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Induced EMF and long steel beam

  1. May 17, 2007 #1
    A 11.8m long steel beam is accidentally dropped by a construction crane from a height of 8.62m. The horizontal component of the Earth's magnetic field over the region is 16.7μT. What is the induced emf in the beam just before impact with the Earth, assuming its long dimension remains in a horizontal plane, oriented perpendicularly to the horizontal component of the Earth's magnetic field?

    E=[(xf*L-xi*L)/(tf-ti)]B

    and that reads Induced EMF equals x final times length minus x initial times length divided by time final minus time initial quantity multiplied by b.

    so i plugged in...

    6.62(11.8)-0(11.8)/1.066

    i get 95.4183 then i take that and multiply it by B which is 1.67E-5 and i get...

    .001593V

    and its wrong. im not sure if im using the right formula. i looked at the other formulas i was given but they dont seem to use the values im given.
     
  2. jcsd
  3. May 17, 2007 #2

    Doc Al

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    Staff: Mentor

    That equation gives you an average EMF during the fall; what you want is the EMF at the moment before impact.

    Hint: Write an equation for induced EMF in terms of instantaneous velocity, not average velocity.
     
  4. May 17, 2007 #3
    thats what i thought i did. to find time i took the height that was given of 8.62m and i took 9.80m/s^2 and divided it by 8.62m and got 1.137s^2 and i took the square root of that to find just seconds and i got 1.06s. then i took 8.62m and divided it by 1.066s to get a velocity of 8.132m/s. but then from there i get lost.
     
  5. May 17, 2007 #4

    Doc Al

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    Staff: Mentor

    Your calculation of time and velocity is incorrect. But even if it were correct, plugging into that EMF equation would give the wrong answer because it uses average velocity. (That equation is meant for things moving at a constant speed--not the case here.)

    Two things to do:
    (1) Come up with the correct equation for motional EMF; hint: It will express EMF in terms of speed (and not distance or time);
    (2) Find the correct speed of the beam using kinematics.
     
  6. May 17, 2007 #5
    ok i did the kinematics and i found the velocity to be 168.952m/s. and i found a formula E=vBL and if i plug in thoes values that i found before E=168.952(1.67E-5)11.8 and E equals .0332. but again i think thats the wrong formula obviously.
     
  7. May 17, 2007 #6

    hage567

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    Homework Helper

    Why do you think it's the wrong one?
     
  8. May 17, 2007 #7
    cause i put in the answer and its the wrong answer..
     
  9. May 17, 2007 #8
    but then again it could be the right formula but i have the wrong numbers
     
  10. May 17, 2007 #9
    i got it..i just didnt take the square root of the velocity. simple overlook.
     
  11. May 17, 2007 #10

    hage567

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    Homework Helper

    OK good, I was just about to point that out. lol
     
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