 –19 C/el) = 0.560 A (4π × 10−7 T ⋅ m/A)(0.560 A) = 2.80 × 10 –6 T 2π (0.0400 m) Since electrons are negative, the conventional current runs from east to west, so the magnetic field above the wire points toward the north EVALUATE: This magnetic field is much less than that of the earth, so any experiments involving such a current would have to be shielded from the earth’s magnetic field, or at least would have to take it into consideration IDENTIFY: We can model the current in the heart as that of a long straight wire It produces a magnetic field around it μ I SET UP: For a long straight wire, B = μ0 = 4π × 10−7 T ⋅ m/A gauss = 10−4 T 2π r Now find the magnetic field: 28.19 28.20 28.21 28.22 2π rB 2π (0.050 m)(1.0 × 10−10 T) = 2.5 × 10−5 A = 25 μ A 4π × 10−7 T ⋅ m/A EVALUATE: By household standards, this is a very small current But the magnetic field around the heart (≈ μ G) is also very small IDENTIFY: The current in the transmission line creates a magnetic field If this field is greater than 5% of the earth’s magnetic field, it will interfere with the navigation of the bacteria μ I SET UP: B = due to a long straight wire 2π r μ I EXECUTE: We know the field is B = (0.05)(5 × 10−5 T) = 2.5 × 10−6 T Solving B = for r gives 2π r μ I 100 A = m r = = (2 × 10−7 T ⋅ m/A) 2π B 2.5 × 10−6 T EVALUATE: If the bacteria are within m (≈ 25 ft) of the cable, its magnetic field may be strong enough to affect their navigation IDENTIFY: The long current-carrying wire produces a magnetic field μ I SET UP: The magnetic field due to a long wire is B = 2π r EXECUTE: First solve for the current, then substitute the numbers using the above equation (a) Solving for the current gives EXECUTE: Solving for the current gives I = μ0 = I = 2π rB/μ0 = 2π (0.0200 m)(1.00 × 10−4 T)/(4π × 10−7 T ⋅ m/A) = 10.0 A (b) The earth’s horizontal field points northward, so at all points directly above the wire the field of the wire would point northward (c) At all points directly east of the wire, its field would point northward EVALUATE: Even though the earth’s magnetic field is rather weak, it requires a fairly large current to cancel this field G μ I IDENTIFY: For each wire B = (Eq 28.9), and the direction of B is given by the right-hand rule 2π r (Figure 28.6 in the textbook) Add the field vectors for each wire to calculate the total field (a) SET UP: The two fields at this point have the directions shown in Figure 28.22a © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-8 Chapter 28 EXECUTE: At point P midway between G G the two wires the fields B1 and B2 due to the two currents are in opposite directions, so B = B2 − B1 Figure 28.22a But B1 = B2 = μ0 I , so B = 2π a (b) SET UP: The two fields at this point have the directions shown in Figure 28.22b EXECUTE: At point Q above the upper G G wire B1 and B2 are both directed out of the page (+ z -direction), so B = B1 + B2 Figure 28.22b B1 = μ0 I μ0 I , B2 = 2π a 2π (3a ) B= μ0 I 2μ I G 2μ I + 13 = ; B = kˆ 2π a 3π a 3π a ( ) (c) SET UP: The two fields at this point have the directions shown in Figure 28.22c EXECUTE: At point R below the lower G G wire B1 and B2 are both directed into the page (− z -direction), so B = B1 + B2 Figure 28.22c μ0 I μ I , B2 = 2π (3a) 2π a μ0 I μ I G 2μ I B1 = + 13 = ; B = − kˆ 2π a 3π a 3π a B1 = ( ) G EVALUATE: In the figures we have drawn, B due to each wire is out of the page at points above the wire and into the page at points below the wire If the two field vectors are in opposite directions the magnitudes subtract © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28.23 28-9 IDENTIFY: The total magnetic field is the vector sum of the constant magnetic field and the wire’s magnetic field μ I SET UP: For the wire, Bwire = and the direction of Bwire is given by the right-hand rule that is 2π r G illustrated in Figure 28.6 in the textbook B = (1.50 × 10−6 T)iˆ G G μ I μ (8.00 A) ˆ EXECUTE: (a) At (0, 0, m), B = B0 − iˆ = (1.50 × 10−6 T)iˆ − i = −(1.0 × 10−7 T)iˆ 2π r 2π (1.00 m) G G μ I μ (8.00 A) ˆ (b) At (1 m, 0, 0), B = B0 + kˆ = (1.50 × 10−6 T) iˆ + k 2π r 2π (1.00 m) G B = (1.50 × 10−6 T)iˆ + (1.6 × 10−6 T)kˆ = 2.19 × 10−6 T, at θ = 46.8° from x to z 28.24 G G μ I μ (8.00 A) ˆ (c) At (0, 0, –0.25 m), B = B0 + iˆ = (1.50 × 10−6 T)iˆ + i = (7.9 × 10−6 T)iˆ 2π r 2π (0.25 m) EVALUATE: At point c the two fields are in the same direction and their magnitudes add At point a they are in opposite directions and their magnitudes subtract At point b the two fields are perpendicular IDENTIFY: The magnetic field is that of a long current-carrying wire μ I SET UP: B = 2π r EXECUTE: B = 28.25 μ0 I (2.0 × 10−7 T ⋅ m/A)(150 A) = = 3.8 × 10−6 T This is 7.5% of the earth’s field 2π r 8.0 m EVALUATE: Since this field is much smaller than the earth’s magnetic field, it would be expected to have less effect than the earth’s field G μ I IDENTIFY: B = The direction of B is given by the right-hand rule 2π r SET UP: Call the wires a and b, as indicated in Figure 28.25 The magnetic fields of each wire at points P1 and P2 are shown in Figure 28.25a The fields at point are shown in Figure 28.25b EXECUTE: (a) At P1, Ba = Bb and the two fields are in opposite directions, so the net field is zero G μ I μ I G (b) Ba = Bb = Ba and Bb are in the same direction so 2π 2π rb B = Ba + Bb = μ0 I ⎛ 1 ⎞ (4π × 10−7 T ⋅ m/A)(4.00 A) ⎡ 1 ⎤ −6 + ⎜ + ⎟= ⎢ ⎥ = 6.67 × 10 T 2π ⎝ rb ⎠ 2π ⎣ 0.300 m 0.200 m ⎦ G B has magnitude 6.67 μ T and is directed toward the top of the page G G G G cm (c) In Figure 28.25b, Ba is perpendicular to and Bb is perpendicular to rb tan θ = and 20 cm θ = 14.04° = rb = (0.200 m)2 + (0.050 m) = 0.206 m and Ba = Bb ⎛ μ I ⎞ 2(4π × 10−7 T ⋅ m/A)(4.0 A)cos14.04° = 7.54 μ T B = Ba cosθ + Bb cosθ = Ba cosθ = ⎜ ⎟ cosθ = 2π (0.206 m) ⎝ 2π ⎠ B has magnitude 7.53 μ T and is directed to the left EVALUATE: At points directly to the left of both wires the net field is directed toward the bottom of the page © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-10 Chapter 28 Figure 28.25 28.26 IDENTIFY: Each segment of the rectangular loop creates a magnetic field at the center of the loop, and all these fields are in the same direction G μ I 2a B is into paper so I is clockwise around the SET UP: The field due to each segment is B = 4π x x + a loop EXECUTE: Long sides: a = 4.75 cm x = 2.10 cm For the two long sides, B = 2(1.00 × 10−7 T ⋅ m/A) I 2(4.75 × 10−2 m) (2.10 × 10 −2 m) (0.0210 m) + (0.0475 m) = (1.742 × 10−5 T/A) I Short sides: a = 2.10 cm x = 4.75 cm For the two short sides, B = 2(1.00 × 10−7 T ⋅ m/A) I 28.27 2(2.10 × 10−2 m) (4.75 × 10 −2 m) (0.0475 m) + (0.0210 m) = (3.405 × 10−6 T/A) I Using the known field, we have B = (2.082 × 10−5 T/A) I = 5.50 × 10−5 T, which gives I = 2.64 A EVALUATE: This is a typical household current, yet it produces a magnetic field which is about the same as the earth’s magnetic field IDENTIFY: The net magnetic field at the center of the square is the vector sum of the fields due to each wire G μ I SET UP: For each wire, B = and the direction of B is given by the right-hand rule that is illustrated 2π r in Figure 28.6 in the textbook EXECUTE: (a) and (b) B = since the magnetic fields due to currents at opposite corners of the square cancel (c) The fields due to each wire are sketched in Figure 28.27 ⎛μ I⎞ B = Ba cos 45° + Bb cos 45° + Bc cos 45° + Bd cos 45° = Ba cos 45° = ⎜ ⎟ cos 45° ⎝ 2π r ⎠ r = (10 cm) + (10 cm) = 10 cm = 0.10 m, so (4π × 10−7 T ⋅ m/A)(100 A) cos 45° = 4.0 × 10−4 T, to the left 2π (0.10 m) EVALUATE: In part (c), if all four currents are reversed in direction, the net field at the center of the square would be to the right B=4 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-20 Chapter 28 Figure 28.57 28.58 IDENTIFY: The presence of the magnetic material causes the net field to be slightly stronger than it would be in air SET UP: K m = Binside /Boutside μ = K m μ0 EXECUTE: (a) K m = 1.5023 T = 1.0015 1.5000 T (b) μ = K m μ0 = (1.0015)(4π × 10−7 T ⋅ m/A) = 1.259 × 10−6 T ⋅ m/A EVALUATE: μ is not much different from μ0 since this is a paramagnetic material 28.59 IDENTIFY: Moving charges create magnetic fields The net field is the vector sum of the two fields A charge moving in an external magnetic field feels a force μ qv sin φ (a) SET UP: The magnetic field due to a moving charge is B = Both fields are into the paper, 4π r μ ⎛ qvsinφ q′v′sinφ ′ ⎞ so their magnitudes add, giving Bnet = B + B′ = ⎜ + ⎟ 4π ⎝ r r ′2 ⎠ EXECUTE: Substituting numbers gives μ ⎡ (8.00 μ C)(9.00 × 104 m/s)sin 90° (5.00 μ C)(6.50 × 104 m/s)sin 90° ⎤ Bnet = ⎢ + ⎥ 4π ⎣⎢ (0.300 m) (0.400 m) ⎦⎥ Bnet = 1.00 × 10−6 T = 1.00 μ T, into the paper G G G (b) SET UP: The magnetic force on a moving charge is F = qv × B, and the magnetic field of charge q′ at the location of charge q is into the page The force on q is G G G G μ qv′ × rˆ ⎛ μ qv' sin φ ⎞ ˆ = ( qv) iˆ × ⎜ F = qv × B′ = (qv)iˆ × ⎟ ( −k ) = 4π r r2 ⎠ ⎝ 4π G where φ is the angle between v ′ and rˆ′ EXECUTE: Substituting numbers gives ⎛ μ0 qq′vv′sinφ ⎞ ˆ ⎜ ⎟j r2 ⎝ 4π ⎠ G μ ⎡ (8.00 × 10−6 C)(5.00 × 10−6 C)(9.00 × 104 m/s)(6.50 × 104 m/s) ⎛ 0.400 ⎞ ⎤ ˆ F= 0⎢ ⎜ ⎟⎥ j 4π ⎣⎢ (0.500 m)2 ⎝ 0.500 ⎠ ⎦⎥ G F = (7.49 × 10−8 N) ˆj 28.60 EVALUATE: These are small fields and small forces, but if the charge has small mass, the force can affect its motion IDENTIFY: Charge q1 creates a magnetic field due to its motion This field exerts a magnetic force on q2 , which is moving in that field G G μ qvG × rG G SET UP: Find B1, the field produced by q1 at the location of q2 B1 = , since rˆ = r/r 4π r G EXECUTE: r = (0.150 m)iˆ + (−0.250 m) ˆj , so r = 0.2915 m G G v × r = [(9.20 × 105 m/s)iˆ] × [(0.150 m) iˆ + (−0.250 m) ˆj ] = (9.20 × 105 m/s)(−0.250 m)kˆ G (4.80 × 10−6 C)(9.20 × 105 m/s)(−0.250 m) ˆ B1 = (1.00 × 10−7 T ⋅ m/A) k = −(4.457 × 10−6 T)kˆ (0.2915 m)3 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-21 G The force that B1 exerts on q2 is G G F2 = q2v2 × B1 = (−2.90 × 10−6 C)(−5.30 × 105 m/s)(−4.457 × 10−6 T) ˆj × kˆ = −(6.850 × 10−6 N)iˆ EVALUATE: If we think of the moving charge q1 as a current, we can use the right-hand rule for the direction of the magnetic field due to a current to find the direction of the magnetic field it creates in the vicinity of q2 Then we can use the cross product right-hand rule to find the direction of the force 28.61 this field exerts on q2 , which is in the −x-direction, in agreement with our result IDENTIFY: Use Eq (28.9) and the right-hand rule to determine points where the fields of the two wires cancel (a) SET UP: The only place where the magnetic fields of the two wires are in opposite directions is between the wires, in the plane of the wires Consider a point a distance x from the wire carrying I = 75.0 A Btot will be zero where B1 = B2 μ0 I1 μ I = 2π (0.400 m − x) 2π x I (0.400 m − x) = I1x; I1 = 25.0 A, I = 75.0 A x = 0.300 m; Btot = along a line 0.300 m from the wire carrying 75.0 A and 0.100 m from the wire carrying current 25.0 A (b) SET UP: Let the wire with I1 = 25.0 A be 0.400 m above the wire with I = 75.0 A The magnetic fields of the two wires are in opposite directions in the plane of the wires and at points above both wires or below both wires But to have B1 = B2 must be closer to wire #1 since I1 < I , so can have Btot = only at points above both wires Consider a point a distance x from the wire carrying I1 = 25.0 A Btot will be zero where B1 = B2 μ0 I1 μ0 I EXECUTE: = 2π x 2π (0.400 m + x) I x = I1(0.400 m + x); x = 0.200 m Btot = along a line 0.200 m from the wire carrying current 25.0 A and 0.600 m from the wire carrying current I = 75.0 A EVALUATE: For parts (a) and (b) the locations of zero field are in different regions In each case the points of zero field are closer to the wire that has the smaller current IDENTIFY: The wire creates a magnetic field near it, and the moving electron feels a force due to this field μ I SET UP: The magnetic field due to the wire is B = , and the force on a moving charge is 2π r F = q vB sin φ EXECUTE: 28.62 EXECUTE: F = q vB sin φ = (evμ0 I sin φ )/2π r Substituting numbers gives 28.63 F = (1.60 × 10−19 C)(6.00 × 104 m/s)(4π × 10−7 T ⋅ m/A)(5.20 A)(sin 90°)/[2π (0.0450 m)] G G F = 2.22 × 10 –19 N From the right-hand rule for the cross product, the direction of v × B is opposite to the current, but since the electron is negative, the force is in the same direction as the current EVALUATE: This force is small at an everyday level, but it would give the electron an acceleration of about 1011 m/s IDENTIFY: Find the force that the magnetic field of the wire exerts on the electron SET UP: The force on a moving charge has magnitude F = q vB sin φ and direction given by the rightG μ I hand rule For a long straight wire, B = and the direction of B is given by the right-hand rule 2π r q vB sin φ F ev ⎛ μ I ⎞ EXECUTE: (a) a = = = ⎜ ⎟ Substituting numbers gives m m m ⎝ 2π r ⎠ a= (1.6 × 10−19 C)(2.50 × 105 m/s)(4π × 10−7 T ⋅ m/A)(13.0 A) (9.11 × 10−31 kg)(2π )(0.0200 m) = 5.7 × 1012 m/s , away from the wire © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-22 Chapter 28 (b) The electric force must balance the magnetic force eE = evB, and μ0 I (250,000 m/s)(4π × 10−7 T ⋅ m/A)(13.0 A) = = 32.5 N/C The magnetic force is directed 2π r 2π (0.0200 m) away from the wire so the force from the electric field must be toward the wire Since the charge of the electron is negative, the electric field must be directed away from the wire to produce a force in the desired direction E = vB = v EVALUATE: (c) mg = (9.11 × 10−31 kg)(9.8 m/s ) ≈ 10−29 N Fel = eE = (1.6 × 10−19 C)(32.5 N/C) ≈ × 10−18 N Fel ≈ × 1011 Fgrav , so we can neglect gravity 28.64 IDENTIFY: The net magnetic field is the vector sum of the fields due to each wire G μ I SET UP: B = The direction of B is given by the right-hand rule 2π r EXECUTE: (a) The currents are the same so points where the two fields are equal in magnitude are equidistant from the two wires The net field is zero along the dashed line shown in Figure 28.64a (b) For the magnitudes of the two fields to be the same at a point, the point must be times closer to the wire with the smaller current The net field is zero along the dashed line shown in Figure 28.64b (c) As in (a), the points are equidistant from both wires The net field is zero along the dashed line shown in Figure 28.64c EVALUATE: The lines of zero net field consist of points at which the fields of the two wires have opposite directions and equal magnitudes Figure 28.64 28.65 IDENTIFY: Find the net magnetic field due to the two loops at the location of the proton and then find the force these fields exert on the proton SET UP: For a circular loop, the field on the axis, a distance x from the center of the loop is μ0 IR B= R = 0.200 m and x = 0.125 m 2( R + x )3/ ⎡ ⎤ μ0 IR EXECUTE: The fields add, so B = B1 + B2 = B1 = ⎢ Putting in the numbers gives 2 3/2 ⎥ ⎣⎢ 2( R + x ) ⎦⎥ B= (4π × 10−7 T ⋅ m/A)(3.80 A)(0.200 m)2 2 3/ [(0.200 m) + (0.125 m) ] = 1.46 × 10−5 T The magnetic force is F = q vB sin φ = (1.6 × 10−19 C)(2,400,000 m/s)(1.46 × 10−5 T)sin 90° = 5.59 × 10−18 N EVALUATE: The weight of a proton is w = mg = 1.6 × 10−24 N, so the force from the loops is much greater 28.66 than the gravity force on the proton G μ qvG × rˆ IDENTIFY: B = 02 4π r G ˆ SET UP: rˆ = i and r = 0.250 m, so v0 × rˆ = v0 z ˆj − v0 y kˆ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-23 G μ q μ q EXECUTE: B = (v0 z ˆj − v0 y kˆ ) = (6.00 × 10−6 T) ˆj v0 y = 0 v0 z = 6.00 × 10−6 T and 4π r 4π r v0 z = 4π (6.00 × 10−6 T)(0.25 m)2 μ0 (−7.20 × 10−3 C) = −521 m/s v0 x = ± v02 − v02y − v02z = ± (800 m/s) − (−521 m/s) = ±607 m/s The sign of v0 x isn’t determined G μ qvG × rˆ μ q G (b) Now r = ˆj and r = 0.250 m B = 02 = (v0 x kˆ − v0 z iˆ) 4π r 4π r μ0 q μ q μ (7.20 × 10−3 C) 800 m/s = 9.20 × 10−6 T v0 x + v02z = v0 = 4π r 4π r 4π (0.250 m) EVALUATE: The magnetic field in part (b) doesn’t depend on the sign of v0 x IDENTIFY: Use Eq (28.9) and the right-hand rule to calculate the magnitude and direction of the magnetic field at P produced by each wire Add these two field vectors to find the net field (a) SET UP: The directions of the fields at point P due to the two wires are sketched in Figure 28.67a B= 28.67 G G EXECUTE: B1 and B2 must be equal and opposite for the resultant field at P to be zero G B2 is to the right so I is out of the page Figure 28.67a B1 = μ0 I1 μ0 ⎛ 6.00 A ⎞ = ⎜ ⎟ 2π r1 2π ⎝ 1.50 m ⎠ B1 = B2 says B2 = μ0 I μ0 ⎛ I ⎞ = ⎜ ⎟ 2π r2 2π ⎝ 0.50 m ⎠ μ0 ⎛ 6.00 A ⎞ μ0 ⎛ I ⎞ ⎜ ⎟= ⎜ ⎟ 2π ⎝ 1.50 m ⎠ 2π ⎝ 0.50 m ⎠ ⎛ 0.50 m ⎞ I2 = ⎜ ⎟ (6.00 A) = 2.00 A ⎝ 1.50 m ⎠ (b) SET UP: The directions of the fields at point Q are sketched in Figure 28.67b EXECUTE: B1 = μ0 I1 2π r1 ⎛ 6.00 A ⎞ −6 B1 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.40 × 10 T ⎝ 0.50 m ⎠ μ I B2 = 2π r2 ⎛ 2.00 A ⎞ −7 B2 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.67 × 10 T ⎝ 1.50 m ⎠ Figure 28.67b G G B1 and B2 are in opposite directions and B1 > B2 so G B = B1 − B2 = 2.40 × 10−6 T − 2.67 × 10−7 T = 2.13 × 10−6 T, and B is to the right (c) SET UP: The directions of the fields at point S are sketched in Figure 28.67c © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-24 Chapter 28 EXECUTE: B1 = μ0 I1 2π r1 ⎛ 6.00 A ⎞ −6 B1 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 2.00 × 10 T 60 m ⎝ ⎠ μ0 I B2 = 2π r2 ⎛ 2.00 A ⎞ −7 B2 = (2 × 10−7 T ⋅ m/A) ⎜ ⎟ = 5.00 × 10 T ⎝ 0.80 m ⎠ Figure 28.67c G G B1 and B2 are right angles to each other, so the magnitude of their resultant is given by B = B12 + B22 = (2.00 × 10−6 T) + (5.00 × 10−7 T) = 2.06 × 10−6 T 28.68 EVALUATE: The magnetic field lines for a long, straight wire are concentric circles with the wire at the G center The magnetic field at each point is tangent to the field line, so B is perpendicular to the line from the wire to the point where the field is calculated IDENTIFY: Find the vector sum of the magnetic fields due to each wire G μ I SET UP: For a long straight wire B = The direction of B is given by the right-hand rule and is 2π r perpendicular to the line from the wire to the point where the field is calculated EXECUTE: (a) The magnetic field vectors are shown in Figure 28.68a a μ I μ0 I μ0 Ia (b) At a position on the x-axis Bnet = sin θ = = , in the positive 2 2 2π r π ( x2 + a ) π x +a x +a x-direction (c) The graph of B versus x/a is given in Figure 28.68b EVALUATE: (d) The magnetic field is a maximum at the origin, x = μ Ia (e) When x a, B ≈ πx Figure 28.68 28.69 IDENTIFY: Apply F = lB sin φ , with the magnetic field at point P that is calculated in Problem 28.68 μ0 Ia SET UP: The net field of the first two wires at the location of the third wire is B = , in the π ( x2 + a ) + x-direction EXECUTE: (a) Wire is carrying current into the page, so it feels a force in the − y -direction ⎛ μ0 Ia ⎞ F μ0 (6.00 A)2 (0.400 m) = IB = I ⎜⎜ = = 1.11 × 10−5 N/m ⎟ 2 ⎟ 2 L ( x a ) ((0 600 m) (0 400 m) ) + + π π ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-25 (b) If the wire carries current out of the page then the force felt will be in the opposite direction as in part (a) Thus the force will be 1.11 × 10−5 N/m, in the + y -direction 28.70 EVALUATE: We could also calculate the force exerted by each of the first two wires and find the vector sum of the two forces IDENTIFY: The wires repel each other since they carry currents in opposite directions, so the wires will move away from each other until the magnetic force is just balanced by the force due to the spring SET UP: Call x the distance the springs each stretch The force of the spring is kx and the magnetic force μ I 2L on each wire is Fmag = 2π x EXECUTE: On each wire, Fspr = Fmag , and there are two spring forces on each wire Therefore μ0 I L μ0 I L , which gives x = 2π x 4π k EVALUATE: Since μ0 /4π is small, x will likely be much less than the length of the wires 2kx = 28.71 G IDENTIFY: Apply ∑ F = to one of the wires The force one wire exerts on the other depends on I so G ∑ F = gives two equations for the two unknowns T and I SET UP: The force diagram for one of the wires is given in Figure 28.71 ⎛ μ I2 ⎞ The force one wire exerts on the other is F = ⎜ ⎟ L, where ⎜ 2π r ⎟ ⎝ ⎠ r = 2(0.040 m)sin θ = 8.362 × 10−3 m is the distance between the two wires Figure 28.71 EXECUTE: ∑ Fy = gives T cosθ = mg and T = mg/ cosθ ∑ Fx = gives F = T sin θ = ( mg/ cosθ )sin θ = mg tan θ And m = λ L, so F = λ Lg tan θ ⎛ μ0 I ⎞ ⎜⎜ ⎟⎟ L = λ Lg tan θ ⎝ 2π r ⎠ I= I= 28.72 λ gr tan θ ( μ0 /2π ) (0.0125 kg/m)(9.80 m/s )(tan 6.00°)(8.362 × 10−3 m) × 10−7 T ⋅ m/A = 23.2 A EVALUATE: Since the currents are in opposite directions the wires repel When I is increased, the angle θ from the vertical increases; a large current is required even for the small displacement specified in this problem IDENTIFY: Consider the forces on each side of the loop SET UP: The forces on the left and right sides cancel The forces on the top and bottom segments of the loop are in opposite directions, so the magnitudes subtract ⎛1 1⎞ ⎛μ I ⎞ ⎛ Il Il ⎞ μ IlI EXECUTE: F = Ft − Fb = ⎜ wire ⎟ ⎜ − ⎟ = wire ⎜ − ⎟ 2π ⎝ rt rb ⎠ ⎝ 2π ⎠ ⎝ rt rb ⎠ ⎞ μ0 (5.00 A)(0.200 m)(14.0 A) ⎛ 1 −5 + ⎜− ⎟ = 7.97 × 10 N The force on the top segment is 2π ⎝ 0.100 m 0.026 m ⎠ away from the wire, so the net force is away from the wire EVALUATE: The net force on a current loop in a uniform magnetic field is zero, but the magnetic field of the wire is not uniform; it is stronger closer to the wire F= © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-26 28.73 Chapter 28 IDENTIFY: Knowing the magnetic field at the center of the ring, we can calculate the current running through it We can then use this current to calculate the torque that the external magnetic field exerts on the ring SET UP: The torque on a current loop is τ = IAB sin φ We can use the magnetic field of the ring, μ0 I , to calculate the current in the ring 2R RBring 2(2.50 × 10−2 m)(75.4 × 10−6 T) EXECUTE: I = = = 3.00 A The torque is a maximum when μ0 4π × 10−7 T ⋅ m/A φ = 90° and the plane of the ring is parallel to the field B= 28.74 28.75 τ max = IAB = (3.00 A)(0.375 T)π (2.50 × 10−2 m)2 = 2.21 × 10−3 N ⋅ m EVALUATE: When the external field is perpendicular to the plane of the ring the torque on the ring is zero G G μ Idl × rˆ IDENTIFY: Apply dB = 4π r SET UP: The two straight segments produce zero field at P The field at the center of a circular loop of μ I μ I radius R is B = , so the field at the center of curvature of a semicircular loop is B = 2R 4R EXECUTE: The semicircular loop of radius a produces field out of the page at P and the semicircular loop of ⎛ μ I ⎞⎛ 1 ⎞ μ I ⎛ a ⎞ radius b produces field into the page Therefore, B = Ba − Bb = ⎜ ⎟⎜ − ⎟ = ⎜1 − ⎟ , out of page ⎝ ⎠⎝ a b ⎠ 4a ⎝ b ⎠ EVALUATE: If a = b, B = IDENTIFY: Find the vector sum of the fields due to each loop μ0 Ia SET UP: For a single loop B = Here we have two loops, each of N turns, and measuring 2( x + a )3/2 the field along the x-axis from between them means that the “x” in the formula is different for each case: EXECUTE: Left coil: x → x + a μ0 NIa ⇒ Bl = 2(( x + a/2) + a )3/ a μ0 NIa ⇒ Br = 2(( x − a /2) + a )3/2 So, the total field at a point a distance x from the point between them is ⎞ μ NIa ⎛ 1 + B= ⎜⎜ ⎟ 2 3/2 2 3/2 ⎝ (( x + a /2) + a ) (( x − a /2) + a ) ⎟⎠ Right coil: x → x − (b) B versus x is graphed in Figure 28.75 Figure 28.75a is the total field and Figure 28.75b is the field from the right-hand coil 3/2 ⎞ μ NIa ⎛ 1 μ0 NIa ⎛ ⎞ μ0 NI (c) At point P, x = and B = + = = ⎜⎜ ⎟ ⎜ ⎟ ⎝ ((a /2) + a )3/2 ((− a /2)2 + a )3/ ⎠⎟ (5a /4)3/ ⎝ ⎠ a ⎛ 4⎞ (d) B = ⎜ ⎟ ⎝5⎠ (e) dB dx 3/ μ0 NI ⎛ ⎞ =⎜ ⎟ a ⎝5⎠ 3/2 μ0 (300)(6.00 A) = 0.0202 T (0.080 m) ⎞ dB μ0 NIa ⎛ −3( x + a /2) −3( x − a /2) = + ⎜⎜ ⎟ At x = 0, 2 5/2 2 5/2 dx ⎝ (( x + a /2) + a ) (( x − a /2) + a ) ⎟⎠ = x =0 d B dx = μ0 NIa ⎛ ⎞ −3(a /2) −3(− a /2) + = ⎜⎜ 2 5/2 2 5/2 ⎟ ((− a /2) + a ) ⎟⎠ ⎝ (( a /2) + a ) μ0 NIa ⎛ −3 6( x + a /2) (5/2) −3 6( x − a /2) (5/2) ⎞ + + + ⎜⎜ ⎟ ⎝ (( x + a /2) + a )5/ (( x + a /2) + a )7/ (( x − a /2) + a )5/ (( x − a /2) + a )7/2 ⎟⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-27 At x = 0, d 2B dx = x =0 μ0 NIa ⎛ −3 6( a /2) (5/2) −3 6( −a /2) (5/2) ⎞ + + + ⎜⎜ ⎟ = 2 5/2 2 7/2 2 5/2 ⎝ (( a /2) + a ) ((a /2) + a ) ((a /2) + a ) ((a /2) + a )7/2 ⎟⎠ EVALUATE: Since both first and second derivatives are zero, the field can only be changing very slowly Figure 28.75 28.76 IDENTIFY: A current-carrying wire produces a magnetic field, but the strength of the field depends on the shape of the wire SET UP: The magnetic field at the center of a circular wire of radius a is B = μ0 I/2a, and the field a μ0 I 2a 4π x x + a EXECUTE: (a) Since the diameter D = 2a, we have B = μ0 I/2a = μ0 I/D (b) In this case, the length of the wire is equal to the diameter of the circle, so 2a = π D, giving a = π D/2, μ I 2(π D/2) μ0 I = and x = D/2 Therefore B = 2 4π ( D/2) D /4 + π D /4 D + π distance x from the center of a straight wire of length 2a is B = 28.77 EVALUATE: The field in part (a) is greater by a factor of + π It is reasonable that the field due to the circular wire is greater than the field due to the straight wire because more of the current is close to point A for the circular wire than it is for the straight wire (a) IDENTIFY: Consider current density J for a small concentric ring and integrate to find the total current in terms of α and R SET UP: We can’t say I = JA = J π R , since J varies across the cross section To integrate J over the cross section of the wire, divide the wire cross section up into thin concentric rings of radius r and width dr, as shown in Figure 28.77 Figure 28.77 EXECUTE: The area of such a ring is dA, and the current through it is dI = J dA; dA = 2π rdr and dI = J dA = α r (2π r dr ) = 2πα r dr R 3I 2π R3 I = ∫ dI = 2πα ∫ r dr = 2πα ( R3/3) so α = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-28 Chapter 28 (b) IDENTIFY and SET UP: (i) r ≤ R Apply Ampere’s law to a circle of radius r < R Use the method of part (a) to find the current enclosed by Ampere’s law path G G G EXECUTE: v∫ B ⋅ dl = v∫ B dl = B v ∫ dl = B(2π r ), by the symmetry and direction of B The current passing through the path is I encl = ∫ dl , where the integration is from to r r I encl = 2πα ∫ r dr = 2πα r 2π ⎛ 3I ⎞ Ir = ⎜ ⎟ r = Thus 3 ⎝ 2π R3 ⎠ R G G v∫ B ⋅ dl = μ0 Iencl ⎛ Ir ⎞ μ Ir B (2π r ) = μ0 ⎜ ⎟ and B = ⎜R ⎟ 2π R ⎝ ⎠ (ii) IDENTIFY and SET UP: r ≥ R Apply Ampere’s law to a circle of radius r > R G G EXECUTE: v∫ B ⋅ dl = v ∫ B dl = B v∫ dl = B(2π r ) I encl = I ; all the current in the wire passes through this path Thus and B = G G v∫ B ⋅ dl = μ0 Iencl μ0 I 2π r EVALUATE: Note that at r = R the expression in (i) (for r ≤ R ) gives B = 28.78 28.79 gives B (2π r ) = μ0 I μ0 I At r = R the 2π R μ I expression in (ii) (for r ≥ R ) gives B = , which is the same 2π R G G μ Idl × rˆ IDENTIFY: Apply dB = 4π r G G SET UP: The horizontal wire yields zero magnetic field since dl × r = The vertical current provides the magnetic field of half of an infinite wire (The contributions from all infinitesimal pieces of the wire point in the same direction, so there is no vector addition or components to worry about.) ⎛μ I ⎞ μ I EXECUTE: B = 12 ⎜ ⎟ = and is directed out of the page ⎝ 2π R ⎠ 4π R EVALUATE: In the equation preceding Eq (28.8) the limits on the integration are to a rather than − a to a and this introduces a factor of 12 into the expression for B IDENTIFY: Apply Ampere’s law to a circular path of radius r SET UP: Assume the current is uniform over the cross section of the conductor EXECUTE: (a) r < a ⇒ I encl = ⇒ B = ⎛ π (r − a ) ⎞ ⎛A ⎞ (r − a ) (b) a < r < b ⇒ I encl = I ⎜ a → r ⎟ = I ⎜ ⎟⎟ = I 2 ⎜ (b − a ) ⎝ Aa →b ⎠ ⎝ π (b − a ) ⎠ B= G G (r − a ) v∫ B ⋅ dl = B 2π r = μ0 I (b2 − a ) and μ0 I ( r − a ) 2π r (b − a ) G G μ0 I 2π r EVALUATE: The expression in part (b) gives B = at r = a and this agrees with the result of part (a) μ I The expression in part (b) gives B = at r = b and this agrees with the result of part (c) 2π b IDENTIFY: The net field is the vector sum of the fields due to the circular loop and to the long straight wire μ I μ I SET UP: For the long wire, B = , and for the loop, B = 2π D 2R (c) r > b ⇒ I encl = I 28.80 gives v∫ B ⋅ dl = B 2π r = μ0 I and B = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-29 EXECUTE: At the center of the circular loop the current I generates a magnetic field that is into the page, so the current I1 must point to the right For complete cancellation the two fields must have the same μ0 I1 μ0 I πD I2 = Thus, I1 = R 2π D 2R EVALUATE: If I1 is to the left the two fields add IDENTIFY: Use the current density J to find dI through a concentric ring and integrate over the appropriate G cross section to find the current through that cross section Then use Ampere’s law to find B at the specified distance from the center of the wire (a) SET UP: magnitude: 28.81 Divide the cross section of the cylinder into thin concentric rings of radius r and width dr, as shown in Figure 28.81a The current through each ring is dI = J dA = J 2π r dr Figure 28.81a EXECUTE: dI = 2I0 π a2 [1 − ( r/a ) ]2π r dr = 4I0 a2 [1 − (r/a ) ]r dr The total current I is obtained by integrating a a ⎛ 4I ⎞ a ⎛ 4I ⎞ ⎡ ⎤ dI over the cross section I = ∫ dI = ⎜ 20 ⎟ ∫ (1 − r /a )r dr = ⎜ 20 ⎟ ⎢ r − r /a ⎥ = I , as was to be 0 ⎝ a ⎠ ⎝ a ⎠⎣2 ⎦0 shown (b) SET UP: Apply Ampere’s law to a path that is a circle of radius r > a, as shown in Figure 28.81b G G v∫ B ⋅ dl = B(2π r ) I encl = I (the path encloses the entire cylinder) Figure 28.81b EVALUATE: G G v∫ B ⋅ dl = μ0 Iencl says B (2π r ) = μ0 I and B = μ0 I 2π r (c) SET UP: Divide the cross section of the cylinder into concentric rings of radius r′ and width dr′, as was done in part (a) See Figure 28.81c The current 4I ⎡ ⎛ r′ ⎞ ⎤ dI through each ring is dI = 20 ⎢1 − ⎜ ⎟ ⎥ r ′ dr ′ a ⎢⎣ ⎝ a ⎠ ⎥⎦ Figure 28.81c EXECUTE: The current I is obtained by integrating dI from r ′ = to r ′ = r: r 4I r ⎡ ⎛ r′ ⎞ ⎤ 4I I = ∫ dI = 20 ∫ ⎢1 − ⎜ ⎟ ⎥ r ′ dr ′ = 20 ⎡ 12 (r′) − 14 (r′) /a ⎤ ⎣ ⎦ 0 a a ⎢⎣ ⎝ a ⎠ ⎥⎦ I= 4I0 a (r /2 − r /4a ) = I 0r ⎛ r2 ⎞ 2− ⎟ ⎜ a ⎜⎝ a ⎟⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-30 Chapter 28 (d) SET UP: Apply Ampere’s law to a path that is a circle of radius r < a, as shown in Figure 28.81d G G v∫ B ⋅ dl = B(2π r ) I encl = I 0r ⎛ r2 ⎞ − ⎟ (from part (c)) ⎜ ⎜ a ⎝ a ⎟⎠ Figure 28.81d G G I 0r μ I r B ⋅ d l = μ I says B (2 π r ) = μ (2 − r /a ) and B = 0 (2 − r /a ) encl v∫ 2π a a μ0 I EVALUATE: Result in part (b) evaluated at r = a: B = Result in part (d) evaluated at 2π a μ I a μ I r = a: B = 0 (2 − a /a ) = 0 The two results, one for r > a and the other for r < a, agree at 2π a 2π a r = a IDENTIFY: Apply Ampere’s law to a circle of radius r G G SET UP: The current within a radius r is I = ∫ J ⋅ dA , where the integration is over a disk of radius r EXECUTE: 28.82 G G a ⎛b ⎞ EXECUTE: (a) I = ∫ J ⋅ dA = ∫ ⎜ e(r − a )/δ ⎟ rdrdθ = 2π b ∫ e( r − a )/δ dr = 2π bδ e(r − a )/δ ⎝r ⎠ a = 2π bδ (1 − e − a /δ ) I = 2π (600 A/m)(0.025 m)(1 − e(0.050/0.025) ) = 81.5 A G G μ I (b) For r ≥ a, v∫ B ⋅ dl = B 2π r = μ0 I encl = μ0 I and B = 0 2π r G G r (r − a )/δ r ⎛ b (r ′− a )/δ ⎞ (c) For r ≤ a, I ( r ) = ∫ J ⋅ dA = ∫ ⎜ e dr = 2π bδ e(r ′− a )/δ ⎟ r ′dr′dθ = 2π b ∫0 e ⎝ r′ ⎠ I ( r ) = 2π bδ (e(r − a )/δ − e − a/δ ) = 2π bδ e− a/δ (er/δ − 1) and I ( r ) = I (d) For r ≤ a, (er/δ − 1) (e a/δ − 1) G G (er/δ − 1) μ0 I (er/δ − 1) and B l π μ μ ⋅ d = B ( r )2 r = I = I B = encl 0 a/δ v∫ (e − 1) 2π r (e a/δ − 1) (e) At r = δ = 0.025 m, B = μ0 I (e − 1) 2πδ (e a / δ − 1) = μ0 (81.5 A) (e − 1) = 1.75 × 10−4 T 2π (0.025 m) (e0.050/0.025 − 1) μ0 I (ea/δ − 1) μ0 (81.5 A) = = 3.26 × 10−4 T 2π a (ea/δ − 1) 2π (0.050 m) μ I μ (81.5 A) At r = 2a = 0.100 m, B = 0 = = 1.63 × 10−4 T 2π r 2π (0.100 m) At r = a = 0.050 m, B = 28.83 EVALUATE: At points outside the cylinder, the magnetic field is the same as that due to a long wire running along the axis of the cylinder IDENTIFY: Use what we know about the magnetic field of a long, straight conductor to deduce the symmetry of the magnetic field Then apply Ampere’s law to calculate the magnetic field at a distance a above and below the current sheet SET UP: Do parts (a) and (b) together © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28-31 Consider the individual currents in pairs, where the currents in each pair are equidistant on either G side of the point where B is being calculated Figure 28.83a shows that for each pair the z-components cancel, and that above the sheet the field is in the – x-direction and that below the sheet it is in the + x -direction Figure 28.83a G G Also, by symmetry the magnitude of B a distance a above the sheet must equal the magnitude of B a G distance a below the sheet Now that we have deduced the symmetry of B , apply Ampere’s law Use a path that is a rectangle, as shown in Figure 28.83b G G v∫ B ⋅ dl = μ0 Iencl Figure 28.83b I is directed out of the page, so for I to be positive the integral around the path is taken in the counterclockwise direction G EXECUTE: Since B is parallel to the sheet, on the sides of the rectangle that have length 2a, G G G v∫ B ⋅ dl = On the long sides of length L, B is parallel to the side, in the direction we are integrating G G around the path, and has the same magnitude, B, on each side Thus v∫ B ⋅ dl = BL n conductors per unit length and current I out of the page in each conductor gives I encl = InL Ampere’s law then gives BL = μ0 InL and B = 12 μ0 In 28.84 EVALUATE: Note that B is independent of the distance a from the sheet Compare this result to the electric field due to an infinite sheet of charge (Example 22.7) IDENTIFY: Find the vector sum of the fields due to each sheet G SET UP: Problem 28.83 shows that for an infinite sheet B = 12 μ0 In If I is out of the page, B is to the left G above the sheet and to the right below the sheet If I is into the page, B is to the right above the sheet and to the left below the sheet B is independent of the distance from the sheet The directions of the two fields at points P, R and S are shown in Figure 28.84 EXECUTE: (a) Above the two sheets, the fields cancel (since there is no dependence upon the distance from the sheets) (b) In between the sheets the two fields add up to yield B = μ0nI , to the right (c) Below the two sheets, their fields again cancel (since there is no dependence upon the distance from the sheets) EVALUATE: The two sheets with currents in opposite directions produce a uniform field between the sheets and zero field outside the two sheets This is analogous to the electric field produced by large parallel sheets of charge of opposite sign © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 28-32 Chapter 28 Figure 28.84 28.85 IDENTIFY and SET UP: Use Eq (28.28) to calculate the total magnetic moment of a volume V of the iron Use the density and atomic mass of iron to find the number of atoms in this volume and use that to find the magnetic dipole moment per atom μatom μ total , so μ total = MV The average magnetic moment per atom is V = μ total /N = MV/N , where N is the number of atoms in volume V The mass of volume V is m = ρV , EXECUTE: M = where ρ is the density ( ρiron = 7.8 × 103 kg/m3 ) The number of moles of iron in volume V is n= m 55.847 × 10−3 kg/mol = ρV 55.847 × 10−3 kg/mol , where 55.847 × 10−3 kg/mol is the atomic mass of iron from Appendix D N = nN A , where N A = 6.022 × 1023 atoms/mol is Avogadro’s number Thus N = nN A = μatom = μatom = ρVN A 55.847 × 10−3 kg/mol ⎛ 55.847 × 10−3 kg/mol ⎞ M (55.847 × 10−3 kg/mol) MV = MV ⎜ ⎟⎟ = ⎜ ρVN A ρ NA N ⎝ ⎠ (6.50 × 104 A/m)(55.847 × 10−3 kg/mol) (7.8 × 103 kg/m3 )(6.022 × 1023 atoms/mol) μatom = 7.73 × 10−25 A ⋅ m = 7.73 × 10−25 J/T 28.86 μ B = 9.274 × 10−24 A ⋅ m , so μatom = 0.0834μ B EVALUATE: The magnetic moment per atom is much less than one Bohr magneton The magnetic moments of each electron in the iron must be in different directions and mostly cancel each other IDENTIFY: Approximate the moving belt as an infinite current sheet SET UP: Problem 28.83 shows that B = 12 μ0 In for an infinite current sheet Let L be the width of the sheet, so n = 1/L EXECUTE: The amount of charge on a length Δx of the belt is ΔQ = LΔxσ , so I = Approximating the belt as an infinite sheet B = μ0 I = μ0vσ ΔQ Δx = L σ = Lvσ Δt Δt G B is directed out of the page, as shown in 2L Figure 28.86 EVALUATE: The field is uniform above the sheet, for points close enough to the sheet for it to be considered infinite Figure 28.86 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Sources of Magnetic Field 28.87 28-33 IDENTIFY: The current-carrying wires repel each other magnetically, causing them to accelerate horizontally Since gravity is vertical, it plays no initial role F μ0 I SET UP: The magnetic force per unit length is = , and the acceleration obeys the equation L 2π d F/L = m/L a The rms current over a short discharge time is I / EXECUTE: (a) First get the force per unit length: 2 F μ0 I μ ⎛ I ⎞ μ ⎛V ⎞ μ ⎛Q ⎞ = = ⎜ ⎟ = ⎜ ⎟ = ⎜ 0⎟ L 2π d 2π d ⎝ ⎠ 4π d ⎝ R ⎠ 4π d ⎝ RC ⎠ 2 Now apply Newton’s second law using the result above: a= F m μ ⎛Q ⎞ = a = λ a = ⎜ ⎟ Solving for a gives L L 4π d ⎝ RC ⎠ μ0Q02 μ0Q02 From the kinematics equation v = v + a t , we have v = at = aRC = x 0x x 4πλ RCd 4πλ R 2C 2d 28.88 ⎛ μ0Q02 ⎞ ⎜ ⎟ 2 ⎛ μ0Q02 ⎞ v0 ⎜⎝ 4πλ RCd ⎟⎠ = = (b) Conservation of energy gives mv0 = mgh and h = ⎜⎜ ⎟⎟ 2g 2g g ⎝ 4πλ RCd ⎠ EVALUATE: Once the wires have swung apart, we would have to consider gravity in applying Newton’s second law IDENTIFY: There are two parts to the magnetic field: that from the half loop and that from the straight wire segment running from − a to a SET UP: Apply Eq (28.14) Let the φ be the angle that locates dl around the ring EXECUTE: Bx ( ring ) = 12 Bloop = − μ0 Ia 4( x + a )3/2 μ0 I dl x μ Iax sin φ dφ and sin φ = 4π ( x + a ) ( x + a )1/2 4π ( x + a )3/ π π π μ0 Iax sinφ dφ μ0 Iax μ0 Iax = B y (ring ) = ∫ dB y ( ring ) = ∫ cosφ = − 0 4π ( x + a )3/2 4π ( x + a )3/2 2π ( x + a )3/ μ0 Ia dB y (ring ) = dB sin θ sin φ = B y (rod ) = Bx = − 2π x ( x + a )1/2 μ0 Ia 4( x + a )3/2 EVALUATE: B y = − , using Eq (28.8) The total field components are: and B y = 2a π x ⎛ x2 ⎞ μ0 Ia3 − = ⎜ ⎟ 2π x( x + a )1/2 ⎜⎝ x + a ⎟⎠ 2π x( x + a )3/2 μ0 Ia Bx By decreases faster than Bx as x increases For very small x, Bx = − μ0 I 4a μ0 I and B y = In this limit Bx is the field at the center of curvature of a semicircle and By is the field of 2π x a long straight wire © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... (i) B0 = μ0nI = μ0 (6000 m −1 )(0.15 A) = 1 .13 × 10−3 T (ii) M = Km − μ0 B0 = 5199 μ0 (1 .13 × 10−3 T) = 4.68 × 106 A/m (iii) B = K m B0 = (5200)(1 .13 × 10−3 T) = 5.88 T G G G G (b) The directions... -direction), so B = B1 + B2 Figure 28.22b B1 = μ0 I μ0 I , B2 = 2π a 2π (3a ) B= μ0 I 2μ I G 2μ I + 13 = ; B = kˆ 2π a 3π a 3π a ( ) (c) SET UP: The two fields at this point have the directions shown... -direction), so B = B1 + B2 Figure 28.22c μ0 I μ I , B2 = 2π (3a) 2π a μ0 I μ I G 2μ I B1 = + 13 = ; B = − kˆ 2π a 3π a 3π a B1 = ( ) G EVALUATE: In the figures we have drawn, B due to each