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A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a 4.5-m-long, 2.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.0 mT magnetic field 1.0 mm from the surface of the wire. What is the material's resistivity

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Answer:

[tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex]

Explanation:

r = Radius = [tex]\dfrac{2}{2}=1\ \text{mm}[/tex]

B = Magnetic field = 3 mT

1 mm = Distance from the surface of the wire

V = Voltage

x = Distance from the probe = [tex]r+1=1+1=2\ \text{mm}[/tex]

R = Resistance

L = Length of wire = 4.5 m

Magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi x}\\\Rightarrow I=\dfrac{B2\pi x}{\mu_0}\\\Rightarrow I=\dfrac{3\times 10^{-3}\times 2\times \pi 2\times 10^{-3}}{4\pi 10^{-7}}\\\Rightarrow I=30\ \text{A}[/tex]

Voltage is given by

[tex]V=IR\\\Rightarrow R=\dfrac{V}{I}\\\Rightarrow R=\dfrac{1.5}{30}\\\Rightarrow R=0.05\ \Omega[/tex]

Resistivity is given by

[tex]\rho=\dfrac{RA}{L}\\\Rightarrow \rho=\dfrac{0.05\times \pi (1\times 10^{-3})^2}{4.5}\\\Rightarrow \rho=3.49\times 10^{-8}\ \Omega\text{m}[/tex]

The resistivity of the material is [tex]3.49\times 10^{-8}\ \Omega\text{m}[/tex].

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