Answer:
B = 6.6825 ×[tex]10^{-5}[/tex] T
Explanation:
Given that: I = 20 A, distance between the wires = 16 cm = 0.16 m, magnetic field at a point 10 cm (0.10 m) from each wire can be determined by;
The magnetic flux density (B) in a straight current carrying wire, is given as;
B = [tex]\frac{U_{0}I }{2\pi r}[/tex]
Where: [tex]U_{0}[/tex] is the permeability of free space = 1.26×[tex]10^{-6}[/tex] m kg/[tex]s^{2}[/tex][tex]A^{-2}[/tex], I is the current through the wires, r is the distance between the wires.
B = ( 1.26×[tex]10^{-6}[/tex] × 20) ÷ (2× [tex]\frac{22}{7}[/tex] × (0.16 - 0.10))
= 25.2 × [tex]10^{-6}[/tex] ÷ 0.3771
B = 6.6825 ×[tex]10^{-5}[/tex] T
