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:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

Answer 2

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