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For what electric field strength would the current in a 2.0mmdiameter nichrome wire be the same as the current in a 1.0mmdiameter aluminum wire in which the electric field strength is0.0080V/m?

Short Answer

Expert verified

The electric field strength is0.104V/m.

Step by step solution

01

Given Information

Nichrome wire diameter=2.0mm

Aluminum wire diameter=1.0mm

Electric field strength=0.0080V/m

02

Explanation

Express the electric field strength as,

J=σE⇒E=Jσ

Electric current density is,

J=IA

Here, the cross-sectional area is a circle,

J=4IÏ€D2.

Combining the expression,

E=4IπσD2,

So, the product EπσD2will be constant for both wires.

Simplify the π,

E1σ1D12=E2σ2D22

The aluminum is represented by 1and the nichrome is represented by 2.

In the case of the nichrome, the electric field is calculated as follows:

E2=σ1σ2D1D22E1

Substitute the expression,

localid="1649072796279" E2=350s6.7s×0.52mm×0.008V/m

localid="1649072915118" =0.104V/m

03

Final Answer

Hence, the electric field strength will be104mV/m.

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