Question

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

Answer 1

This is the question and the answer

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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