Question

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

Answer 1

`3.49* 10^(-8)\ \Omega\text{m}`

Step-by-step explanation:

r = Radius =
`(2)/(2)=1\ \text{mm}`

B = Magnetic field = 3 mT

1 mm = Distance from the surface of the wire

V = Voltage

x = Distance from the probe =
`r+1=1+1=2\ \text{mm}`

R = Resistance

L = Length of wire = 4.5 m

Magnetic field is given by

`B=(\mu_0I)/(2\pi x)\\\Rightarrow I=(B2\pi x)/(\mu_0)\\\Rightarrow I=(3* 10^(-3)* 2* \pi 2* 10^(-3))/(4\pi 10^(-7))\\\Rightarrow I=30\ \text{A}`

Voltage is given by

`V=IR\\\Rightarrow R=(V)/(I)\\\Rightarrow R=(1.5)/(30)\\\Rightarrow R=0.05\ \Omega`

Resistivity is given by

`\rho=(RA)/(L)\\\Rightarrow \rho=(0.05* \pi (1* 10^(-3))^2)/(4.5)\\\Rightarrow \rho=3.49* 10^(-8)\ \Omega\text{m}`

The resistivity of the material is
`3.49* 10^(-8)\ \Omega\text{m}`.