Respuesta :
The average current density in the wire is given by:
[tex] J=\frac{I}{A} [/tex]
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:
[tex] A=\pi r^2 [/tex]
where r is the radius of the wire. In this problem, [tex] r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m [/tex], so the cross-sectional area is
[tex] A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2 [/tex]
and the average current density is
[tex] J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2 [/tex]
The average current density in the given wire is [tex]\boxed{7.63\times{10^5}{\text{ A/}}{{\text{m}}^{{\text{2}}}}}[/tex].
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 [tex]\vec A[/tex] (also called as flux density) is flowing in space through a particular area vector [tex]\overrightarrow {ds}[/tex] then the flux in the region [tex]s[/tex] is given by,
[tex]\text{flux}=\iint\limits_s{\vec A\cdot\overrightarrow{ds}}[/tex]
Now, in the given question, electric current [tex]I[/tex] 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 [tex]\overrightarrow j[/tex].
Therefore, the flux equation for current in the region [tex]s[/tex] is written as,
[tex]I=\iint\limits_s{\vec j.\vec{ds} }[/tex]
Or
[tex]\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}[/tex]
Or total current density is given by
[tex]\boxed{j=\frac{I}{A}}[/tex]
[tex]\begin{aligned}j&=\frac{I}{{\pi {r^2}}}\\&=\dfrac{I}{{\pi {{\left( {\dfrac{d}{2}} \right)}^2}}}\\\end{aligned}[/tex]
Here, [tex]r[/tex] and [tex]d[/tex] are the radius and the diameter of the wire through which current is flowing,
Now, substituting [tex]2.4\,{\text{A}}[/tex] for [tex]I[/tex], [tex]2.0{\text{ mm = 2}}{\text{.0}} \times {\text{1}}{{\text{0}}^{ - 3}}{\text{ m}}[/tex] for [tex]d[/tex].
We have,
[tex]\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}[/tex]
Thus, the average current density in the given wire is [tex]\boxed{7.63\times{10^5}{\text{ A/}}{{\text{m}}^{{\text{2}}}}}[/tex].
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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.
