Answer:
[tex]1.8\cdot 10^{-7} \Omega m[/tex]
Explanation:
The magnetic field produced by the wire is given by:
[tex]B=\frac{\mu_0 I}{2 \pi r}[/tex]
where
B = 3.5 mT = 0.0035 T is the magnetic field
[tex]\mu_0[/tex] is the vacuum permeability
[tex]I[/tex] is the current
r = 1.0 mm = 0.001 m is the distance from the wire
Solving for I, we find the current in the wire
[tex]I=\frac{2 \pi r B}{\mu_0}=\frac{2 \pi (0.0035 T)(0.001 m)}{4 \pi \cdot 10^{-7} H/m}=17.5 A[/tex]
Now we can find the resistance of the wire, R, by using Ohm's law:
[tex]R=\frac{V}{I}=\frac{1.5 V}{17.5 A}=0.086 \Omega[/tex]
And the resistance is related to the material's resistivity ([tex]\rho[/tex]) by
[tex]R=\rho \frac{L}{A}[/tex]
where
L = 6.0 m is the length of the wire
[tex]A= \pi r^2 = \pi (d/2)^2= \pi (0.004 m/2)^2=1.26\cdot 10^{-5}m^2[/tex] is the cross-sectional area of the wire
Solving for [tex]\rho[/tex], we find
[tex]\rho = \frac{RA}{L}=\frac{(0.086 \Omega)(1.26\cdot 10^{-5} m^2)}{6.0 m}=1.8\cdot 10^{-7} \Omega m[/tex]
The material's resistivity is set at : 1.8 * 10⁻⁷ Ωm
Given data :
Length of wire = 6 m
wire diameter = 4 mm
voltage = 1.5 V
magnetic field produced by wire; B = [tex]\frac{u_{o}I }{2\pi r}[/tex] ---- ( 1 )
where ; r = 0.001 m , u₀ = vacuum permeability, I = current
First step : Determine the current ( I )
from equation ( 1 )
[tex]I = \frac{2\pi Br}{u_{o} }[/tex] -------- ( 2 )
where : B = 0.0035T , r = 0.001 m, u₀ = 4[tex]\pi[/tex] * 10⁻⁷ H/m
insert values into equation ( 2 )
I ( current ) = 17.5 A
Next step : determine the resistance of the wire
we will apply Ohm's law
R = Voltage / current
= 1.5 / 17.5 = 0.086 Ω
Also :
Resistance can be expressed as
R = p [tex]\frac{L}{A}[/tex] ---- ( 3 )
where : p = resistivity
making resistivity subject of the relation
p = [tex]\frac{RA}{L}[/tex] ----- ( 4 )
Where : A = area of wire = 1.26 * 10⁻⁵ m², L = 6 m, R = 0.086 Ω
insert values into equation 4
P = 1.8 * 10⁻⁷ Ωm
Hence we can conclude that The material's resistivity is set at : 1.8 * 10⁻⁷ Ωm .
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