Answer:
1.04μT
Explanation:
Due to both wires have opposite currents, the magnitude of the total magnetic field is given by
[tex]B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}[/tex]
I: electric current = 10A
mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2
r1: distance from wire 1 to the point in which B is measured.
r2: distance from wire 2.
The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m
By replacing in the formula you obtain:
[tex]B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T[/tex]
hence, the magnitude of the magnetic field is 1.04μT
