Four very long, current carrying wires 1, 2, 3 and 4 in the same plane intersect to form a square 40 cm on each side, as shown in figure. Find the magnitude of the current I in wire 2, so that the mag

Question

Four very long, current carrying wires 1, 2, 3 and 4 in the same plane intersect to form a square 40 cm on each side, as shown in figure. Find the magnitude of the current I in wire 2, so that the magnetic field at the centre of the square is zero. Assume that there are no cross-contacts between any pair of wires. Give your answer in ampere.

TestHub · Accepted Answer

At the centre, the four fields must cancel: $B_{1} + B_{2} + B_{3} + B_{4} = 0$

$\frac{μ _{0} I _{1}}{2 πa} \otimes + B_{2} + \frac{μ _{0} I _{3}}{2 πa} ⊙ + \frac{μ _{0} I _{4}}{2 πa} \otimes = 0$

Taking inward as positive with a = 20 cm = 0.2 m: $B_{2} = \frac{μ _{0}}{2 πa} (I_{3} - I_{1} - I_{4}) = \frac{μ _{0}}{2 πa} (20 - 10 - 8) = \frac{μ _{0}}{2 πa} (2 A)$

B2 comes out positive (inward), so the current in wire 2 must be downward.

$\frac{μ _{0} I}{2 πa} = \frac{μ _{0}}{2 πa} (2 A) \Rightarrow I = 2 A$

Answer: 2.00

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