hw10 - Homework 10 Solutions Spring 2010 Required Problems...

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Unformatted text preview: Homework 10 Solutions Spring 2010 Required Problems Chapter 28 7. Since the magnetic field from a current carrying wire circles the wire, the individual field at point P from each wire is perpendicular to the radial line from that wire to point P. We define 1 B r as the field from the top wire, and 2 B r as the field from the bottom wire. We use Eq. 28-1 to calculate the magnitude of each individual field. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 7 4 1 1 7 5 2 2 4 10 T m A 35A 1.17 10 T 2 2 0.060m 4 10 T m A 35A 7.00 10 T 2 2 0.100m I B r I B r π μ π π π μ π π---- = = = = = = g g We use the law of cosines to determine the angle that the radial line from each wire to point P makes with the vertical. Since the field is perpendicular to the radial line, this is the same angle that the magnetic fields make with the horizontal. ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 1 1 2 2 2 1 2 0.060 m 0.130 m 0.100 m cos 47.7 2 0.060 m 0.130 m 0.100 m 0.130 m 0.060 m cos 26.3 2 0.100 m 0.130 m θ θ-- +- = = +- = = Using the magnitudes and angles of each magnetic field we calculate the horizontal and vertical components, add the vectors, and calculate the resultant magnetic field and angle. ( 29 ( 29 ( 29 4 o 5 o 5 net 1 1 2 2 cos cos 1.174 10 T cos47.7 7.00 10 T cos26.3 1.626 10 T x B B B θ θ--- = =6- =- ( 29 ( 29 ( 29 4 o 5 o 4 net 1 1 2 1 sin sin 1.17 10 T sin 47.7 7.00 10 T sin 26.3 1.18 10 T y B B B θ θ--- = + = + T = ( 29 ( 29 2 2 2 2 5 4 4 , , 4 , 1 1 5 , 4 4 1.626 10 T 1.18 10 T 1.19 10 T 1.18 10 T =tan tan 82.2 1.626 10 T 1.19 10 T @ 82.2 1.2 10 T @ 82 net x net y net y net y B B B B B θ--------- = + = + = = = = ♠ B r 18. The magnetic field at the loop due to the long wire is into the page, and can be calculated by Eq. 28-1. The force on the segment of the loop closest to the wire is towards the wire, since the currents are in the same direction. The force on the segment of the loop farthest from the wire is away from the wire, since the currents are in the opposite direction. Because the magnetic field varies with distance, it is more difficult to calculate the total force on the left and right segments of the loop. Using the right hand rule, the force on each small piece of the left segment of wire is to the left, and the force on each small piece of the right segment of wire is to the right. If left and right small pieces are chosen that are equidistant from the long wire, the net force on those two small pieces is zero. Thus the total force on the left and right segments of wire is zero, and so only the parallel segments need to be considered in the calculation. Use Eq. 28-2....
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hw10 - Homework 10 Solutions Spring 2010 Required Problems...

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