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# Ch28Word - Chapter 28 p.1 CHAPTER 28 – Sources of...

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Unformatted text preview: 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 /4p)2 I / r = (10 –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 = ( μ /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 ( μ I 2 /2p d ) L = μ I 1 I 2 L /2p d = (4p × 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 = μ 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. U s e W o r d 6 .0 c o r l a te r to v i e w M a c i n to s h p i c tu r e . 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 = ( μ /4p)2 I / r < 0.01 B Earth ; (10 –7 T ∙ m/A)2 I /(1.00 m) < 0.010(5.5 × 10 –5 T), which gives I < 2.8 A . 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 /4p)2 I / r 1 = (10 –7 T ∙ m/A)2(25 A)/(0.120 m) = 4.17 × 10 –5 T. B 2 = ( μ /4p)2 I B / r 2 = (10 –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 θ 1 , which gives cos θ 1 = 0.956, θ 1 = 17.1°; r 1 2 = r 2 2 + d 2 – 2 r 2 d cos θ 2 ; (12.0 cm) 2 = (5.0 cm) 2 + (15.0 cm) 2 – 2(5.0 cm)(15.0 cm) cos θ 2 , which gives cos θ 2 = 0.707, θ 2 = 45.0°; From the vector diagram, we have B = B 1 (– cos θ 1 i + sin θ 1 j ) + B 2 (cos θ 2 i + sin θ 2 j ) = (4.17 × 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 ....
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Ch28Word - Chapter 28 p.1 CHAPTER 28 – Sources of...

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