# hw07 - PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS...

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Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 7. 28.35, 29.30, 29.36, 29.38, 29.87; Q30.02 28.35. a) The magnetic force on the wire must produce an upward force equal to the weight of the wire if the tension in the leads is to equal zero. The magnetic force is described by the equation, B L i F B × = . Since L and B are perpendicular to each other, the magnitude of the magnetic force is given by, F B = iLB . Then, we must have, iLB = mg , so T) m)(0.440 620 . ( ) m/s kg)(9.8 0130 . ( 2 = = LB mg i i = 0.467 A b) B L i F B × = . According to the right hand rule for the vector product, the current must be directed from left to right to produce an upward force with the magnetic field given. From left to right 29.30. Since all the currents are parallel, the forces between each pair of wires is attractive. From the symmetry, it should be clear that the magnitude of the net force on each wire is the same. Number the wires, starting with the lower left corner and proceeding clockwise. Then the force per length on wire number 1, as shown in the diagram, is the vector sum of 14 13 12 F F F + + ....
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## This note was uploaded on 09/17/2008 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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hw07 - PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS...

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