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 = 2.35 ✕ 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 11.5 μm and is at a radial distance of 1.20 mm?

Answer 1

Answer:

18.1 x 10^-6 A  

Explanation:

A cylindrical wire carries a current density of J(r) = Br, where B = 2.35 x 10^5 A/m^3, to find the current within a certain area we multiply the current density with the are of this area:  

I = J*A  

for a ring with r distance from the center and width Δr, where Δr<<  r, the area is:  

A= 2$\pi$rΔr

where 2$\pi$r is the circumference and Δr is the width, substitute to get:  

I=J(2$\pi$rΔr)

I=2$\pi$Br^2Δr

substitute with the given values to get:

I= 2$\pi$(2.35 x 10^5)(1.2 x 10^-3)^2(11.5 x 10^-6)

 = 18.1 x 10^-6 A  

Answer 2

The current contained within the width of a thin ring concentric is 18.1 x 10⁻⁶A  

What is Current?

This is defined as electric charges moving through an electric conductor or space.

Parameters

Current density of J(r) = Br, where B = 2.35 x 10⁵ A/m³.

I = Jₓ A  

where I is current, A is area and J is current density

A= 2rΔr

where 2r = circumference, Δr = width,

Substitute the values into the equation.

I=J(2rΔr)

I=2Br^2Δr

I= 2(2.35 x 10⁵)(1.2 x 10⁻³)^2(11.5 x 10⁻⁶)

= 18.1 x 10^-6 A  

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