Question

In the rainy season, the Amazon flows fast and runs deep. In one location, the river is 25 m deep and moves at a speed of 4.5 m/stoward the east. The earth's 50 μT magnetic field is parallel to the ground and directed northward.

If the bottom of the river is at 0 V, what is the potential (magnitude and sign) at the surface?

Answer 1

Imagine the water split into a set of vertical 'columns', side by side. Each column is conductor moving east perpendicular to a north magnetic field.

Just use the standard formula for induced emf in a moving conductor:
E = Blv:
. .= 50x10⁻⁶ x 25x 4.5
. .= 5.625x10⁻³ V
This equals the potential difference between top and bottom of the water.

To get the direction use a 'hand rule' such as the right hand dynamo rule:
- forefinger (field) points north
- thumb (motion) points east.

You will find the second finger point upwards. That means an induced conventional current would flow upwards (though there is no complete circuit for this to occur in this situation).

A conventional current is a flow of positive charges. Since a complete circuit is not avilable, an equilibrium is established where free positive charge (e.g. H+ ions) flow upward and build up to a steady concentration at the surface; and free negative ions (e.g. OH-) flow down and build up at the base.

So the top becomes more positive than the bottom.

Since the potential difference between the top and bottom of the water is
5.63x10⁻³ V, and the bottom is taken as 0V, the top is at a potential of +5.63x10⁻³V (+5.63mV).