Answer:
[tex]1.04 * 10^{-7} T[/tex]
Explanation:
By using Ampere's law formula we have:
[tex]B = \frac{\mu_0 I}{2\pi r}[/tex]
where [tex]\mu_0 = 4\pi10^{-7} Tm/A[/tex] is the permeability constant. r = 10 m is the distance from the wire, and I is the current magnitude. As there are 2 wires carrying currents northward and 1 carrying current southward:
[tex]B = \frac{\mu_0 I_{N1}}{2\pi r} + \frac{\mu_0 I_{N2}}{2\pi r} - \frac{\mu_0 I_{S}}{2\pi r}[/tex]
[tex]B = \frac{\mu_0}{2\pi r}(I_{N1} + I_{N2} - I_S)[/tex]
[tex]B = \frac{4 \pi 10^{-7}}{2\pi *10}(18.9 + 8.6 - 22.3) = 1.04 * 10^{-7} T[/tex]
Answer:
The resultant magnetic field is -1.04x10⁻⁷T
Explanation:
Given:
i₁ = current through the first wire = 22.3 A (south)
i₂ = current through the second wire = 18.9 A (north)
i₃ = current through the third wire = 8.6 A (north)
R = distance from the cables = 10 m
The magnetic field through the first wire is equal:
[tex]B_{1} =\frac{\mu _{o}i_{1} }{2\pi R} =\frac{4\pi x10^{-7}*22.3 }{2\pi 10} =4.46x10^{-7} T[/tex]
The magnetic field through the second wire is equal:
[tex]B_{2} =\frac{\mu _{o}i_{2} }{2\pi R} =\frac{4\pi x10^{-7}*18.9 }{2\pi 10} =3.78x10^{-7} T[/tex]
The magnetic field through the third wire is equal:
[tex]B_{3} =\frac{\mu _{o}i_{3} }{2\pi R} =\frac{4\pi x10^{-7}*8.6 }{2\pi 10} =1.72x10^{-7} T[/tex]
The resultant magnetic field is equal:
[tex]B=B_{1} -B_{2} -B_{3} =4.46x10^{-7} -3.78x10^{-7} -1.72x10^{-7} =-1.04x10^{-7} T[/tex]
