Question

A 0.63-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 27 m/s, and the emf induced across its length is 5.5 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

Let the length of the aluminum bar is L = 0.63 m

Let the velocity of the aluminum bar is V = 27 m/s

Emf induced into the aluminum bar is = 5.5 × 10^-4 V

This emf is induced due to change in magnetic flux due to horizontal component of the Earth's magnetic field B_H as the bar is falling in the earth's magnetic field.

Induced emf is given by = B L V

Here B = B_H

So horizontal component of the Earth's magnetic field is

B_H = / Lv = ( 5.5 × 10^-4 ) / ( 0.63 × 27 )

= 0.3233 × 10^-4 T

= 3.233 × 10^-5 T

So the magnetic field B_H = 3.233 × 10^-5 T

B) Stretch the index finger of right hand along the direction in which bar or the conductor is falling ( downwards ) and middle finger towards north ( magnetic field direction ). Automatically your thumb will point towards east which is the direction of magnetic force. We have used the right hand rule here.

So the force is acting along the east direction.

Due to this positive charges move towards east and negative charges towards west. So the east end of the bar becomes positive and west end negative.