Ch28Word - Chapter 28 p.1 CHAPTER 28 Sources of Magnetic...

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Chapter 28 p.1 CHAPTER 28 – Sources of Magnetic Field 1. The magnetic field of a long wire depends on the distance from the wire: B = ( µ 0 /4p)2 I / r = ( 1 0 –7 T · m/A)2(65 A)/(0.075 m) = 1.7 × 10 –4 T . When we compare this to the Earth’s field, we get B / B Earth = (1.7 × 10 –4 T)/(5.5 × 10 –5 T) = 3.1 × . 2. We find the current from B = ( 0 /4p)2 I / r ; 5 . 5 × 10 –5 T = (10 –7 T · m/A)2 I /(0.25 m), which gives I = 69 A . 3. The two currents in the same direction will be attracted with a force of F = I 1 ( 0 I 2 /2p d ) L = 0 I 1 I 2 L /2p d = ( 4 p × 10 –7 T · m/A)(35 A)(35 A)(45 m)/2p(0.060 m) = 0.18 N attraction . 4. Because the force is attractive, the second current must be in the same direction as the first. We find the magnitude from F / L = 0 I 1 I 2 /2p d 8 . 8 × 10 –4 N/m = (4p × 10 –7 T · m/A)(22 A) I 2 /2p(0.070 m), which gives I 2 = 14 A upward . 5. The magnetic field will be tangent to the circle with the current at the center. C E B C D I B E B D 6. The magnetic field produced by the wire must be less than 1.0% of the magnetic field of the Earth. We find the current from B = ( 0 /4p)2 I / r < 0.01 B Earth ; ( 1 0 –7 T · m/A)2 I /(1.00 m) < 0.010(5.5 × 10 –5 T), which gives I < 2.8 A .
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Chapter 28 p.2 7. We find the direction of the field for each wire from the tangent to the circle around the wire, as shown. For their magnitudes, we have B 1 = ( µ 0 /4p)2 I / r 1 = ( 1 0 –7 T · m/A)2(25 A)/(0.120 m) = 4.17 × 10 –5 T. B 2 = ( 0 /4p)2 I B / r 2 = ( 1 0 –7 T · m/A)2(25 A)/(0.050 m) = 1.00 × 10 –4 T. We use the property of the triangle to find the angles shown: r 2 2 = r 1 2 + d 2 – 2 r 1 d cos θ 1 ; (5.0 cm) 2 = (12.0 cm) 2 + (15.0 cm) 2 – 2(12.0 cm)(15.0 cm) cos 1 , which gives cos 1 = 0.956, 1 = 17.1°; r 1 2 = r 2 2 + d 2 – 2 r 2 d cos 2 ; (12.0 cm) 2 = (5.0 cm) 2 + (15.0 cm) 2 – 2(5.0 cm)(15.0 cm) cos 2 , which gives cos 2 = 0.707, 2 = 45.0°; From the vector diagram, we have B = B 1 (– cos 1 i + sin 1 j ) + B 2 (cos 2 i + sin 2 j ) = ( 4 . 1 7 × 10 –5 T)(– cos 17.1° i + sin 17.1° j ) + (1.00 × 10 –4 T)(cos 45.0° i + sin 45.0° j ) = (3.1 × 10 –5 T) i + (8.3 × 10 –5 T) j . We find the direction from t a n α = B y / B x = (8.30 × 10 –5 T)/(3.09 × 10 –5 T) = 2.68, = 70.1°. We find the magnitude from B = B x /cos = (3.09 × 10 –5 T)/cos 70.1° = 8.9 × 10 –5 T, 70° above horizontal . 8. The magnetic field from the wire at a point south of a downward current will be to the west, with a magnitude: B wire = ( 0 /4p)2 I / r = ( 1 0 –7 T · m/A)2(40 A)/(0.20 m) = 4.0 × 10 –5 T. Because this is perpendicular to the Earth’s field, we find the direction of the resultant field, and thus of the compass needle, from t a n = B wire / B Earth = (4.0 × 10 –5 T)/(0.45 × 10 –4 T) = 0.889, or = 42° W of N . 9. The magnetic field to the west of a wire with a current to the north will be up, with a magnitude: B wire = ( 0 /4p)2 I / r = ( 1 0 –7 T · m/A)2(22.0 A)/(0.200 m) = 2.20 × 10 –5 T. The net downward field is B down = B Earth sin 40° – B wire = (5.0 × 10 –5 T) sin 40° – 2.20 × 10 –5 T = 1.01 × 10 –5 T. The northern component is B north = B Earth cos 40° = 3.83 × 10 –5 T. We find the magnitude from B = [( B down ) 2 + ( B north ) 2 ] 1/2 = [(1.01 × 10 –5 T) 2 + (3.83 × 10 –5 T) 2 ] 1/2 = 4.0 × 10 –5 T .
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch28Word - Chapter 28 p.1 CHAPTER 28 Sources of Magnetic...

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