Two long fixed parallel wires, A and B , are 10cm apart in air and carry 40 A an
ID: 1762515 • Letter: T
Question
Two long fixed parallel wires, A and B, are 10cm apart in air and carry 40 A and 20 A respectively, in oppositedirections. Determine the resultant field (a) on a line midwaybetween the wires and parallel to them and (b) on a line 8.0 cmfrom wire A and 18 cm from wire B (c) What is theforce per meter on a third long wire, midway between A andB and in their plane, when it carries a current of 5.0 Ain the same direction as the current in A.
1) (a) 8.0 x 10-5 T (b) 1.2 x 10-4 T (c)1.2 x 10-3 N/m, toward B
2) (a) 2.4 x 10-4 T (b) 7.8 x 10-5 T (c)1.2 x 10-3 N/m, toward A
3) (a) 2.4 x 10-6 T (b) 7.8 x 10-7 T (c)0
4) (a) 1.2 x 10-4 T (b) 5.6 x 10-6 T (c)2.1 x 104 N/m, toward A
Explanation / Answer
You have to calculate the B field from each wire and add orsubtract them appropriately. When the point is between the wires,the B fields add. When the point is on the same side of bothwires, the B fields subtract. . (a) Each B field is given by k I/ r where k = 2 x 10-7 . So... B field from A = 2 x 10-7 *40 / 0.05 = 1.60 x 10-4 . from B = 2 x 10-7 * 20 / 0.05 = 0.80 x 10-4 . Total B field = 2.40 x10-4 Tesla . (b) B field from A = 2 x10-7 * 40 / 0.08 = 1.00 x10-4 . from B = 2 x 10-7 * 20 / 0.18 = 0.222 x 10-4 . Total B field = 7.8 x10-5 Tesla . (c) Force per meter = B field *current = 2.40 x 10-4 * 5 = 1.20 x10-3 Newtons towardA. . CORRECT ANSWER IS #2 . Total B field = 2.40 x10-4 Tesla . (b) B field from A = 2 x10-7 * 40 / 0.08 = 1.00 x10-4 . from B = 2 x 10-7 * 20 / 0.18 = 0.222 x 10-4 . Total B field = 7.8 x10-5 Tesla . (c) Force per meter = B field *current = 2.40 x 10-4 * 5 = 1.20 x10-3 Newtons towardA. . CORRECT ANSWER IS #2 . from B = 2 x 10-7 * 20 / 0.18 = 0.222 x 10-4 . Total B field = 7.8 x10-5 Tesla . (c) Force per meter = B field *current = 2.40 x 10-4 * 5 = 1.20 x10-3 Newtons towardA. . . CORRECT ANSWER IS #2Related Questions
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