# Induced Emf

1. Mar 20, 2005

### waywardtigerlily

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?

2. Mar 20, 2005

### whozum

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. Mar 20, 2005

### waywardtigerlily

That doesn't seem to be working, anymore ideas? Am i even going about this problem the right way?

4. Mar 20, 2005

### whozum

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. Mar 20, 2005

### Gamma

$$\phi = 14.2* B x$$

Absolute value of emf, $$v = d\phi \dt = 14.2 *B dx/dt$$
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:
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