In the figure below, a long straight wire carries a current i_1 = 320 A and...

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Answer 1

Answer #1

magnetic force on upper part of wire

F1 = uo*i1*i2*L/(2*pi*a) j

magnetic force on lower part of wire

F2 = -uo*i1*i2*L/(2*pi*(a+b)) j

Fnet = F1 + F2

Fnet = (uo*i1*i2*L/(2*pi))*(1/a - 1/(a+b))

Fnet = (4*pi*10^-7*32*17*0.3/(2*pi))*(1/0.013 - 1/(0.013+0.08)) j

Fnet = 0.00216 j <<<<=======answer

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Answer 2

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