The wire made of Nichrome
The ρ resistivity of an object is the ratio of the magnitude of the electric field to the current density
Can be formulated:
[tex]\large{\boxed{\bold{\rho\:=\:\frac{E}{J} }}}[/tex]
Current density (J) itself is the ratio of currents per unit area of cross-section
Can be formulated:
[tex]\large{\boxed{\bold{J\:=\:\frac{I}{A} }}}[/tex]
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
[tex]\rho\:=\:\frac{E}{J}[/tex]
[tex]\rho\:=\:\frac{E.A}{I}[/tex]
[tex]\rho\:=\:\frac{0.0075\frac{10^{-6}}{4} }{3.9.10^{-3}}}[/tex]
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
