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Fittoniya [83]
3 years ago
7

If the current density in a wire or radius R is given by J-k+5,0F wire? R, what is the current in the wire?

Physics
1 answer:
Deffense [45]3 years ago
8 0

Answer:

I = R^{2}(K+5)

Explanation:

Given :

J = k+5

Now selecting a thin ring in the wire of radius "r" and thickness dr.

Current through the thin ring is

dI = J X 2πrdr

dI = (K+5) x 2πrdr

Now integrating we get

I = \int_{0}^{R} = (K+5).2\pi rdr

I = (K+5) 2π\int_{0}^{R} rdr

I = (K+5) 2π \frac{R^{2}}{2}

I = R^{2}(K+5)

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