 # For what electric field strength would the current in a2.0-mm-diameter nichrome wire be the same as the current in a1.0-mm-diameter aluminum wire in which the electric field strengthis 0.0080 N/C? Jaya Legge 2021-04-02 Answered
For what electric field strength would the current in a2.0-mm-diameter nichrome wire be the same as the current in a1.0-mm-diameter aluminum wire in which the electric field strengthis 0.0080 N/C?
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Let us consider ${A}_{n}$ is the area of the nichromewire.
Let us consider ${A}_{a}$ is the area of the aluminum wire.
Similarly let us consider ${\rho }_{n}$ resistivity of nichrome wire.
Similarly let us consider ${\rho }_{a}$ resistivity of aluminum wire.
Similarly let us consider ${E}_{n}$ is the electric field intensity of nichrome wire.
Similarly let us consider ${E}_{a}$ is the electric fieldintensity of aluminum wire.
we know that current density
$J=\frac{E}{\rho }$
$⇒\frac{I}{A}=\frac{E}{\rho }$
$⇒I=\frac{AE}{\rho }$
Given ${I}_{n}={I}_{a}$
therefore $\frac{{A}_{n}{E}_{n}}{{\rho }_{n}}=\frac{{A}_{a}{E}_{a}}{{\rho }_{a}}$
$⇒{E}_{n}=\frac{{A}_{a}}{{A}_{n}}×\frac{{\rho }_{n}}{{\rho }_{a}}×{E}_{a}$
Given that ${E}_{a}=0.0080$ N/C

###### Not exactly what you’re looking for? Jeffrey Jordon

Diameter of nichrome wire ${D}_{ni}=2.0mm$

Diameter of aluminum wire ${D}_{al}=1.0mm$

Resistivity of nichrome ${\rho }_{ni}=150×{10}^{-8}\mathrm{\Omega }\cdot m$

Resistivity of aluminum ${\rho }_{al}=2.82×{10}^{-8}\mathrm{\Omega }\cdot m$

Electric field strength of Aluminum wire

Electric field strength of nichrome wire ${E}_{ni}=?$

The relation between diameter of the wire and electric field strength is

obtained from using the following relations

$R=\rho \frac{L}{A}$

V= I R

$A=\pi \frac{{D}^{2}}{4}$

The relation between electric field strength and diameter of the wire is

$E\alpha l\frac{\rho }{{D}^{2}}$

Therefore we have

$I=kE\frac{{D}^{2}}{\rho }$

Where k is constant of proportionality

For nichrome we have

${I}_{ni}=k{E}_{ni}\frac{{D}_{ni}^{2}}{{\rho }_{ni}}\dots \left(1\right)$

and For aluminum we have

${I}_{al}=k{E}_{al}\frac{{D}_{al}^{2}}{{\rho }_{al}}\dots \left(2\right)$

Here it is given that current passes through both the wires are same, so that we have

${I}_{ni}={I}_{al}$

therefore

${E}_{ni}\frac{{D}_{ni}^{2}}{{\rho }_{ni}}=\left[{E}_{al}\frac{{D}_{al}^{2}}{{\rho }_{al}}\right]$

${E}_{ni}=\left[\frac{{E}_{al}{D}_{al}^{2}{\rho }_{ni}}{{D}_{ni}^{2}{\rho }_{al}}\right]\dots \left(3\right)$

Now substituting the given data, we get

${E}_{ni}=\left[\frac{\left(0.0080\right)\left(1.0×{10}^{-3}{\right)}^{2}\left(150×{10}^{-8}\right)}{\left(2.0×{10}^{-3}{\right)}^{2}\left(2.82×{10}^{-8}\right)}\right]$

=0.1063 V/m