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Course: PHYS 1200, Spring 2008
School: RPI
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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 7. 28.35, 29.30, 29.36, 29.87; Q30.02 SPRING 2006 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, FB iL B . Since L and B are perpendicular to each other, the magnitude of the magnetic force is given by, FB = iLB....

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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 7. 28.35, 29.30, 29.36, 29.87; Q30.02 SPRING 2006 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, FB iL B . Since L and B are perpendicular to each other, the magnitude of the magnetic force is given by, FB = iLB. Then, we must have, mg (0.0130 kg)(9.8 m/s2 ) iLB = mg, so i i = 0.467 A LB (0.620 m)(0.440 T) b) FB iL 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 F12 F13 F14 . The magnitudes of F12 and F14 are both equal to F12 i2 , where a is the 2 a 2 0i square edge, and i is the current in each wire. The magnitude of F13 is F13 , 2 d F14 0 where d equals the length of the diagonal of the square, and d 2 0i F13 . 2 a 2 a 2 . Then, The x and y components of the force on wire 1 are equal in magnitude. Fx = F14 + F13 cos , and Fy = F12 + F13 cos . For either, Fx , y Fx , y i2 2 a 0 0 i2 2 a 2 cos 45 i2 2 a 0 2 2 a 2 2 0 4 i2 i2 2 a 0 i2 4 a 0 3 0i 2 4 a 3(4 In unit notation, vector the force becomes, F 10 7 H/m)(15.0 A) 2 4 (0.085 m) 7.94 10 N/m . (7.94 10 4 N/m)^ (7.94 10 4 N/m)^ i j 29.36. B ds i , where i is the current enclosed by the loop. (4 a) For path 1: B ds 0 (3.0 A 5.0 A) 0 ( 2.0 A) 0 10 7 T m/A)( 2.0 A) B ds 2.5 10 6 T m (4 10 7 T m/A) ( 13.0 A) B ds 1.6 10 5 T m b) For path2: B ds 0 ( 3.0 A 5.0 A 5.0 A) 0 ( 13.0 A) 29.87. F ev B a) If v is directed toward the wire, F will be in the same direction as the current. v and B are perpendicular to each other, so the magnitude of F is i (4 10 -7 T m/A)(50 A) F evB ev 0 (16 10 19 C)(10 10 7 m/s) . . 2 r 2 (0.05 m) F = 3.2 10 -16 N b) If v is in the direction of the current, F will be away from the wire. Again, v and B are perpendicular to each other, so the magnitude of F is the same as in part a) -16 F = 3.2 10 N c) If v is perpendicular to the two directions already chosen, it will be parallel to the magnetic field, so F = 0. F=0 Q30.02. Since loops 1 and 3 are symmetric about the wire, the net flux through each is zero. Above the wire, there is flux coming out of the page and through the loops, but below the wire there is an equal amount of flux going into the page and through the loops in the opposite direction. Only loop 2 has net flux passing through it. a) Since i is constant, there is no change of flux, and no current is induced in any loop. 1 = 2 = 3. b) With a changing current, the flux through 2 changes with time, so a current is induced in 2. However, the flux through 1 and 3 remains zero so no current is induced in them. 2 > 1 = 3.
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