# Find an expression for the magnetic field magnitude B at the following distances, measured from the axis.

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 = 2br5, where b is a constant.

Find an expression for the magnetic field magnitude B at the following distances, measured from the axis. (Use any variable or symbol stated above along with the following as necessary: μ0.)

(a) r1 < R
B =

(b) r2 > R
B =
asked Feb 19, 2012 in Physics

Use Ampere's Law to solve this question:

Surface integral (Magnetic field*ds) = μ0 * current

Current density is J = 2br^5

Plugging in Current density the equation becomes:

Magnetic field *distance= μ0 * Integral (J*dA)

distance= 2pi*r1

Magnetic field * (2pi*r1) = μ* Integral from r1 to 0, (J*2pi*r dr)

further more replace J, with 2br^5

Magnetic field * (2pi*r1) = μ* Integral from r1 to 0, ((2br^5)*2pi*r dr)

solving the integral we get:

Magnetic field * (2pi*r1) = (μ*4*pi*b *r1^7) /7

Magnetic field = (μ*4*pi*b *r1^7) / (7(2pi*r1^1))

Magnetic field = (μ*2*b *r1^6) / (7)

Part B is solved similarly:

Magnetic field *distance= μ0 * Integral (J*dA)

distance= 2pi*r2  (*this is now r2)

Magnetic field * (2pi*r2) = μ* Integral from R to 0, (J*2pi*r dr)

plug in J:

Magnetic field * (2pi*r2) = μ* Integral from R to 0, ((2br^5)*2pi*r dr)

solving the integral we get:

Magnetic field * (2pi*r2) = (μ0 4pi*b  R^7) / 7

Magnetic field  = (μ4pi*b  R^7) / (7 * (2pi*r2))

Magnetic field  = (μ0 2*b  R^7) / (7 * (r2))

answered Mar 8, 2012 by ~Expert~ (919 points)
selected Mar 9, 2012