I_1 = 6.00 A and I_2 = 8.00 A are currents in straight conductors. I_1 runs from

Question

I_1 = 6.00 A and I_2 = 8.00 A are currents in straight conductors. I_1 runs from top to bottom in the plane of the page. I_2 is directed out of the page and exits the page at a point located on the x axis 10.0 cm from I_1. Determine the magnitude and direction of the net magnetic field at the point P that is located 3.00 cm to the left of I_1 on the x axis. Determine the magnitude and direction of the net magnetic field at the point Q that is located 4.00 cm to the right of I_1 on the x axis. Determine the magnitude and direction of the net magnetic field at the point R that is 5.00 cm from I_2 in the y direction.

Explanation / Answer

magnetic field due to current carrying wire - k * I / r

magnetic field at P due to wire 1 = 2 * 10^-7 * 6 / 0.03^2

magnetic field at P due to wire 2 = 2 * 10^-7 * 8 / 0.13^2

net mangitude = sqrt((2 * 10^-7 * 6 / 0.03^2)^2 + (2 * 10^-7 * 8 / 0.13^2)^2)

net magnitude of magnetic field at P = 0.001336 T

direction = tan^-1((2 * 10^-7 * 6 / 0.03^2) / (2 * 10^-7 * 8 / 0.13^2))

direction = 85.94 degree

magnetic field at Q due to wire 1 = 2 * 10^-7 * 6 / 0.04^2

magnetic field at Q due to wire 2 = 2 * 10^-7 * 8 / 0.06^2

net mangitude = sqrt((2 * 10^-7 * 6 / 0.04^2)^2 + (2 * 10^-7 * 8 / 0.06^2)^2)

net magnitude of magnetic field at P = 0.00087 T

direction = tan^-1((2 * 10^-7 * 6 / 0.04^2) / (2 * 10^-7 * 8 / 0.06^2))

direction = 59.35 degree

magnetic field at R due to wire 1 = 2 * 10^-7 * 6 / 0.1^2

magnetic field at R due to wire 2 = 2 * 10^-7 * 8 / 0.05^2

net mangitude = sqrt((2 * 10^-7 * 6 / 0.1^2)^2 + (2 * 10^-7 * 8 / 0.05^2)^2)

net magnitude of magnetic field at P = 0.000651 T

direction = tan^-1((2 * 10^-7 * 6 / 0.1^2) / (2 * 10^-7 * 8 / 0.05^2))

direction = 10.62 degree

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