The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

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The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

r 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?

Physics

2 answers:

Softa [3K] · Answer 1 · 2026-03-22T00:21:40+03:00

J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.

serg [3.5K] · Answer 2 · 2026-03-22T00:19:02+03:00

18.1 x 10^-6 A. A cylindrical wire exhibits a current density described as J(r) = Br. Given B = 2.35 x 10^5 A/m^3, the current across a specified area is calculated by multiplying the current density with that area. For a ring at a radial distance r, with an infinitesimal width of Δr, the area is given as A= 2rΔr, where 2r represents the circumference and Δr is the width. Therefore, we can express I=J(2rΔr) leading to I=2Br²Δr. Substituting the assigned values results in I= 2(2.35 x 10^5)(1.2 x 10^-3)^2(11.5 x 10^-6), yielding a result of 18.1 x 10^-6 A.