Q9P - Fundamentals Of Physics

Question

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the centre of the wire’s cross section as J(r) = Br, where r is in meters, J is in amperes per square meter, and $\mathit{B}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}\mathbf{ }}\mathit{A}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$ .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 $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{\mu m}$ and is at a radial distance of 1.20 mm ?

Step-by-Step Solution

Answer

Current contained within the width of a thin ring concentric with the wire is $18.1 \mu A$ .

Step 1: The given data

a) Current density, J(r) = Br

b) agnetic field,$B=2\times {10}^{5}A/m^3$

c) Radius of the wire is $R=2 mm$

d) Radial width of the ring, $∆r=10 \mu m$

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

Step 2: Understanding the concept of the flow of current and its density

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,  $i=\int \overrightarrow{J}.\overrightarrow{dA}$                     ...(i)

The cross-sectional area of the circle,  $A={\mathrm{\pi r}}^2$                                                      ...(ii)

Step 3: Calculation of the current contained within the concentric ring

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

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

$dA=2\mathrm{\pi rdr}$

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

$$\begin{gathered} i=\int 2{\mathrm{\pi r}}^2\mathrm{Bdr} \\ ∆\mathrm{i}=2{\mathrm{\pi r}}^2\mathrm{B}∆\mathrm{r} \\ =2\mathrm{\pi }{\left(1.20\times {10}^{-3}\mathrm{m}\right)}^2\left(2\times {10}^5\mathrm{A}/{\mathrm{m}}^3\right)\left(10\times {10}^{-6}\mathrm{m}\right) \\ =1.809\times {10}^{-5}\mathrm{A} \\ \approx 1.809\times {10}^{-5}\mathrm{A} \\ =18.1 \mathrm{\mu A} \end{gathered}$$

Hence, the required value of the current is $18.1 \mathrm{\mu A}$ .