The resistivity of gold is 2.44 x 10-8 ohms. m at room temperature. a gold wire that is 0.9 mm in diameter and 14 cm long carries a current of 940 ma. what is the electric field in the wire?
answer

0.090 v/m

0.028 v/m

0.046 v/m

0.0090 v/m

0.036 v/m

The electric field in the gold wire is 0.036V/m.

Hence, Option E) 0.036V/m is the correct answer.

Given the data in the question;

Electric field in the wire;

First we find the cross-sectional area of the gold wire;

Area;

We substitute in our values

Next, we calculate the current density( J ), which is the ratio of Current to Area

Current density( J ) = Current  / Area

We substitute our values into the equation

Now, We know that, using the expression for electric field (E) inside a current conductor:

Where E is electric field, is the resistivity of gold and the current density.

We substitute our values into the equation

The electric field in the gold wire is 0.036V/m.

Hence, Option E) 0.036V/m is the correct answer.

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E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

E=0.036 V/m

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .