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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 = (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 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 ( µ 0 I 2 /2p d ) L = µ 0 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 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 = ( µ 0 /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 .
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