921_Physics ProblemsTechnical Physics

# 921_Physics ProblemsTechnical Physics - Chapter 32 P32.78 P...

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Chapter 32 261 P32.78 P = IV I V == × × 100 10 500 10 9 3 . . W 200 10 V A 3 From Ampere’s law, Br I 2 0 πµ bg = enclosed or B I r = µ π 0 2 enclosed . (a) At ra 00200 . m , I enclosed A 3 . FIG. P32.73 and B = ×⋅ × 4 10 5 00 10 20 0 2 0 0 00500 500 73 TmA A m T m T ej e j . . .. . (b) At rb m , II enclosed A × 3 . and B = × 4 10 5 00 10 0 5 0 0 200 A m m T e j . . . (c) Uu d V rdr I dr r I b a a b a b = = F H G I K J zz z af 2 0 0 2 0 2 2 24 4 A AA ln U = × × F H G I K J = 4 10 5 00 10 1 000 10 4 229 10 229 2 3 6 A m cm 2.00 cm J M J e j e j . ln . (d) The magnetic field created by the inner conductor exerts a force of repulsion on the current in the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small rectangular section of the outer cylinder of length A and width w . It carries a current of
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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