A truck is carrying a steel beam of length 14.5 m on a freeway. An accident causes the beam to be dumped off the truck and slide horizontally along the ground at a speed of 21.5 m/s. The velocity of the center of mass of the beam is northward while the length of the beam maintains an east–west orientation. The vertical component of the Earth's magnetic field at this location has a magnitude of 31.0 μT. What is the magnitude of the induced emf between the ends of the beam?

Magnitude of E.M.F = 0.009664 V

Explanation:

We are given that;

Length of steel beam = 14.5m

Velocity of beam = 21.5 m/s

Magnitude of magnetic field = 31.0 μT = 31 x 10^(-6) T

Now, equation of the motional E.M.F induced in the beam is given as;

ε = BLv

Where, B is the magnitude of the magnetic field which is perpendicular to the length of the beam

L is the length of the steel beam

V is the velocity of the steel beam

Substituting the relevant values, we obtain;

ε = 31 x 10^(-6) x 14.5 x 21.5 = 9664.25 x 10^(-6) = 0.009664 V

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