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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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  1. jon
    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<r<2R 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

      Thanks for the reply, jon!

      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 $

      If you have any questions, please feel free to ask.

  2. jon
    26.08.2011 at 22:05

    Thank you for replying. So you’re saying that I_{enc} is only a fraction of the current found bet. R & 2R?

    • 26.08.2011 at 22:47

      Yes, I_{enc} is the current enclosed by the integration path. To find I_{enc} for R<r<2R 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).
      P.S. For latex support information read http://en.support.wordpress.com/latex/

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