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Fall 2011 Homework 14 Solutions

# Fall 2011 Homework 14 Solutions - Physics 21 Fall 2011...

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Physics 21 Fall, 2011 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. 28-22 Two long, parallel transmission lines, L = 38 . 0 cm apart, carry currents I 1 = 23 . 0-A and I 2 = 78 . 0-A. Find all locations where the net magnetic field of the two wires is zero if these currents are (a) in the same direction or (b) in opposite directions. (a) (b) L x L x I 1 I 2 I 2 I 1 (a) Panel (a) of the diagram shows the two wires end on when the currents are in the same direction. L is the distance between the wires. The magnetic field lines due to each wire separately are shown by the concentric circles (dashed for I 1 , solid for I 2 ). The direction of the field follows from the right hand rule and is shown at selected points by an arrow next to each circle. By looking at the directions of the two fields in various locations, it’s easy to see that for case (a), the vector sum of the fields can only be zero on the line between the wires. The magnitude of the field B a distance R from a long wire with current I is B = μ 0 I/ (2 πR ). The vector field will be zero at a point on the line between the wires a distance x from the left wire and L - x from the right wire, where the magnitudes of the fields are equal. Then

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