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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 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 = 2.34 × 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 13.1 μm and is at a radial distance of 1.50 mm?

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oladayoademosu

Answer:

2.54 μA

Explanation:

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

Since J(r) = Br, in the radial width of 13.1 μm, dr = 13.1 μm. r = 1.50 mm. We have a differential current dI. We remove the first integral 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 evaluate at r = 1.50 mm = 1.50 × 10⁻³ m and dr = 13.1 μm = 0.013 mm = 0.013 × 10⁻³ m

I = (2πBr²)dr = 2π × 2.34 × 10 A/m³ × (1.50 × 10⁻³ m)² × 0.013 × 10⁻³ m  =  2544.69 × 10⁻⁹ A = 2.54 × 10⁻⁶ A = 2.54 μA

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