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anonymous

  • one year ago

Hi everyone, I having physics exam tomorrow. I will post my question on the comment section below

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  1. anonymous
    • one year ago
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    This is the question and the answer

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  2. anonymous
    • one year ago
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    But i dont understand the math on the disappeared natural log part

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  3. IrishBoy123
    • one year ago
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    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.

  4. IrishBoy123
    • one year ago
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    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.

  5. Michele_Laino
    • one year ago
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    I think that your answers are correct! I got these answers: \[\Large \begin{gathered} \Phi = \frac{{{\mu _0}Ia}}{{2\pi }}\ln \left( {\frac{{r + b/2}}{{r - b/2}}} \right) \hfill \\ \hfill \\ current = \frac{{{\mu _0}Iab}}{{2\pi R}}\frac{1}{{{r^2} - \frac{{{b^2}}}{4}}} \hfill \\ \end{gathered} \]

  6. Michele_Laino
    • one year ago
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    the first derivative with respect to time can be written using the chain rule, like this: \[\Large \frac{{d\Phi }}{{dt}} = \frac{{d\Phi }}{{dr}} \times \frac{{dr}}{{dt}}\]

  7. Michele_Laino
    • one year ago
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    furthermore, we have: \[\Large \frac{d}{{dr}}\left( {\ln r} \right) = \frac{1}{r}\]

  8. anonymous
    • one year ago
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    thank you guys... Sorry i can only give 1 medal..

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