phys7 - Physics 231 Exam III Nov 29 2004 Soc Sec Name 1 One...

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Physics 231 Exam III Nov. 29, 2004 Soc. Sec # Name 1. One night while asleep you fi nd you are having a nightmare. You fi nd that you are in the middle of the French revolution. Unfortunately you are being led up to the guillotine. You notice that the blade runs in an east-west direction, parallel to the ground. You remember that the Earth’s magnetic fi eld runs south to north. For this scenario take the Earth’s magnetic fi eld to be 0.25 Tesla. You notice that there is a current generator nearby that is miraculously hooked to the blade. The guillotine blade has a mass of 500 gms and has a length of 0.75 meter. What must the magnitude and direction (east or west) of the current be through the blade such that your head remains connected to your shoulders? The blade of the guillotine can be treated as a current carrying wire. The force on a current carrying wire in a magnetic fi eld is given by −→ F = I −→ l × −→ B Since the blade and the magetic fi eld are perpendicular to each other, this then reduces to F = I l B . This force will be needed to be directed upwards to counter the gravitational force acting downwards on the blade. So we have I l B = m g I = m g l B = (0 . 5)(9 . 8) (0 . 75)(0 . 25) = 26 . 13 Amps To get the force being upwards, the current would hvae to run West to East. 1
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Soc. Sec # Name 2. A sliding bar, of length 2.5 meters, rests on conducting rails as shown. The conducting rails have a net resistance of 12 ohms. a. Suppose that B = 3.5 Tesla and is directed as shown and that it is constant in time. What is the magnitude and direction of the induced current in the closed loop if the bar moves to the right with a constant velocity of 1 4 m/sec? Since the area is increasing, the fl ux through the loop is also increasing. The induced emf and current are given by ε = d Φ B dt = B dA dt = B d dt ( l x ) = B l v = (3 . 5)(2 . 5)(0 . 25) = 2 . 1875 V olts I = ε R = B l v R = 2 . 1875 12 = 0 . 183 Amps Since the fl ux through the loop is increasing into the paper due to the motion of the rod to the right, the fl ux that will be generated by the current in the loop needs to be out of the paper. To get this fl ux out of the paper, the induced current will need to be counterclockwise .
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  • Spring '08
  • Bingham
  • Physics, Magnetic Field, ε

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