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Chapter 27 - 27.7

7. The cross-sectional area of wire is given by A = πr 2 ,where r is its radius (half its thickness). The

magnitude of the current density vector is J = i/A = i/πr 2 ,so

r = i

10 4 A / m 2 ) =1 . 9

10 − 4 m .

The diameter of the wire is therefore d =2 r =2(1 . 9

10 − 4 m) = 3 . 8

10 − 4 m.

πJ = 0 . 50A

π (440