HW8solutions-10

# HW8solutions-10 - Physics 2213 Homework 8 Solutions Spring...

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Physics 2213 Homework 8 Solutions Spring 2010 28.60. We must find the vector sum of the magnetic fields due to each wire. For a long straight wire B 0 I 2 r . The direction of B is given by the right-hand rule and is perpendicular to the line from the wire to the point where then field is calculated. (a) The magnetic field vectors are shown in Figure (a) below. (b) At a position on the x -axis 0 0 0 net 22 2 2 2 2 2 sin , 2 ( ) I I a Ia B r x a x a x a in the positive x -direction. (c) The graph of B versus x / a is given below (Figure (b)). (d) The magnetic field is a max at the origin, x = 0. (e) When 0 2 ,. Ia x a B x (f) Ampere’s law states that C encl I l d B 0 . The magnetic field about a cylindrical conductor carrying a current should be constant on a circle centered on the conductor and should be tangent to the circle. (See the figure.) So Bdl l d B and the line integral around C is C C C r B dl B l d B ) 2 ( . Ampere’s law then gives I I r B encl 0 0 ) 2 ( and then r I B 2 0 . B 1x =B 1 sin B I r C l d

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Physics 213 Homework 8 Solutions Spring 2009 #1. [Wire & Loop] (a) The magnetic field due to the current I 2 in each segment of the loop points into the page everywhere along each of the other three segments. The directions of the forces on each segment due to the other three segments are therefore: Force on segment ab due to I 2 Force on segment bc due to I 2 Force on segment cd due to I 2 Force on segment da due to I 2 These forces try to expand the loop. The force on a wire segment of length L and current I due to magnetic field B is F = IL× B . The magnetic field B 1 due to the straight wire is pointing into the page everywhere along the loop. The magnitude of the field at a distance d below the wire is B 1 (d) = μ o I 1 2 d . The vertical sections of the loop (labeled abcd in the diagram at right) experience equal and opposite forces: F da,1 = - F bc,1 , and these do not contribute to the net force on the loop. Segment ab of the loop experiences a force of magnitude: | F ab,1| = B 1 (y) I 2 D = μ o I 1 I 2 D 2 y in the up direction ( ), and segment cd experiences a force of magnitude: | F cd,1| = B 1 (y+H) I 2 D = μ o I 1 I 2 D 2 (y+H) in the down direction ( ). The net force on the loop has magnitude: F net = | F ab,1| - | F cd,1| = μ o I 1 I 2 D 2 1 y - 1 y+H = μ o I 1 I 2 D H 2 y(y+H) The direction of F net is up . (b) We need to find the distance y at which the magnetic force F net from part (a) exactly balances the weight of the loop W : F net + W = 0 or μ o I 1 I 2 D H 2 y(y+H) = mg.
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HW8solutions-10 - Physics 2213 Homework 8 Solutions Spring...

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