224002511ˆˆˆjj(1.8810N)j.312ilFdddµµππ−⎛⎞=+==×⎜⎟⎝⎠G(c) F3= 0 (because of symmetry). (d)442ˆ(1.8810N)jFF−=−= −×GG, and (e)451ˆ(4.69 10 N)j−=− =−×. 38. We label these wires 1 through 5, left to right, and use Eq. 29-13. Then, (a) The magnetic force on wire 1 is ( )()212425 4
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This note was uploaded on 06/08/2010 for the course PHYS 112 taught by Professor Strong during the Spring '10 term at UAA.