A long straight conducting rod carries a current I with a no

A long straight conducting rod carries a current I with a non-uniform current density J = ar^2, and has a radius R. The value of the constant is 28.0 A/mm^4 and the radius of the rod is 4.90 mm. Determine the magnitude of the magnetic field at the following points. r_1 = R/2 r_2 = 2R

Solution

here Applying Ampere\'s Law,

integral of B.dL = u0 Iinside

a) for r1 = R/2

dL = 2 pi r1

For Iinside:

lets suppse a ring of width dr of radius r.

then J = a r^2

dI = J. A = (a r^2) ( 2 pi r dr) = 2 pi a r^3 dr

dI = 2pi a r^3 dr

Integrating from 0 to r

I = 2 pi a r^4 / 4 = 0.5 pi a r^4

here r = R/2 = 4.90/2 = 2.45 mm

Iinside = 0.5 x pi x 28 x 2.45^4 = 504.42pi

B ( 2pi (2.45 x 10^-3 ) ) = 4pi x 10^-7 x 504.42pi

B = 0.13 T

b) for r2 = 2R

Iinside = 0.5 pi a R^4 = 0.5 x pi x 28 x 4.90^4 = 8070.72pi

B (2 x pi x 2 x 4.90 x 10^-3) = 4 x pi x 10^-7 x 8070.72pi   

B = 0.517 T