anonymous
  • anonymous
Hi everyone, I having physics exam tomorrow. I will post my question on the comment section below
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
This is the question and the answer
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anonymous
  • anonymous
But i dont understand the math on the disappeared natural log part
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IrishBoy123
  • IrishBoy123
which part? guessing the problem, due to the fact that \(\Phi = \Phi(r)\), Faraday's \(d \Phi / dt\) is chained ruled as \( d \Phi / dr \times d r / dt\) in order to reflect both the rate at which the flux will change due to velocity v but also that fact that \(\Phi\) will change due to a change dr.

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IrishBoy123
  • IrishBoy123
the log disappears because you first split out the log's denom ad numerator to make differentiation easy - then you diff and recombine those fractions in the solution.
Michele_Laino
  • Michele_Laino
I think that your answers are correct! I got these answers: \[\Large \begin{gathered} \Phi = \frac{{{\mu _0}Ia}}{{2\pi }}\ln \left( {\frac{{r + b/2}}{{r - b/2}}} \right) \hfill \\ \hfill \\ current = \frac{{{\mu _0}Iab}}{{2\pi R}}\frac{1}{{{r^2} - \frac{{{b^2}}}{4}}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
the first derivative with respect to time can be written using the chain rule, like this: \[\Large \frac{{d\Phi }}{{dt}} = \frac{{d\Phi }}{{dr}} \times \frac{{dr}}{{dt}}\]
Michele_Laino
  • Michele_Laino
furthermore, we have: \[\Large \frac{d}{{dr}}\left( {\ln r} \right) = \frac{1}{r}\]
anonymous
  • anonymous
thank you guys... Sorry i can only give 1 medal..

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