9_Ch 19 College Physics ProblemCH19 Magnetism

9_Ch 19 College Physics ProblemCH19 Magnetism - sin90 BIL k...

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Magnetism 149 19.15 () ( ) () ( ) sin 0.300 T 10.0 A 5.00 m sin 30.0 7.50 N FB I L θ == ° = 19.16 (a) The magnitude is () () ( ) ( ) -4 3 0.60 10 T 15 A 10.0 m sin 90 9.0 10 N FB I L θ == × ° = × B G is perpendicular to . Using the right hand rule, the orientation of is found to be F G F G 15 above the horizontal in the northward direction ° . (b) () ( ) () -4 3 sin 0.60 10 T 15 A 10.0 m sin 165 2.3 10 N I L θ == × °
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Unformatted text preview: ( ) sin90 BIL k mg µ ° = , or ( ) ( ) ( ) ( ) ) ( )( 1.00 g cm 9.80 1.50 A 1.00 g = = 2 2 3 0.20 m s 1 kg 10 cm 0.131 T sin90 10 g 1 m k m L B I = ° 19.18 To have zero tension in the wires, the magnetic force per unit length must be directed upward and equal to the weight per unit length of the conductor. Thus, m mg BI L L = = F G , or ( ) ( ) ( ) 2 0.040 kg m 9.80 m s 0.109 A 3.60 T m L g I B = = = From the right hand rule, the current must be to the right if the force is to be upward when the magnetic field is into the page. ± ±² ±²...
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