16_Chapter 19 HomeworkCH19 Magnetism

16_Chapter 19 HomeworkCH19 Magnetism - 156 CHAPTER 19 19.35...

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Unformatted text preview: 156 CHAPTER 19 19.35 Treat the lightning bolt as a long, straight conductor. Then, the magnetic field is µ I ( 4π × 10 B= 0 = 2π r 19.36 −7 T ⋅ m A ) ( 1.00 × 10 4 A ) 2π ( 100 m ) = 2 .00 × 10 −5 T = 20.0 µ T Model the tornado as a long, straight, vertical conductor and imagine grasping it with the right hand so the fingers point northward on the western side of the tornado. (that is, at the observatory’s location) The thumb is directed downward, meaning that the conventional current is downward or negative charge flows upward . The magnitude of the current is found from B = µ0 I 2π r as I= 19.37 2π rB µ0 = 2π ( 9.00 × 10 3 m ) ( 1.50 × 10 −8 T ) 4π × 10 -7 T ⋅ m A = 675 A From B = µ0 I 2π r , the required distance is µ0 I ( 4π × 10 T ⋅ m A ) ( 20 A ) = = 2 .4 × 10 −3 m = 2 .4 mm −3 2π B 2π ( 1.7 × 10 T ) -7 r= 19.38 Assume that the wire on the right is wire 1 and that on the left is wire 2. Also, choose the positive direction for the magnetic field to be out of the page and negative into the page. (a) At the point half way between the two wires, µ I µI µ Bnet = −B1 − B2 = − 0 1 + 0 2 = − 0 ( I1 + I 2 ) 2π r 2π r1 2π r2 ( 4π × 10 T ⋅ m A ) (10.0 A ) = − 4.00 × 10 2π ( 5.00 × 10 m ) −7 =− -2 −5 T or Bnet = 40.0 µ T into the page Bnet = + B1 − B2 = (b) At point P1 , Bnet = ( 4π × 10 µ0 2π I1 I 2 − r1 r2 T ⋅ m A ) 5.00 A 5.00 A 0.100 m − 0.200 m = 5.00 µ T out of page 2π −7 ...
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This document was uploaded on 10/20/2011.

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