Question

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the center of the wire's cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and B = 2.08 × 105 A/m3. This function applies out to the wire's radius of 2.00 mm. How much current is contained within the width of a thin ring concentric with the wire if the ring has a radial width of 12.0 μm and is at a radial distance of 1.35 mm?

Answer 1

Magnitude of the current density in a certain cylindrical wire is given as -

J(r) = B r

For a ring with distance 'r' from the center and width 'r'.

At r << r, we have

A = (2 r) r

We know that, I = J A = (B r) (2 r) r

I = 2 B r²r

where, B = constant term = 2.08 x 10⁵ A/m³

r = radial width of ring = 12 x 10^-6 m

r = radial distance of ring = 1.35 x 10^-3 m

then, we get

I = [2 (3.14) (2.08 x 10⁵ A/m³) (1.35 x 10^-3 m)² (12 x 10^-6 m)]

I = 0.0000285 A

converting A into A :

I = 28.5 A