Question

A long cylindrical conductor of radius 54cm carries a current. The current density is not uniform over the cross section of the wire, instead is is given as J = 7r. How much current is enclosed in an Amperian loop of radius 29? Note: If we had a uniform current, then we could simply multiply a current density by an area (pi r2) to get the current. Since we are dealing with a non-uniform current, we must instead take an integral for the form I = integral[J(r) * 2 pi r dr]. A quick double-check on the form of that integral is to replace J(r) with a constant current density to recognize that it reduces to density times area. The integral is needed each time we have a non-constant current.

Answer 1

we know, A = pi*r^2

dA = 2*pi*r*dr

we know, J = dI/dA

==> dI = J*dA

= J*2*pi*r*dr

total current enclosed, integral dI = integral J*2*pi*r*dr

I = integral J*2*pi*r*dr

= integral 7*r*2*pi*r*dr

= integral 14*pi*r^2*dr

= 14*pi integral r^2*dr

= 14*pi*(r^3/3) (from r = 0 m to r = 0.29 m )

= 14*pi*(0.29^3/3 - 0^3/3)

= 0.358 A <<<<<<<----------------Answer