Physics, 12.08.2020 05:01 60129

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current

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Answer from: smagallanes

Explanation:

From the question we are told that

The radius is $r = 1.4 \ mm = 1.4 *10^{-3} \ m$

The current is $I = 4.5 \ A$

Generally the electric field is mathematically represented as

$E = \frac{J}{\sigma }$

Where $\sigma$ is the conductivity of aluminum with value $\sigma = 3.5 *10^{7} \ s/m$

J is the current density which mathematically represented as

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

Here A is the cross-sectional area which is mathematically represented as

$A = \pi r^2$

$A = 3.142 * (1.4*10^{-3})^2$

$A = 6.158*10^{-6} \ m^2$

$J = \frac{ 4.5 }{6.158*10^{-6}}$

J = 730757 A/m^2

$E = \frac{ 730757}{3.5*10^{7} }$

$E = 0.021 \ N/C$

Answer from: Quest

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Answer from: Quest

Answer: 1.4 seconds which is the third optionexplanation: current is defined as the flow of charge per timetherefore, the equation that relates current, charge and time can be expressed as follows: i = δq / δtwhere: i is the current given = 1.8 ampereδq is the charge given = 2.5 coulombδt is the time required for the flow of the given charges to give the 1.8 ampere.substitute with the givens in the above equation to get the time as follows: 1.8 = 2.5 / δtδt = 2.5 / 1.8 = 1.38888 which is approximately 1.4 secondshope this was !

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What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that ca...

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