Home > 0877 > GR0877 Problem 92

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!

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Want to get a full PDF with solutions? Read THIS.

The comments appending particular solutions are due to the respective users. Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 2008 by ETS.

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

I see we cannot post equations using latex.
—————–
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.
—————-
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).