Current in Seawater (Purcell 7.12)

[NullSpace0]

Homework Statement

This is Purcell 7.12. Seawater moves at 1 meter/second, there is a vertical B-field of .35 gauss, and the conductivity of water is .04 (1/ohm*cm). What is the current in Amps/meter/meter? If you moved water in a bottle at this speed, would there be a current induced?

Homework Equations

The Attempt at a Solution

I used E=(v/c)B since I know that the induced electric field comes from velocity and B-field which are perpendicular to one another... can you talk of induced electric fields like this?

Then I get E=(35/c) esu/cm^2. J=σE, so J=42.4 esu/sec/cm^2. This converts to 1.4e-4 Amps/m^2. Is that the right number?

I'm not sure how to answer the second part... why would it matter is the water is in a bottle or in a large body of water? All that matters it the conductivity and relative motion, so the current would flow, yes?

 

[haruspex]

why would it matter is the water is in a bottle or in a large body of water? All that matters it the conductivity and relative motion, so the current would flow, yes?

A current requires a circuit. Where's the return path in each case?

 

[NullSpace0]

Presumably the current in the ocean would return at some point on the surface of the water or maybe below the surface of the water?

I guess for a bottle of water there would be no path for the current to return on, though.

 

[haruspex]

Presumably the current in the ocean would return at some point on the surface of the water or maybe below the surface of the water?

Or even through the ocean floor.

 

[paranormal]

I would like to clarify the units and calculations used in this problem. First, the unit of velocity used is meters per second, not centimeters per second. This affects the calculation of the induced electric field, which should be E=(v/c)B = (1/299792458) * 0.35 = 1.17e-9 V/m.

Secondly, the units for conductivity should be in siemens per meter, not ohms per centimeter. This affects the calculation of current, which should be J= σE = (0.04 S/m) * (1.17e-9 V/m) = 4.68e-11 A/m^2.

As for the second part of the question, the size of the container does not affect the induced current as long as the water is moving at the same velocity and the conductivity remains constant. However, the shape and material of the container can affect the strength and direction of the induced current. For example, a conductive container can act as a Faraday cage, shielding the contents from external electric fields.