A long wire lies along the x-axis and carries a current in the positive x-direct

Question

A long wire lies along the x-axis and carries a current in the positive x-direction with magnitude I_x = 90 A. A second long wire lies along the y-axis and carries a current in the positive y-direction with magnitude I_y = 45 A. A third wire lies along the z-axis and carries a current in the positive z-direction with magnitude I_z = 50 A. Determine the magnitude of the net magnetic field in the xy-plane at the point r = (0.030i + 0.040j)m. (Answer 2.5 G)

4. A long wire lies along the r-axis and carries a current in the positive r-direction with magnitude 90 A. A second long wire lies along the y-axis and carries a current in the positive y-direction with magnitude 1 45 A. A third wire lies along the z-axis and carries a current in the positive z-direction with magnitude I-50 A. Determine the magnitude of the net magnetic field in the ry-plane at the point r = (0.0301 + 0.040) m. (Answer 2.5 G)

Explanation / Answer

magnetic field at the point z = b due to the current in x direction
= 0i/(2b) in -y direction

magnetic field at the point z = b due to the current in y direction
= 0i/(2b) in +x direction

so the magnetic field at the point z = b
= 0i/(2b) (i - j)

applying this

Bz = u0 ix/(2 pi rx) = (4*3.14e-7)*(90)/((2)*(3.14)*(0.040)) = 0.00045 TBz = -u0 iy/(2 pi ry) = -(4*3.14e-7)*(45)/((2)*(3.14)*(0.030)) = -0.0003 TBxy = u0 iz/(2 pi r) = (4*3.14e-7)*(50)/((2)*(3.14)*(0.050)) = 0.0002 TThe net magnetic field is:B = sqrt((0.00045-0.0003)^2 + (0.0002)^2) =  2.5 G

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