Question

A 12.3 m long steel beam is accidentally dropped by a construction crane from a height of 9.71 m. The horizontal component of the Earth's magnetic field over the region is 17.1 μT. 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 1

Firstly you need to find the velocity of the bar just before it hits the ground. Use
V_f² =V_i² + 2*a*d (Vi=0)
V_f = (2*a*d)^0.5
= (2*9.8*9.71)^0.5
= 13.795 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
= 13.795 * (17.1 * 10^-6) * 12.3
= 0.0029 Volts