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 ! 2.00 $ 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 10.0 mm and is at a radial distance of 1.20 mm?

Answer 1

18.1 × 10⁻⁶ A = 18.1 μA

Step-by-step explanation:

The current I in the wire is I = ∫∫J(r)rdrdθ

Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.

So, dI = J(r)rdrdθ

dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²

Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m

I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA