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Induced Emf

  • #1
A 14.2m long steel beam is accidentally dropped by a construction crane from a height of 9.53m. The horizontal component of the EArth's magnetic field over the region is 16.4 uT. 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?

ok, so I know E= chg flux/chg in time

chg flux = Bl(chg x)

but how do i get a time?
 

Answers and Replies

  • #2
2,209
1
You could try a kinematic approach by solving the position equation for time.

x(t) = x_0+v_0*t+.5gt^2.
x_0 = v_0 = 0
x(t) = .5gt^2 solved for t becomes:
t=sqrt(2*x(t)/g) where x(t) = 9.53 and g = 9.8ms^-2
 
  • #3
That doesn't seem to be working, anymore ideas? Am i even going about this problem the right way?
 
  • #4
2,209
1
In all honesty I didnt pay much attention during E&M. Sorry. I believe Faraday's and Lenz's Laws are what your looking for though.
 
  • #5
356
11
[tex] \phi = 14.2* B x [/tex]

Absolute value of emf, [tex] v = d\phi \dt = 14.2 *B dx/dt[/tex]
Induced emf when the pole hit the ground is obtaind by substituting the velocity of the pole when it hits the ground which = sqrt (2gh) where h=9.53m

Note:
Lately I have not been able to view LaTex Formula. I hope the above shows ok in your browser.
 

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