317. Spira in campo magnetico

[Physicsguy51]

No, unfortunately, this is not the correct solution. Most likely, the error is due to your confusion between the diameter of the loop which, therefore, has radius and the diameter of the thin conducting wire which, therefore, has radius .

Ok this is my last attempt for now, I'll retry in a few hours again.
Is the answer:

[Tarapìa Tapioco]

Sorry, but the result is still wrong. I suggest you write down, from time to time, the steps you took to arrive at the different solutions so that I can realize the errors.

[Physicsguy51]

what am I missing? a numerical factor or something completely different? is my method correct?

[Tarapìa Tapioco]

In your first message, you carried out a correct approach with an end result very close to the official one. I recommend that you go from there.

[Physicsguy51]

Ok, so in my first result we got
Now, we need to get rid of the mass and the resistance . So, we can assume the resistance to be: where r is radius of wire cross section used to make loop and D is diameter of loop. Mass density is simply where V is volume. So, volume can be written as: and substituting, we get final answer as:



Now, if this is wrong, I give up. I am probably making some stupid error.

[Tarapìa Tapioco]

Now the answer is correct. You can post the following problem.

[Physicsguy51]

Oh my god.. finally.. now how do I post the next challenge? Is there any tag I should use or just normally post it?

[Tarapìa Tapioco]

When entering text, you must mark the title with the number following the previous problem (in this case, 318). Of course, it would be desirable for you to post a problem for which you already know the solution well.

[Physicsguy51]

Sure.

[Higgs]

Ricollegandomi stamani ho visto la soluzione corretta di Physicsguy51 che però secondo me non è chiarissima rispetto al raggiungimento del valore costante di regime della velocità v. A me tornava lo stesso risultato corretto con il seguente ragionamento. La potenza della forza peso, motore di tutto il processo, è mgv. Avviene, a causa della corrente indotta nella spira, una dissipazione di potenza termica che risulta proporzionale a . Pertanto è ragionevole ritenere che la potenza dissipata raggiunga a un certo momento quella del peso. Da questo momento non c'è più incremento di energia cinetica cioè di v. Il valore è quello riportato nella soluzione di PhYsicsguy51.