GREPhysics.NET: GR9677 #88

GR9677 #88

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Electromagnetism $\Rightarrow$ }Ampere Law

Recall Ampere's Law $\vec{B} \cdot d\vec{l} = \mu_0 I_{in}$ , where $I_{in}$ is the current enclosed by the loop dl.

Apply it to the region between a and b, $B(2\pi r) = \mu_0 I \frac{\pi r^2 l}{\pi R^2 l} \Rightarrow B = \mu_0 I \frac{r}{2 R^2 \pi}$ , which gives a linearly increasing field, and thus choices (D) and (E) and (A) are eliminated.

Choices (B) and (C) remain.

Apply Ampere's Law to the region outside of the outer sheath. For $r>c$ , one has $I_{in}=0 \Rightarrow B(2\pi r) = 0 \Rightarrow B=0$ . Choice (B) shows the behavior of zero-field outside the sheathed coax cable. Choose that.

Alternate Solutions

evanb
2008-07-02 13:59:18

At r = 0, the field must be zero, because an infinitely-small amperian loop contains no current.

At r=c, the field must be zero, because the inner current and the outer current cancel out.

The only graph that satisfies these two contraints is B.

Reply to this comment

Comments

ssp
2008-09-07 18:13:00

I used the condition they give us s << l... If that is true (c), (d) and (e) could approach infinity, so they are out. The in 2 in (a) seems very unphysical to be honest because we are only talking about the portion of a trajectory... So I picked (B)

Reply to this comment

evanb
2008-07-02 13:59:18

Reply to this comment

sawtooth
2007-10-30 02:24:29

For first stage, apply Ampere's law in a circle with radius 0Reply to this comment

newton
2007-09-27 21:58:17

Amperes law, $\int$ B.dl= $\mu_{0}$ $\I_{enc}$

For r=C, $\I_{enc}$ = $\I_{in}$ - $\I_{out}$ =I-I=0

Reply to this comment

welshmj
2007-08-02 20:44:49

I think the part where you are finding the magnetic field between a and b is actually the magnetic field where r < a not the magnetic field where a

newton
2007-09-27 22:13:05

Yes i too think so.The magnetic field linearly increases in the region 0 to a.
In the region between a and b it is linearly decreasing, B= $\mu_0$ I/2 $\pi$ r

vsravani
2008-11-04 19:55:07

Yes, the linearly increasing field is between 0 to r and not between a-b. The error needs to be corrected.
FYI,
Between a and b, B is of the form
$\frac{1}{r}\(r^4-a^4)$

Reply to this comment

yosun
2005-11-16 23:51:57

astro_allison: subscripts $x_2$ and superscripts $x^2$ both work. however, the $n$ is due to a file-conversion error; it happens whenever a subscript or superscript ends with the letter n, as in _{n} or ^{n}. and, it's only a file-conversion error... user comment posts should not produce the error if you enter in your equation right.

please post a typo-alert (in the comments section of the solution) if you find other solutions with mal-formed equations.

Reply to this comment

astro_allison
2005-11-16 23:13:17

see, i can't even get it to work!

Reply to this comment

astro_allison
2005-11-16 23:13:03

I think your subscripts aren't working for I $\_{n}$ this is in several problems. fyi

Reply to this comment

Post A Comment!

You are replying to:

For first stage, apply Ampere's law in a circle with radius 0

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

Chat Archives | Delete left banner ad | Donate

(Click to view chat.)

Hate being Anonymous? Login or Register

Upgrade to Poser 7 Now

Huge Textbook Selection, Low Prices – Phat Campus.

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / 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 © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone