Two long, parallel copper wires of diameters 2.3 mm carry currents of 7.0 a in opposite directions. (a) assuming that their central axes are separated by width w = 19 mm, calculate the magnetic flux per meter of wire that exists in the space between those axes. (b) what percentage of this flux lies inside the wires? (c) repeat part (a) for parallel currents.

(a). = 5.43

(b). 13

(C) 0

Explanation:

part A:

φ(outside) = ∬B(outside) dS

note that;

φ(sum) = 2[φ(outside) + φ(inside).

then, we say

φ(outside) = ∫ μI/ 2πr dr (taking boundaries at R - d/2 and d/2.

μI/2πr dr= μI/2πr ln 2R- d/d

= 1.26× 1) ^-6 ×10 ln 40-2.5/2.5.

= 5.43 μWb.

magnetic flux in the conductor can be calculated from magnetic induction as integral of d/2 to R

please note that dS= 1. dr

φ (inside) = B(inside) dS

∫2μIr/ π d^2 dV.

μI/πd2 r^2 (at boundary d/2 and 0)

μI/4π

= 1.26×10^-6 ×10/4π.

= 1.0032 μWb .

where B(inside) = μI(inside)/ 2πr and I= Ir^2/ (d/2) ^2.

φ(sum)= φ(outside + inside)

=2(5.42 + 1.0032

= 13μWb

(c). is zero