GREPhysics.NET: GR9677 #88

GR9677 #88

Problem

GREPhysics.NET Official Solution

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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.

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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)

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evanb
2008-07-02 13:59:18

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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

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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)$

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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.

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astro_allison
2005-11-16 23:13:17

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

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astro_allison
2005-11-16 23:13:03

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

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Amperes law, $\int$ B.dl= $\mu_{0}$ $\I_{enc}$ For r=C, $\I_{enc}$ = $\I_{in}$ - $\I_{out}$ =I-I=0

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