A long horizontal wire carries 23.0 A of current due north. The Earth's field 23.0 cm due west of the wire points north but downward, 46 ∘ below the horizontal, and has magnitude of 5.0×10−5T. Part APart complete What is the magnitude of the net magnetic field 23.0 cm due west of the wire?

Answer:The magnitude of net magnetic field is [tex]3.85\times10^{-5}\ T[/tex]Explanation:Given that,Current = 23.0 ADistance = 23.0 cmAngle = 46°Magnetic field [tex]B= 5.0\times10^{-5}\ T[/tex]We need to calculate the magnetic field due to the wireUsing formula of magnetic field[tex]B=\dfrac{\mu_{0}I}{2\pi r}[/tex]Put the value into the formula[tex]B_{w}=\dfrac{4\pi\times10^{-7}\times23.0}{2\pi\times23.0\times10^{-2}}[/tex][tex]B_{w}=0.00002\ T[/tex][tex]B_{w}=2.0\times10^{-5}\ T[/tex]We need to calculate the horizontal component of the magnetic field Using formula of magnetic field[tex]B_{net}x=B_{e}\cos\theta[/tex]Put the value into the formula[tex]B_{net}x=5.0\times10^{-5}\times\cos45[/tex][tex]B_{net}x=0.0000353\ T[/tex][tex]B_{net}x=3.53\times10^{-5}\ T[/tex]We need to calculate the  component of the magnetic field Using formula of magnetic field[tex]B_{net}y=B_{w}-B\sin\theta[/tex]Put the value into the formula[tex]B_{net}y=2.0\times10^{-5}-(5.0\times10^{-5})\times\sin45[/tex][tex]B_{net}y=-0.0000153\ T[/tex][tex]B_{net}y=-1.53\times10^{-5}\ T[/tex]We need to calculate the magnitude of net magnetic field Using formula of magnetic field[tex]B_{net}=\sqrt{(B_{net}x)^2+(B_{net}y)^2}[/tex]Put the value into the formula[tex]B_{net}=\sqrt{(3.53\times10^{-5})^2+(-1.53\times10^{-5})^2}[/tex][tex]B_{net}=0.0000385\ T[/tex][tex]B_{net}=3.85\times10^{-5}\ T[/tex]Hence, The magnitude of net magnetic field is [tex]3.85\times10^{-5}\ T[/tex]