Question

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?

Answer 1

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)