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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Expert Answer

Arham Warner
Answered 2021-04-04 Author has 102 answers

Let us consider An is the area of the nichromewire.
Let us consider Aa is the area of the aluminum wire.
Similarly let us consider ρn resistivity of nichrome wire.
Similarly let us consider ρa resistivity of aluminum wire.
Similarly let us consider En is the electric field intensity of nichrome wire.
Similarly let us consider Ea is the electric fieldintensity of aluminum wire.
we know that current density
J=Eρ
IA=Eρ
I=AEρ
Given In=Ia
therefore AnEnρn=AaEaρa
En=AaAn×ρnρa×Ea
Given that Ea=0.0080 N/C

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Jeffrey Jordon
Answered 2021-10-01 Author has 2495 answers

Diameter of nichrome wire Dni=2.0mm

Diameter of aluminum wire Dal=1.0mm

Resistivity of nichrome ρni=150×108Ωm

Resistivity of aluminum ρal=2.82×108Ωm

Electric field strength of Aluminum wire Eal=0.0080 V/m

Electric field strength of nichrome wire Eni=?

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

obtained from using the following relations

R=ρLA

V= I R

A=πD24

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

EαlρD2

Therefore we have

I=kED2ρ

Where k is constant of proportionality

For nichrome we have

Ini=kEniDni2ρni(1)

and For aluminum we have

Ial=kEalDal2ρal(2)

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

Ini=Ial

therefore

EniDni2ρni=[EalDal2ρal]

Eni=[EalDal2ρniDni2ρal](3)

Now substituting the given data, we get

Eni=[(0.0080)(1.0×103)2(150×108)(2.0×103)2(2.82×108)]

=0.1063 V/m

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