Question

Electric field creates a 3.9 ma current in a 1.0-mm-diameter wire. what material is the wire made of?

Answer 1

Thank you for posting you question here. I think the question is incomplete. I manage to check the complete question elsewhere:

A 0.0075 V/m electric field creates a 3.9 mA current in a 1.0 mm diameter wire. What is the wire made of? 

a) Aluminum (resistivity: 2.82 * 10 ^ -8) 
b) Nichrome (resistivity: 1.1 * 10 ^ -6) 
c) Copper (resistivity: 1.68 * 10 ^ -8) 
d) Iron (resistivity: 1.0 * 10 ^ -7) 

the answer is B which is Nichrome (resistivity: 1.1 * 10 ^ -6) 

Answer 2

The wire made of Nichrome  

Further explanation

The ρ resistivity of an object is the ratio of the magnitude of the electric field to the current density

Can be formulated:

$\large{\boxed{\bold{\rho\:=\:\frac{E}{J} }}}$

Current density (J) itself is the ratio of currents per unit area of cross-section

Can be formulated:

$\large{\boxed{\bold{J\:=\:\frac{I}{A} }}}$

Complete questions from the assignments above can be stated as follows:

A 0.0075 V / m electric field creates a 3.9 mA current in a 1.0 mm diameter wire. What is the wire made of?

a) Aluminum (resistivity: 2.82 * 10 ^ -8)

b) Nichrome (resistivity: 1.1 * 10 ^ -6)

c) Copper (resistivity: 1.68 * 10 ^ -8)

d) Iron (resistivity: 1.0 * 10 ^ -7)

If we see the question above then to find the resistivity of the material we must find the wire cross-section area first

Known value

E = 0.0075 V / m

I = 3.9 mA = 3.9.10⁻³ A

d = 1.0 mm = 1.10⁻³ m

A = 1 / 4 π.d²

A = 1 / 4. π. (10⁻³)²

A = 1/4 . π . 10⁻⁶

so that

$\rho\:=\:\frac{E}{J}$

$\rho\:=\:\frac{E.A}{I}$

$\rho\:=\:\frac{0.0075\frac{10^{-6}}{4} }{3.9.10^{-3}}}$

ρ = 1.5.10⁻⁶ ohm . m

From this value the material which approaches its resistivity is Nichrome with resistivity: 1.1 . 10⁻⁶ ohm.m

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Keywords: the magnetic field, electric current, resistivity , Nichrome