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ssm_ch29 - Chapter 29 7(a If the currents are parallel the...

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Chapter29 7 (a) If the currents are parallel, the two magnetic fields are in opposite directions in the region between the wires. Since the currents are the same, the net field is zero along the line that runs halfway between the wires. There is no possible current for which the field does not vanish. If there is to be a field on the bisecting line the currents must be in opposite directions. Then the fields are in the same direction in the region between the wires. (b) At a point halfway between the wires, the fields have the same magnitude, ¹ 0 i= 2 ¼r . Thus the net field at the midpoint has magnitude B = ¹ 0 i=¼r and i = ¼rB ¹ 0 = ¼ (0 : 040 m)(300 £ 10 ¡ 6 T) 4 ¼ £ 10 ¡ 7 T ¢ m = A = 30 A : 15 Sum the fields of the two straight wires and the circular arc. Look at the derivation of the expression for the field of a long straight wire, leading to Eq. 29–6. Since the wires we are considering are infinite in only one direction, the field of either of them is half the field of an infinite wire. That is, the magnitude is ¹ 0 i= 4 ¼R , where R is the distance from the end of the wire to the center of the arc. It is the radius of the arc. The fields of both wires are out of the page at the center of the arc. Now find an expression for the field of the arc at its center. Divide the arc into infinitesimal segments. Each segment produces a field in the same direction. If ds is the length of a segment, the magnitude of the field it produces at the arc center is ( ¹ 0 i= 4 ¼R 2 ) ds . If μ is the angle subtended by the arc in radians, then is the length of the arc and the net field of the arc is ¹ 0 iμ= 4 ¼R . For the arc of the diagram, the field is into the page. The net field at the center, due to the wires and arc together, is B = ¹ 0 i 4 ¼R + ¹ 0 i 4 ¼R ¡ ¹ 0 4 ¼R = ¹ 0 i 4 ¼R (2 ¡ μ ) : For this to vanish, μ must be exactly 2 radians. 19 Each wire produces a field with magnitude given by B = ¹ 0 i= 2 ¼r , where r is the distance from the corner of the square to the center. According to the Pythagorean theorem, the diagonal of the square has length p 2 a , so r = a= p 2 and B = ¹ 0 i= p 2 ¼a . The fields due to the wires at the upper left and lower right corners both point toward the upper right corner of the square. The Chapter 29 177
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fields due to the wires at the upper right and lower left corners both point toward the upper left corner. The horizontal components cancel and the vertical components sum to B net = 4 ¹ 0 i p 2 ¼a cos 45 ± = 2 ¹ 0 i ¼a = 2(4 ¼ £ 10 ¡ 7 T ¢ m = A)(20 A) ¼ (0 : 20 m) = 8 : 0 £ 10 ¡ 5 T : In the calculation cos 45 ± was replaced with 1 = p 2. In unit vector notation ~ B = (8 : 0 £ 10 ¡ 5 T) ˆ j. 21 Follow the same steps as in the solution of Problem 17 above but change the lower limit of integration to ¡ L , and the upper limit to 0. The magnitude of the net field is B = ¹ 0 iR 4 ¼ Z 0 ¡ L dx ( x 2 + R 2 ) 3 = 2 = ¹ 0 iR 4 ¼ 1 R 2 x ( x 2 + R 2 ) 1 = 2 ¯ ¯ ¯ ¯ 0 ¡ L = ¹ 0 i 4 ¼R L p L 2 + R 2 = 4 ¼ £ 10 ¡ 7 T ¢ m = A)(0 : 693 A) 4 ¼ (0 : 251 m) 0 : 136 m p (0 : 136 m) 2 + (0 : 251 m) 2 = 1 : 32 £ 10 ¡ 7 T : 31 The current per unit width of the strip is i=w and the current through a width dx is ( i=w )
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