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SM_chapter30 - 30 Sources of the Magnetic Field CHAPTER...

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30 Sources of the Magnetic Field CHAPTER OUTLINE 30.1 The Biot-Savart Law 30.2 The Magnetic Force Between Two Parallel Conductors 30.3 Ampère’s Law 30.4 The Magnetic Field of a Solenoid 30.5 Gauss’s Law in Magnetism 30.6 Magnetism in Matter 30.7 The Magnetic Field of the Earth ANSWERS TO QUESTIONS *Q30.1 Answers (b) and (c). *Q30.2 (i) Magnetic field lines line in horizontal planes and go around the wire clockwise as seen from above. East of the wire the fi eld points horizontally south. Answer (b). (ii) The same. Answer (b). *Q30.3 (i) Answer (f). (ii) Answer (e). Q30.4 The magnetic fi eld created by wire 1 at the position of wire 2 is into the paper. Hence, the magnetic force on wire 2 is in direction down × into the paper = to the right, away from wire 1. Now wire 2 creates a magnetic fi eld into the page at the location of wire 1, so wire 1 feels force up × into the paper = left, away from wire 2. *Q30.5 Newton’s third law describes the relationship. Answer (c). *Q30.6 (a) No (b) Yes, if all are alike in sign. (c) Yes, if all carry current in the same direction. (d) no Q30.7 Ampère’s law is valid for all closed paths surrounding a conductor, but not always convenient. There are many paths along which the integral is cumbersome to calculate, although not impossible. Consider a circular path around but not coaxial with a long, straight current- carrying wire. Q30.8 The Biot-Savart law considers the contribution of each element of current in a conductor to determine the magnetic field, while for Ampère’s law, one need only know the current passing through a given surface. Given situations of high degrees of symmetry, Ampère’s law is more convenient to use, even though both laws are equally valid in all situations. Q30.9 Apply Ampère’s law to the circular path labeled 1 in the picture. Since there is no current inside this path, the magnetic field inside the tube must be zero. On the other hand, the current through path 2 is the current carried by the conductor. Therefore the magnetic field outside the tube is nonzero. 173 FIG. Q30.9 FIG. Q30.4
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174 Chapter 30 *Q30.10 (i) Answer (b). (ii) Answer (d), according to B NI = µ 0 . (iii) Answer (b). (iv) Answer (c). *Q30.11 Answer (a). The adjacent wires carry currents in the same direction. *Q30.12 Answer (c). The magnetic flux is Φ B BA = cos . θ Therefore the flux is maximum when the field is perpendicular to the area of the loop of wire. The flux is zero when there is no component of magnetic field perpendicular to the loop—that is, when the plane of the loop contains the x axis. *Q30.13 Zero in each case. The fields have no component perpendicular to the area. *Q30.14 (a) Positive charge for attraction. (b) Larger. The contributions away from + and toward – are in the same direction at the midpoint. (c) Downward (d) Smaller. Clockwise around the left-hand wire and clockwise around the right-hand wire are in opposite directions at the midpoint.
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