841_PartUniversity Physics Solution

# 841_PartUniversity Physics Solution - Sources of Magnetic...

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Sources of Magnetic Field 28-19 E VALUATE : The magnetic field lines for a long, straight wire are concentric circles with the wire at the center. The magnetic field at each point is tangent to the field line, so B ! is perpendicular to the line from the wire to the point where the field is calculated. 28.60. I DENTIFY : Find the vector sum of the magnetic fields due to each wire. S ET U P : For a long straight wire 0 2 I B r μ π = . 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. E XECUTE : (a) The magnetic field vectors are shown in Figure 28.60a. (b) At a position on the x -axis 0 0 0 net 2 2 2 2 2 2 2 sin , 2 ( ) μ I μ I a μ Ia B π r π x a π x a x a θ = = = + + + in the positive x -direction. (c) The graph of B versus / x a is given in Figure 28.60b. E VALUATE : (d) The magnetic field is a maximum at the origin, x = 0. (e) When 0 2 , . μ Ia x a B π x >> Figure 28.60 28.61. I DENTIFY : Apply sin F IlB φ = , with the magnetic field at point P that is calculated in problem 28.60. S ET U P : The net field of the first two wires at the location of the third wire is 0 2 2 ( ) Ia B x a μ π = + , in the + x -direction. E XECUTE : (a) Wire is carrying current into the page, so it feels a force in the -direction y . ( ) 2 5 0 0 2 2 2 2 (6 00 A) (0 400 m) 1.11 10 N/m. ( ) (0 600 m) (0 400 m) F μ Ia μ . . IB I L π x a π . . = = = = × + + (b) If the wire carries current out of the page then the force felt will be in the opposite direction as in part (a). Thus the force will be 5 1.11 10 N m, × in the + y -direction. E VALUATE : We could also calculate the force exerted by each of the first two wires and find the vector sum of the two forces. 28.62. I DENTIFY : The wires repel each other since they carry currents in opposite directions, so the wires will move away from each other until the magnetic force is just balanced by the force due to the spring. S ET U P : The force of the spring is kx and the magnetic force on each wire is F mag = 2 0 2 ° I L x π . E XECUTE : Call x the distance the springs each stretch. On each wire, F spr = F mag , and there are two spring forces on each wire. Therefore 2 kx = 2 0 2 ° I L x π , which gives x = 2 0 2 ° I L k π . E VALUATE : Since 0 /2 μ π is small, x will likely be much less than the length of the wires. 28.63. I DENTIFY : Apply = 0 F ! to one of the wires. The force one wire exerts on the other depends on I so = 0 F ! gives two equations for the two unknowns T and I . S ET U P : The force diagram for one of the wires is given in Figure 28.63. The force one wire exerts on the other is 2 0 , 2 I F L r μ π = where 3 2(0.040 m)sin 8.362 10 m r θ = = × is the distance between the two wires. Figure 28.63

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28-20 Chapter 28 E XECUTE : 0 gives cos and /cos y F T mg T mg θ θ = = = 0 gives sin ( /cos )sin tan x F F T mg mg θ θ θ θ = = = = And , so tan m L F Lg λ λ θ = = 2 0 tan 2 I L Lg r μ λ θ π = 0 tan ( /2 ) gr I λ θ μ π = 2 3 7 (0.0125 kg/m)(9.80 m/s) (tan 6.00 )(8.362 10 m) 23.2 A 2 10 T m/A I ° × = = × E VALUATE : Since the currents are in opposite directions the wires repel. When I is increased, the angle θ from the vertical increases; a large current is required even for the small displacement specified in this problem.
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