Question

A 14.9 m long steel beam is accidentally dropped by a construction crane from a height of 13.4 m. The horizontal component of the Earth’s magnetic field over the region is 15.4 µT. The acceleration of gravity is 9.8 m/s 2 . 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? Answer in units of mV.

Answer 1

I'm not sure if this is correct, but i will give it a shot.

Firstly you need to find the velocity of the bar just before it hits the ground. Use 1 of the kinematic equations
Vf^2 =Vi^2 + 2*a*d (Vi=0)
Vf = sqrt(2*a*d)
= sqrt(2*9.8*13.4)
= 16.206 m/s

Now you use the equation for the Lorentz force, which is the force on the electrons in the beam.
F = q * v * B

where q is the charge
v is the velocity
B is the magnetic field strength

Then, the EMF is defined as the work done per unit charge

EMF = w / q

For this question the work is the work done to move an electron from one end of the beam to the other, which is defined as
w = F * d = (q * v * B) * l where l is the length of the beam

Therefore the EMF becomes
EMF = (q * v * B * l) / q
simplified to
EMF = v * B * l
= 16.206 * (15.4 * 10^-6) * 14.9
= 0.00371 Volts

= 3.71 mV

I hope this helps.

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