Two long thin parallel wires a distance d = 14.5 cm apart carry 25-A currents (I

Question

Two long thin parallel wires a distance d = 14.5 cm apart carry 25-A currents (I) in the same direction. The magnetic field is measured at point P, a distance d1 = 13.7 cm from the lower wire and a distance d2 = 6.7 cm from the upper wire.
a. Find the x and y components of the magnetic field at P due to the current in the lower wire only.

B1,x =

B1,y =

b. Find the x and y components of the magnetic field at P due to the current in the upper wire only.

B2,x =

B2,y =

c. Determine the magnitude |B| of the total magnetic field at P.

|B| =

Please include solution as well as answer. Thank you!

Explanation / Answer

Sorry for not using equation editor, for some reason it will not let me. The magnetic field due to a wire is given as u*I/(2*pi*r), where u is mu not, I is the current through the wire, and r is the distance from the wire to the point. This can be proven via the Biot Savart law or Ampere's law. Using that equation the magnitude of magnetic field due to each wire can be calculated, and the direction can be calculated via the right hand rule (point your thumb in the direction of the current and the way your hand wraps is the way the magnetic field is pointing). I do not have anough time to finish answering, but the exact angle that the magnetic field is pointing in can be found via the law of cosine. Then the x and y components can be found by multiplying the amplitude of the magnetic field by cos or sin depending on the compenents. The total magnitude of the magnetic field can be found by taking the sqrt((x1+x2)^2+(y1+y2)^2))

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