A 1.0 times 10 to the two volt potential difference is applied across a 10.0-metre length of wire with a diameter of 4.621 millimetres. The magnitude of the current density produced is 2.0 times 10 to the eighth amps per metre squared. What is the resistivity of the wire?
We can call the potential difference applied across the wire 1.0 times 10 to the two volts 𝑣. The length of the wire 10.0 metres, we can call capital 𝐿. We’re told the wire has a diameter of 4.621 millimetres, which we’ll called 𝑑. And the current density running through the wire is given as 2.0 times 10 to the eighth amps per metre squared. We’ll call that 𝐽. We want to know the resistivity of the wire, which we’ll call 𝜌.
We can start off by recalling a relationship between resistivity and resistance. The electrical resistance 𝑅 of a wire is equal to the resistivity 𝜌 of the material the wire is made of times the length of the wire divided by the cross-sectional area of the wire 𝐴.
Rearranging this equation to solve for 𝜌, we see it’s equal to the resistance of the wire multiplied by its cross-sectional area divided by its length 𝐿. 𝐿 is given to us in the problem statement. And 𝐴 we can solve for using the diameter of the wire 𝑑. Assuming the wire has a circular cross section, we recall that the area of the circle is 𝜋 times its radius squared.
In our case, 𝐴 equals 𝜋 times 𝑑 over two quantity squared or 𝜋 over four times 4.621 times 10 to the negative third metres squared. Knowing the length of the wire 𝐿 and being able to calculate its cross-sectional area 𝐴, we now want to solve for its resistance 𝑅. To do that, we can recall Ohm’s law, which says that the voltage 𝑣 across a circuit is equal to the current in the circuit multiplied by its resistance 𝑅. 𝑣 being equal to 𝐼 times 𝑅 implies that 𝑅 is equal to 𝑣 divided by 𝐼.
We’re told the potential difference across the circuit in the problem statement, but we don’t know the current involved 𝐼. However, we can recall that current density 𝐽 is equal to 𝐼 divided by cross-sectional area 𝐴. 𝐽 being equal to 𝐼 divided by 𝐴 implies that 𝐼 is equal to 𝐽 times 𝐴. 𝐽 is given to us in the problem statement and 𝐴 is a known value. When we plug in for these values to solve for 𝐼 and enter them on our calculator, we find that the current is approximately 3354 amps.
Knowing that current, we can now return to our Ohm’s law equation to solve for the resistance 𝑅. 𝑅 is equal to 𝑣 over 𝐼 or 1.0 times 10 to the two volts divided by 3354 amps. Calculating this fraction, we find that 𝑅 is approximately 0.0298 ohms.
We are now ready to return to our equation for resistivity in terms of resistance 𝑅. We’ve calculated 𝑅 and 𝐴 and our given 𝐿. So we’re ready to plug in and solve for 𝜌. When we do, we find that 𝜌 is equal to two significant figures to 5.0 times 10 to the negative eighth ohm metres. That’s the resistivity of this wire.