chapt29 - Chapter 29: Magnetic Fields Due to Currents...

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Chapter 29: Magnetic Fields Due to Currents Problem 3 A surveyor is using a magnetic compass 6.1 m below a power line in which there is a steady current of 100 A. (a) What is the magnetic field at the site due to the power line? (b) Will this interfere seriously with the compass reading? The horizontal component of Earth's magnetic field at the site is 20 μ T. Answer (a) 3.28 μ T. (b) Yes. Solution (a) The magnetic field at a distance r from the center of a long straight wire carrying current i is B = μ o i 2 π r Therefore, 20 ft from the power line the magnetic field due to the power line will be B = 4 π × 10 7 T-m/A *100 A 2 * 6.1 m = 3.28 μ T . (b) This is a significant field compared to the field of Earth. If the two fields were perpendicular to each other then the resultant field would make an angle of arctan(3.28/20) = 9.3 ° with the earth's field - meaning that the compass would be off by 9.3 ° . Problem 7 Two long straight wires are parallel and 8.0 cm apart. They are to carry equal currents such that the magnetic field halfway between them has magnitude 300 μ T. (a) Should the currents be in the same or opposite directions? (b) How much current is needed? Answer (a) opposite; (b) 30 A. (a) Referring to Diagram 29-7, if the currents are in the same direction then the magnetic field contributions at a point half-way between the two wires will be opposite directions and add up to zero since the two contributions will have the same size. (b) If the two currents are in opposite directions then the total magnetic field at the midpoint will be twice the field due to one wire. B = 2 o i 2 r = o i r Therefore the current in each wire must be
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29-2 Chapter 29: Magnetic Fields Due to Currents i = π rB μ o = 4.00 cm ( ) 300 T ( ) 4 × 10 -7 T m A = 30 A Problem 11 In Fig. 29-42, a current I = 10 A is set up in a long hairpin conductor formed by bending a piece of wire into a semicircle of radius R = 5.0 mm. Point b is midway between the straight sections and so distant from the semicircle that each straight section cam be approximated as being an infinite wire. What are the (a) magnitude and (b) direction (into or out of the page) of B at a and the (c) magnitude and (d) direction of B at b? i Figure 29-42 Answer (a) 1.03 mT, (b) out of the page. (c) 0.80 mT, (d) out of the page. Solution (a) The B field at a is due to two "half infinite" long wires and the semi-circle. The field due to each of these pieces is directed out of the figure so the magnitude of the field is just the sum of the magnitudes of the three contributions B = B /2 + B /2 + B semicircle B = μ o i 4 π R + μ o i 4 π R + μ o i 4R = μ o i 4R 2 π + 1 B = 4 π × 10 7 T - m/A ( ) * 10 A ( ) 4 * 5.0 mm ( ) 2 π + 1 B = 1.03 mT. (b) The right hand rule reveals that the magnetic field is directed out of the page. (c) Calculating the B field at point b is just like the calculation at point a except that the contribution due to the semicircular part will be negligible due to its great distance from point b and the field is due to two infinitely long straight wires (not "half-infinite" in this case).
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This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

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chapt29 - Chapter 29: Magnetic Fields Due to Currents...

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