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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

This is the question and the answer

2. anonymous

But i dont understand the math on the disappeared natural log part

3. 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.

4. 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.

5. Michele_Laino

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

6. Michele_Laino

the first derivative with respect to time can be written using the chain rule, like this: $\Large \frac{{d\Phi }}{{dt}} = \frac{{d\Phi }}{{dr}} \times \frac{{dr}}{{dt}}$

7. Michele_Laino

furthermore, we have: $\Large \frac{d}{{dr}}\left( {\ln r} \right) = \frac{1}{r}$

8. anonymous

thank you guys... Sorry i can only give 1 medal..

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