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Question

Figure shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points A1, A2, A3.
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Solution

For point A1,
Magnitude of current in wires, I = 10 A
Separation of point A1 from the wire on the left side, d = 2 cm
Separation of point A1 from the wire on the right side, d' = 6 cm



In the figure
Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.
P (marked red) denotes the wire carrying current in a plane going into the paper.
Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.
Also from the figure, we can see that
PA4 = QA4 A4A3P = A4A3Q = 90° A4PA3 = A4QA3 = 45° PA4A3 = QA4A3 = 45° PA4Q = 90°
The magnetic field at A1 due to current in the wires is given by
B=μ0I2πd-μ0I2πd' ...(1)

B= 2×10-7×102×10-2-2×10-7×106×10-2 = 1-13×10-4 = 0.67×10-4 T

Similarly, we get the magnetic field at A2 using eq. (1).
B= 2×10-7×101×10-2+2×10-7×103×10-2 =83×10-4 T =2.67×10-4 T

Now,
Magnetic field at A3:
B = 2×10-7×102×10-2+2×10-7×102×10-2 = 2×10-4 T

Magnetic field at A4:
Separation of point A4 from the wire on the left side, d = 22+22 =22 cm
Separation of point A4 from the wire on the right side, d' =22+22 =22 cm
Thus, the magnetic field at A4 due to current in the wires is given by
B = 2×10-7×1022×10-22+2×10-7×1022×10-22 =1×10-4 T

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