In this problem, $I = 20 \, A$ and $r = a = 20 \, cm = 0.2 \, m$.
The magnetic field at the position of wire 4 due to the current in wire 1 is:
$B_1 = \frac{(4\pi \times 10^{-7})(20)}{2\pi (0.2)} = 2 \times 10^{-5} \, T$
Using the right-hand rule, the direction
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