Question

A 0.73-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 20 m/s, and the emf induced across its length is 7.9 x 10-4 V. Assuming the horizontal component of the Earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the Earth's magnetic field, and (b) state whether the east end or the west end of the bar is positive.

Answer 1

Answer:

A)  B = 5.4 10⁻⁵ T, B) the positive side of the bar is to the West

Explanation:

A) For this exercise we must use the expression of Faraday's law for a moving body

            fem = $- \frac{d \phi }{dt}$

            fem = $- \frac{d (B l y}{dt}= - B l v$- d (B l y) / dt = - B lv

            B = $- \frac{fem}{l \ v}$

we calculate

             B = - 7.9 10⁻⁴ /(0.73 20)

             B = 5.4 10⁻⁵ T

B) to determine which side of the bar is positive, we must use the right hand rule

the thumb points in the direction of the rod movement to the south, the magnetic field points in the horizontal direction and the rod is in the east-west direction.

Therefore the force points in the direction perpendicular to the velocity and the magnetic field is in the east direction; therefore the positive side of the bar is to the West