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The figure shows a closed surface that has a line integral of the magnetic field of 1.00x10-6 Tm. If h = 19.0 A, ½ 9.00 A and

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Answer #1

using Ampere circuital law

surface line integral of magnetic foeld over a closed loop = uo ( i2 + i3)

10^-6 = (4* 3.14 * 10^-7)( - 9 + i3)

i3 = 9.8 A, into the page

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Comment in case any doubt, will reply for sure.. Goodluck

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