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Thread: Current Makes Magnetic Field

  1. #1
    Member pflo's Avatar
    Apr 2009
    Albuquerque, NM

    Current Makes Magnetic Field

    A student brought me this problem and I'd like to help him solve it, but it has been so long since I've looked at these sorts of things. Can anyone help me out some with this?

    A long hollow cylindrical conductor of outer radius "b" and inner radius "a" carries a current whose current density (current/area) is $\displaystyle J=J_o\frac{b}{r}$ where "r" is the distance from the axis of the conductor. Determine an expression for the magnetic field within the conductor (a<r<b) in terms of $\displaystyle \mu_o, J_o, $ a, b, and r.

    In the attached picture, the current would be travelling into the paper throughout the area from the circle with radius a to the circle with radius b.

    I started with this: $\displaystyle J=\frac{I}{A}=J_o\frac{b}{r}$ and so I figure: $\displaystyle I=J_o\frac{b}{r}*A$

    Here is my best shot:
    $\displaystyle \oint{B*dr} = \mu_o*I$
    $\displaystyle \oint{B*dr} = \mu_o*J_o\frac{b}{r}*A$
    $\displaystyle B\oint{dr} = \mu_o*J_o\frac{b}{r}*(\pi b^2-\pi a^2)$
    $\displaystyle B*[2\pi r]_{a}^{b} = \mu_o*J_o\frac{b}{r}*(\pi b^2-\pi a^2)$
    $\displaystyle B*2\pi(b-a) = \mu_o*J_o\frac{b}{r}*(\pi b^2-\pi a^2)$
    $\displaystyle B*2\pi(b-a)=\frac{\mu_o*J_ob\pi (b^2-a^2)}{r}$
    $\displaystyle B=\frac{\mu_o*J_ob(b^2-a^2)}{2r(b-a)}$

    Is this a decent approach?

    I'm wondering if I'm mixed up on my limits of integration. Should they be zero to r instead (this is what my student thinks)? How can I account for the current passing though the entire area (and not just at r)?
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  2. #2
    Senior Member
    Mar 2010
    ( I suppose that the current is going along the axis, not along the circle).
    It is my opinion I may be not right.

    $\displaystyle J(r<a)=0$
    $\displaystyle J(r>b)=0$

    $\displaystyle \oint B dl\:=\:\mu_0I$
    $\displaystyle B\:2\pi r \:=\:\mu_0\int_{a}^{r}J(x)dS\:=\:\mu_0\int_{a}^{r} J_0 \frac{b}{x} \:\:2\pi xdx\:=\:2\pi\mu_0 J_0 b\int_{a}^{r} dx\:=\:2\pi\mu_0 J_0 b(r-a)$

    $\displaystyle B\:=\:\mu_0 J_0 b \frac{(r-a)}{r} \:\:\:a<r<b$

    $\displaystyle B=0 \:\:r<a$

    $\displaystyle B\:=\:\mu_0 J_0 b \frac{(b-a)}{r} \:\:\:r>b$
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