Question

A 0.74-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 26 m/s, and the emf induced across its length is 5.4 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

Part A. emf induced in a rod moving through magnetic field is given by:

Emf = B*L*V

B = Magnetic field of earth = Emf/(L*V)

Using given values:

B = 5.4*10^-4/(0.74*26)

B = 2.8*10^-5 T = horizontal component of Earth's magnetic field

Part B.

Assuming east is +i, North is +j, and Up is +k

Given that Velocity is downward, So -k

Earth's magnetic field point in north, So +j

Now magnetic force is givne by:

F = q*VxB

And charge q is positive, So

VxB = (-k)x(j) = +i, So

Force on charge is towards east, So east end of the bar is positive

Let me know if you've any query.