Is my solution correct? (Electromagnetism problem)

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The discussion revolves around solving an electromagnetism problem involving energy dissipation in a loop. The initial calculation yielded an energy dissipation of -60 j/m^3, which was later clarified to be in the correct units of J m^-3. The participants confirmed that the final expression for energy dissipation should include a frequency factor, leading to a result of -3000 J sec^-1 m^-3. However, it was noted that the negative sign should be omitted since energy is dissipated as heat. The final understanding emphasizes the importance of unit conversion and the correct interpretation of energy dissipation in the context of the problem.
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Homework Statement


29lcoyq.jpg

H1=10
H2=-10
B1=1.5
B2=-1.5

Homework Equations


(change)Uloop = (change)H2*B2 + (change)H1*B1

The Attempt at a Solution


=(H1-H2)(B2-B1)
=(10-(-10))(-1.5-1.5)
=-60 j/m^3
 
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You're on the right track but you have computed the energy dissipated in 1 cycle (but with the wrong units). Now convert the energy per cycle to the time rate of energy dissipation.
 
rude man said:
You're on the right track but you have computed the energy dissipated in 1 cycle (but with the wrong units). Now convert the energy per cycle to the time rate of energy dissipation.

So, it sould be something like ((H1-H2)(B2-B1))*f ?

Where did I go wrong with units?
Is it j/m^2 due to area (not volume)?
 
rude man said:
You're on the right track but you have computed the energy dissipated in 1 cycle (but with the wrong units). Now convert the energy per cycle to the time rate of energy dissipation.

loki4000 said:
So, it should be something like ((H1-H2)(B2-B1))*f ?
Yes. A lot like! :-)
Where did I go wrong with units?
Is it j/m^2 due to area (not volume)?
You didn't go wrong. I did. The per-cycle energy dissipation units are J m-3 just as you said. Except the final answer is of course in J sec-1m-3.
 
rude man said:
Yes. A lot like! :-)

You didn't go wrong. I did. The per-cycle energy dissipation units are J m-3 just as you said. Except the final answer is of course in J sec-1m-3.
So, final answer sould be:
=((H1-H2)(B2-B1))*f
=-60 * 50
=-3000 j sec^-1 m^-3
Makes sense. Thanx for help :)
 
loki4000 said:
So, final answer sould be:
=((H1-H2)(B2-B1))*f
=-60 * 50
=-3000 j sec^-1 m^-3
Makes sense. Thanx for help :)
OK, except no minus sign. Energy is dissipated in heat.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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