HW7solutions-10 - Physics 2213 HW #7 Solutions Spring 2010...

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Physics 2213 HW #7 – Solutions Spring 2010 F B 30 o 60 o B down up N S #1 [Magnetic Force on a Cable] (a) The force on the cable due to the Earth’s magnetic field is FI LB = ± ±± . The direction will be due South and 30 o below the horizontal (see diagram on right). The magnitude of this force is F B =ILBsin90 o (90 o is the angle between B ± and L ± .) =(120A)(30m)(0.5x10 -4 T)(1)=0.18N (b) The weight of the cable is given by F g =mg= ± Vg= ± ( ² r 2 L)g=(8.9x10 3 kg/m 3 )( ² )(0.0025m) 2 ³ (30m)(9.8N/kg) =51.4N The magnetic force is much smaller than the cable’s weight, so it can be ignored when designing the cables. 27.19. [Fusion Reactor] In part (a), apply conservation of energy to the motion of the two nuclei. In part (b) apply 2 /. qvB mv R = (a) Let point 1 be when the two nuclei are far apart and let point 2 be when they are at their closest separation. Apply conservation of energy to the motion of the two nuclei. 11 2 2 KUKU +=+ . 12 0, UK == so KU = . Now U 2 =ke 2 /r , while 2 1 1 2 2 Km v = ± (because there are two deuterium nuclei, each of mass m , moving at speed v ) so 22 mv ke r = . 19 6 27 15 (1.602 10 C) 8.4 10 m s (3.34 10 kg)(1.0 10 m) kk ve mr ± ² = ² ²² (3% of the speed of light!) (b) Recall that a charged particle moving perpendicular to a uniform magnetic field will travel in a circle with 2 = Then 27 6 19 (3.34 10 kg)(8.4 10 m/s) 0.14 T (1.602 10 C)(1.25 m) mv B qr ± ± = ² . 27.75. [Swinging Loop] The setup is shown to the right. For the loop to be in equilibrium the net torque on it must be zero. The total torque about the pivot point A is the sum of the torque from gravity and the torque from the magnetic force. Use B ± μ = ² ± to calculate the torque due to the magnetic field and use rF = ² ± for the torque due to the gravity force. Note that, if the angle with the yz -plane were zero, the gravitational torque would be zero, and the magnetic torque would dominate. If the angle were 90 o , the magnetic torque would vanish, and gravity would dominate. Therefore, there must be an equilibrium angle in between, and this angle will increase with B . First, we compute the gravitational torque: sin (0.400 m)sin30.0 mg mgr mg ±² ° The torque is clockwise; mg ± is directed into the paper.
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- 2 - For the loop to be in equilibrium the torque due to B ± must be counterclockwise (opposite to mg ± ± ) and it must be that . Bm g ±± = . B = ± B ± ² μ For this torque to be counterclockwise ( B ± directed out of the paper), B ± must be in the -direction. y + sin sin 60.0 B BI A B == ° () gives sin60.0 0.0400 m sin30.0 g IAB mg = ° ( ) 3 0.15 g/cm 2 8.00 cm 6.00 cm 4.2 g 4.2 10 kg m ± =+ = = ² ( ) 32 0.080 m 0.0600 m 4.80 10 m A ± ² ( ) 0.0400 m sin30.0 sin60.0 mg B IA ° = ° ( ) 9.80 m/s 0.0400 m sin30.0 0.024 T (8.2 A)(4.80 10 m )sin60.0 B ± ± ² ° ² ° 27.28. [Crossed E & B Fields] (a) For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction.
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HW7solutions-10 - Physics 2213 HW #7 Solutions Spring 2010...

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