Strength of a magnetic field around a wire with current

  • Thread starter gigli
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Homework Statement


Three parallel wires are arranged in a partial square configuration with point P as the fourth corner. In two of the wires at opposite corners of the square, a 5.00×10^-4 A current flows toward you, and in the third wire opposite the point P, a 7.50×10−4 A current flows away from you. The sides of the square are 5.36cm. What is the strength of the magnetic field at the point P, at the "missing corner" of the square? (T)

attachment.php?attachmentid=15058&stc=1&d=1218936789.jpg


Homework Equations


I believe I should be using:
(Magnetic field)=[(Permeability of free space)*(current)]/[2*pi*(distance from wire)]
The permeability of free space = (4*pi*10^-7)


The Attempt at a Solution


B1+B2-B3=BP
[(5E-4)((4*pi)E-7)]/(2*pi*.0536) = B1 = B2 = 1.9788436E-9 T

distance from wire3 and point P = r = sqrt(.0536^2+.0536^2) = .0758018469 m
[(7.5E-4)((4*pi)E-7)]/(2*pi*r) = B3 = 1.97884361E-9 T

So, B1+B2-B3 = 1.752E-9 T
which is not correct unfortunately. Do the parallel wires do something weird I am unaware of? I thought only the distance from the wire and the current determined the magnitude of the magnetic field, and the opposing currents caused opposing magnetic fields. Help!
 

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Answers and Replies

  • #2
Redbelly98
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It might help to indicate the B-field vectors at P, due to each individual wire. Be sure to draw each vector in the correct direction.
 
  • #3
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I am still having trouble solving this one. Anyone have any advice?
 
  • #4
Redbelly98
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It's vector addition, where you must account for the angles of the vectors (B1, B2, and B3) you are adding up.

Simple addition & subtraction (i.e. "B1+B2-B3") does not work here.
 

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