A 3.2-mm copper wire carries a 15-A current (uniform across its cross section). Determine the magnetic field (a) at the surface of the wire (b) inside the wire, 0.64 mm below the surface, (c) and outside the wire 3.2 mm from the surface.
Ampere's law
integral B.dl = uo*i*l
B*2*pi*r = uo*i*Aenc/Atot
B = uo*i*pi*r^2/(pi*d^2/4)*(2*pi*r)
1. r = d/2
B = uo*i/(2*pi*d/2) = 4*10^-7*15/0.0032 = 0.001875 T
2. r = 0.64*10^-3
B = 2*u0*i*r/pi*d^2 = 2*4*10^-7*15*0.64*10^-3/(0.0032^2)
B = 0.00075 T
3. r = 0.0064
B = 2*4*10^-7*15*0.0064*/(0.0032^2)
B = 0.0075 T
Comment below if you find any mistake or have any doubts.
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