CH-29 - Chapter 29 1(a The magnitude of the magnetic field...

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Unformatted text preview: Chapter 29 1. (a) The magnitude of the magnetic field due to the current in the wire, at a point a distance r from the wire, is given by B i r 2 . With r = 20 ft = 6.10 m, we have B 4 100 2 33 10 33 6 T m A A m T T. c hb g b g . . (b) This is about one-sixth the magnitude of the Earth’s field. It will affect the compass reading. 2. The straight segment of the wire produces no magnetic field at C (see the straight sections discussion in Sample Problem 29-1). Also, the fields from the two semi-circular loops cancel at C (by symmetry). Therefore, B C = 0. 3. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must be directed due south. Since B i r 2 , i rB 2 2 39 10 4 16 6 m T T m A A. b gc h (b) The current must be from west to east to produce a field which is directed southward at points below it. 4. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in segments AH and JD do not contribute to the field at point C . Using Eq. 29-9 (with = ± ) and the right-hand rule, we find that the current in the semicircular arc H J contributes 1 4 i R (into the page) to the field at C . Also, arc D A contributes 2 4 i R (out of the page) to the field there. Thus, the net field at C is 1 2 1 1 (4 T m A)(0.281A) 1 1 1.67 T. 4 4 0.0315m 0.0780m i B R R -7-6 p 10 10 (b) The direction of the field is into the page. 1165 CHAPTER 29 5. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with P do not contribute to the field at that point. Using Eq. 29-9 (with = ) and the right-hand rule, we find that the current in the semicircular arc of radius b contributes 4 i b (out of the page) to the field at P . Also, the current in the large radius arc contributes 4 i a (into the page) to the field there. Thus, the net field at P is 1 1 (4 T m A)(0.411A)(74 /180 ) 1 1 4 4 0.107m 0.135m 1.02 T. i B b a ± װװ -7-7 p 10 10 (b) The direction is out of the page. 6. (a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the current in the straight segments collinear with C do not contribute to the field at that point. Eq. 29-9 (with = ± ) indicates that the current in the semicircular arc contributes 4 i R to the field at C . Thus, the magnitude of the magnetic field is (4 T m A)(0.0348A) 1.18 T. 4 4(0.0926m) i B R -7-7 p 10 10 (b) The right-hand rule shows that this field is into the page....
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This note was uploaded on 04/19/2008 for the course PHYS 1100, 1200 taught by Professor Many during the Fall '08 term at Rensselaer Polytechnic Institute.

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CH-29 - Chapter 29 1(a The magnitude of the magnetic field...

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