Question

Item 8 Four very long, current-carrying wires in the same plane intersect to form a square with side lengths of 49.0cm , as shown in the figure (Figure 1). The currents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I. Part A Find the magnitude of the current I that will make the magnetic field at the center of the square equal to zero. Express your answer in amperes. Part B What is the direction of the current I?

Answer 1

Note that the currents are equidistant from the center of the square.

Thus, for the B field to be 0 at the center, the currents must just sum up to 0.

Note that

8.0 A produces a B field into the page at the center.
20.0 A produces a B field out of the page at the center.
10.0 A produces a B field into the page at the center.

Thus,

Inet = 0 = -8.0 A + 20.0 A - 10.0 A + I

Thus,

I = 2 A   [ANSWER, PART A]

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PART B:

As the sign of its B field must be + [out of the page], it must point UPWARD. [ANSWER]

Answer 2

The magnetic field at center is because of all the current carrying wires. Thus B - B +B, +B, +B The magnetic field by a current carrying wire is given by, But B 0, thus μο (20 A) O (10A) (8 A) 4(4 I.-20 A-10 A-8 A I4-2 A The direction of the current is in downward direction