Induced EMF and long steel beam

[Boozehound]

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. I am not sure if I am using the right formula. i looked at the other formulas i was given but they don't seem to use the values I am given.

 

[Doc Al]

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.

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.

 

[Boozehound]

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.

 

[Doc Al]

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.

 

[Boozehound]

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 that's the wrong formula obviously.

 

[hage567]

Why do you think it's the wrong one?

 

[Boozehound]

cause i put in the answer and its the wrong answer..

 

[Boozehound]

but then again it could be the right formula but i have the wrong numbers

 

[Boozehound]

i got it..i just didnt take the square root of the velocity. simple overlook.

 

[hage567]

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