Question

A 2.3 mm -diameter copper wire carries a 38 A current (uniform across its cross section).

a) Determine the magnetic field at the surface of the wire.

b) Determine the magnetic field inside the wire, 0.50 mm below the surface.

c)Determine the magnetic field outside the wire 2.5 mm from the surface.

Answer 1

copper wire carries a 38 A current across its cross section

a) magnetic field at the surface of the wire

B = Uo*i/(2*pi*r)

=> B = 4*pi*10^-7*38/(2*pi*1.15*10^-3) = 6.608*10^-3 T

b)  magnetic field inside the wire, R = 0.50 mm below the surface

Here current enclosed I = (i/pi*r^2)*pi*r1^2 = i*r1^2/r^2 = 38*(0.65/1.15)^2 = 12.139 A

where r1 = 1.15 - 0.50 = 0.65 mm , r = 1.15 mm

=> B = Uo*I/(2*pi*r) = 4*pi*10^-7*12.139/(2*pi*0.65*10^-3) = 3.73*10^-3 T

c) magnetic field outside the wire R = 2.5 mm , r = 1.15+2.5 = 3.65

=> B = Uo*i/(2*pi*r) =  4*pi*10^-7*38/(2*pi*3.65*10^-3) = 2.08*10^-3 T