How Do You Calculate Power Loss in a Mercury Stream with Applied Magnetic Field?

[Hakins90]

Homework Statement

The question is R12 of the book here - http://books.google.co.uk/books?id=...hl=en&sa=X&oi=book_result&resnum=10&ct=result


The attempt at a solution

Voltage induced, $$\epsilon = \frac{d\phi}{dt} = B\frac{dA}{dt} = BaV$$ (Using area swept out per second by a line of charge technique)

Resistance, $$R = \frac{\rho l}{A} = \frac{\rho a}{bl}$$

Power $$P = V^2/R = (BaV)^2 \frac{bl}{\rho a} = \frac{B^2V^2abl}{\rho}$$

Now I'm thinking that the power lost in the resistance of the mercury is equal to the kinetic energy lost by the mercury which flows per second. But i can't get to the final equation it wants.

Can anybody help?

Thank you

 

[Jhero]

for your question. To solve this problem, we first need to understand the physical situation described. The problem is asking about a horizontal pipe filled with mercury, through which a magnetic field is being applied. As the mercury flows through the pipe, it experiences a force due to the interaction between the magnetic field and the charged particles in the mercury. This force causes the mercury to resist the flow and creates a voltage across the pipe.

To solve for the power lost in the resistance of the mercury, we can use Ohm's Law: V = IR. In this case, the voltage (V) is the induced voltage due to the magnetic field, the current (I) is the rate of flow of the mercury (aV), and the resistance (R) is given by R = \frac{\rho l}{A} = \frac{\rho a}{bl}. This gives us the equation V = \frac{\rho a}{bl} (aV) = \frac{\rho a^2V}{bl}.

To find the power lost in the resistance, we can use the equation P = VI = V^2/R. Substituting in our equation for V, we get P = \frac{\rho a^2V^2}{bl} = \frac{B^2V^2abl}{\rho}. This is the final equation that the problem is asking for.

To understand why this equation makes sense, we can think about the physical meaning of each term. The first term, B^2V^2, represents the power lost due to the induced voltage from the magnetic field. The second term, abl, represents the kinetic energy lost by the mercury as it flows through the pipe. The third term, \rho, represents the resistance of the mercury. So our final equation is saying that the power lost in the resistance is equal to the sum of the power lost due to the induced voltage and the kinetic energy lost by the mercury.

I hope this helps to clarify the problem. Let me know if you have any further questions. Good luck with your studies!