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## GR0877 Problem 92

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (E) According to Ampere’s law $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}$, where $I_{enc}$ is the total current enclosed by the integration path: $I_{enc} = \int \mathbf{J} \cdot d\mathbf{a}$, $\mathbf{J}$ is the current density. For $r < R$ the enclosed current is zero, so $B = 0$. For $R < r < 2R$ the enclosed current is $\displaystyle I_{enc} = J \int_R^r 2 \pi r dr = \frac{I}{3 \pi R^2} \cdot 2 \pi \cdot \frac{r^2 - R^2}{2} = \frac{I (r^2 - R^2)}{3 R^2}$ and from Ampere’s law $B \cdot 2 \pi r = \mu_o I_{enc}$, so $\displaystyle B(r) = \frac{\mu_0 I (r^2 - R^2)}{6 \pi r R^2} = \frac{\mu_0 I}{6 \pi R^2} \left( r - \frac{R^2}{r} \right)$ which shows a linear character with respect to $r$ on $R < r < 2R$. For $r>2R$ the enclosed current is $I_{enc} = I$ and $\displaystyle B = \frac{\mu_0 I}{2 \pi r}$ or $B \sim \frac{1}{r}$. The only graph that reflects such a behavior of $B$ is (E).

Found a typo? Comment!

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Categories: 0877 Tags: , , , ,
1. 26.08.2011 at 11:58

I see we cannot post equations using latex.
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Trying again.
Thanks for the solutions. I think you have a mistake in problem 92. I found the magnetic field for region $R to be $B= \mu_0 I (r^2-R^2)/(8 \pi R^2 r)$
I got this by setting up since we’re dealing with uniform current density.
The mistake is in your denominator of $3 \pi R^2$, but I think it should be $4 \pi R^2$ bec. we’re comparing the enclosed current to the entire current over the wire.
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The setting up equation was $J = I_{enc}/(\pi (r^2-R^2)) = \frac{I}{4 \pi R^2}$

• 26.08.2011 at 21:25

However, I see no mistake here. The current flows between the radii $R$ and $2R$ and the cross-section is $\pi \left[ (2R)^2 - R^2 \right] = 3 \pi R^2$
P.S. I edited your replies. To insert equations please write “latex [your code]” between two dollar signs \$

Thank you for replying. So you’re saying that $I_{enc}$ is only a fraction of the current found bet. $R$ & $2R$?
Yes, $I_{enc}$ is the current enclosed by the integration path. To find $I_{enc}$ for $R one must find $J = \frac{the~whole~current}{the~area~of~cross-section} = \frac{I}{\pi((2R)^2 - R^2)} = \frac{I}{3 \pi R^2}$ (since its uniform). $I_{enc} = J \int_R^r 2 \pi r dr = \frac{I (r^2 - R^2)} {3 R^2}$. Note also the limiting cases for this formula: for $r =R$ one has $I_{enc} = 0$ and for $r=2R$ the enclosed current equals the total $I_{enc}=I$ (in this case the integration path encloses the whole cylindrical wire).