91Ó°ÊÓ

a) Current density, J(r) = Br

b) agnetic field,$B=2×{10}^{5}A/m^3$

c) Radius of the wire is$R=2mm$

d) Radial width of the ring,$∆r=10\mathrm{μ}m$

e) Radial distance,$r=1.20mm$$r=1.20mm$

The current density is the current across the unit area at a given point in the conductor. We have to use the relation between current and the current density to find the current contained within the width of a thin ring.

Formulae:

The equation of the current flowing through a small area,$i=∫\stackrel{→}{J}.\stackrel{→}{dA}$ ...(i)

The cross-sectional area of the circle, $A={\mathrm{Ï€°ù}}^2$ ...(ii)

We have, the given value of the current density as:

The differential cross-sectional area value using equation (ii) can be given as follows:

$dA=2\mathrm{Ï€°ùdr}$

Substituting these above values in the equation (i), we can get the contained current within the width of the concentric ring as follows:

$$\begin{gathered} i=∫2{\mathrm{Ï€°ù}}^2\mathrm{Bdr} \\ ∆\mathrm{i}=2{\mathrm{Ï€°ù}}^2\mathrm{B}∆\mathrm{r} \\ =2\mathrm{Ï€}{\left(1.20×{10}^{-3}\mathrm{m}\right)}^2\left(2×{10}^5\mathrm{A}/{\mathrm{m}}^3\right)\left(10×{10}^{-6}\mathrm{m}\right) \\ =1.809×{10}^{-5}\mathrm{A} \\ ≈1.809×{10}^{-5}\mathrm{A} \\ =18.1\mathrm{μ´¡} \end{gathered}$$

Hence, the required value of the current is$18.1\mathrm{μ´¡}$ .