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# hw14 - Physics 21 Fall 2009 Solution to HW-14 Cancelling...

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Physics 21 Fall, 2009 Solution to HW-14 Cancelling Magnetic Field Four very long, current- carrying wires in the same plane intersect to form a square with side lengths of 39.0 cm, as shown in the figure. The cur- rents running through the wires are 8.0 A, 20.0 A, 10.0 A, and I . Find the magnitude and direction of the current I that will make the magnetic field at the center of the square equal to zero. The field point at the center of the square is equidistant from all four wires. Let this distance be d = 1 2 × 0 . 39 m. We just have to keep track of the direction of each field using the right hand rule. Let out of the page be plus, and let I > 0 correspond to up: B out of page = µ 0 2 πd ( 10 + I 8 + 20) Solving, we get I = 2 A, and the minus sign means I is directed downward. Wire and Square Loop A square loop of wire with side length a carries a current I 1 . The center of the loop is located a distance d from an infinite wire carrying a current I 2 . The infinite wire and loop are in the same plane; two sides of the square loop are parallel to the wire and two are perpendicular as shown. (a) What is the magnitude F of the net force on the loop? (b) The magnetic moment μ of a current loop is defined as the vector whose magnitude equals the area of the loop times the magnitude of the current ﬂowing in it ( µ = IA ), and whose direction is perpendicular to the plane in which the current ﬂows.

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hw14 - Physics 21 Fall 2009 Solution to HW-14 Cancelling...

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