A long, cylindrical conductor of radius R carries a current I as shown in the fi

Question

A long, cylindrical conductor of radius R carries a current I as shown in the figure below. The current density J, however, is not uniform over the cross-section of the conductor but is a function of the radius according to J = 5br, where b is a constant. Find an expression for the magnetic field magnitude B at the following distances, measured from the axis. (Use the following variables as necessary: ?0, r1, r2, b, R .) (a) r1 < R B = (b) r2 > R B = http://www.webassign.net/serpse8/30-p-036.gif

Explanation / Answer

Use Ampere's Law to solve this question: Surface integral (Magnetic field*ds) = ?0 * current Current density is J = 2br^5 Plugging in Current density the equation becomes: Magnetic field *distance= ?0 * Integral (J*dA) distance= 2pi*r1 Magnetic field * (2pi*r1) = ?0 * Integral from r1 to 0, (J*2pi*r dr) further more replace J, with 2br^5 Magnetic field * (2pi*r1) = ?0 * Integral from r1 to 0, ((2br^5)*2pi*r dr) solving the integral we get: Magnetic field * (2pi*r1) = (?0 *4*pi*b *r1^7) /7 Magnetic field = (?0 *4*pi*b *r1^7) / (7(2pi*r1^1)) Magnetic field = (?0 *2*b *r1^6) / (7) Part B is solved similarly: Magnetic field *distance= ?0 * Integral (J*dA) distance= 2pi*r2 (*this is now r2) Magnetic field * (2pi*r2) = ?0 * Integral from R to 0, (J*2pi*r dr) plug in J: Magnetic field * (2pi*r2) = ?0 * Integral from R to 0, ((2br^5)*2pi*r dr) solving the integral we get: Magnetic field * (2pi*r2) = (?0 4pi*b R^7) / 7 Magnetic field = (?0 4pi*b R^7) / (7 * (2pi*r2)) Magnetic field = (?0 2*b R^7) / (7 * (r2))

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