Question

If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?

Answer 1

The average current density in the wire is given by:

$J=\frac{I}{A}$

where I is the current intensity and A is the cross-sectional area of the wire.

The cross-sectional area of the wire is given by:

$A=\pi r^2$

where r is the radius of the wire. In this problem, $r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m$, so the cross-sectional area is

$A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2$

and the average current density is

$J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2$

Answer 2

The average current density in the given wire is $\boxed{7.63\times{10^5}{\text{ A/}}{{\text{m}}^{{\text{2}}}}}$.

Further Explanation:

In mathematical physics, the flux density is the ‘flux defined over a differential area vector’ in the region of space. So, if a vector field $\vec A$ (also called as flux density) is flowing in space through a particular area vector $\overrightarrow {ds}$ then the flux in the region $s$ is given by,

$\text{flux}=\iint\limits_s{\vec A\cdot\overrightarrow{ds}}$

Now, in the given question, electric current $I$ is defined as the flux of charge particles travelling through a unit area and mathematically, the vector field or the flux density assigned with it is named as current density $\overrightarrow j$.

Therefore, the flux equation for current in the region $s$ is written as,

$I=\iint\limits_s{\vec j.\vec{ds} }$

Or

$\begin{aligned}{\text{Total}}\,{\text{current}}\left( I \right)&={\text{sum}}\,{\text{of}}\,{\text{current}}\,{\text{density}}\,{\text{over}}\,{\text{the}}\,{\text{region}}\,{\text{of}}\,{\text{space}}\,{\text{where}}\,{\text{current}}\,{\text{flows}}\\I&=\left( {{\text{total}}\,{\text{current}}\,{\text{density}}} \right)\left(\text{area through which current is flowing}}}) \\I&=jA\\\end{aligned}$

Or total current density is given by

$\boxed{j=\frac{I}{A}}$

$\begin{aligned}j&=\frac{I}{{\pi {r^2}}}\\&=\dfrac{I}{{\pi {{\left( {\dfrac{d}{2}} \right)}^2}}}\\\end{aligned}$

Here, $r$ and $d$ are the radius and the diameter of the wire through which current is flowing,

Now, substituting $2.4\,{\text{A}}$ for $I$, $2.0{\text{ mm = 2}}{\text{.0}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ m}}$ for $d$.

We have,

$\begin{aligned}j&=\frac{{2.4{\text{A}}}}{{\pi {{\left( {\frac{{2 \times {{10}^{ - 3}}{\text{m}}}}{2}} \right)}^2}}}\\&=\frac{{2.4{\text{A}}}}{{\pi {{({{10}^{ - 3}}{\text{m}})}^2}}}\\&=0.763 \times {10^6}{\text{A}}{{\text{m}}^{{\text{ - 2}}}}\\&=7.63 \times {10^5}{\text{A}}{{\text{m}}^{{\text{ - 2}}}}\\\end{aligned}$

Thus, the average current density in the given wire is $\boxed{7.63\times{10^5}{\text{ A/}}{{\text{m}}^{{\text{2}}}}}$.

 

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Answer Details:

Grade: College

Subject: Physics

Chapter: Electrodynamics

Keywords:

Cylindrical wire, diameter, 2.0mm, current, average current density, 2.4A, flowing, flux density, charged particles, total, vector field.