20_InstSolManual_PDF_Part18

20_InstSolManual_PDF_Part18 - By symmetry each segment of...

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20.61. Set Up: The radius R of the circle is related to the circumference by The direction of is given by applying the right-hand rule to the current direction. Solve: The rela- tion between the direction of I and the direction of is illustrated in Figure 20.61. Figure 20.61 Reflect: The field at the center of the circular turn is larger than the field at a point a distance R from a long straight wire that carries the same current. 20.62. Set Up: Let wire 1 be the inner wire with diameter 20.0 cm and let wire 2 be the outer wire with diameter 30.0 cm. To produce zero net field, the fields and of the two wires must have equal magnitudes and opposite directions. At the center of a wire loop The direction of is given by the right-hand rule applied to the current direction. Solve: gives and The directions of and of its field are shown in Figure 20.62. Since is directed into the page, must be directed out of the page and is counterclockwise. Figure 20.62 20.63. Set Up: The magnetic field at the center of a circular loop is
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Unformatted text preview: By symmetry each segment of the loop that has length contributes equally to the field, so the field at the center of a semicircle is that of a full loop. Since the straight sections produce no field at P, the field at P is Solve: The direction of is given by the right-hand rule: is directed into the page. Reflect: For a quarter-circle section of wire the magnetic field at its center of curvature is B 5 m I 8 R B S B S B 5 m I 4 R . B 5 m I 4 R 1 2 D l B 5 m I 2 R . I 1 I 2 B 1 B 2 1 2 I 2 B S 2 B S 1 I 1 I 2 5 1 R 2 R 1 2 I 1 5 1 15.0 cm 10.0 cm 2 1 12.0 A 2 5 18.0 A. m I 1 2 R 1 5 m I 2 2 R 2 B 1 5 B 2 B 2 5 m I 2 R 2 . B 1 5 m I 2 R 1 , B S B 5 m I 2 R . B S 2 B S 1 B I I B S B 5 1 4 p 3 10 2 7 T # m / A 21 15.0 A 2 2 1 5.73 3 10 2 2 m 2 5 1.64 3 10 2 4 T. R 5 C 2 p 5 0.360 m 2 p 5 5.73 3 10 2 2 m. B S C 5 2 p R . C 5 0.360 m B 5 m I 2 R . 20-18 Chapter 20...
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