# Question Video: Resistance and Resistivity of Conductors Physics • 9th Grade

A copper wire with a resistance of 12.8 mΩ has a cross-sectional area of 1.15 × 10⁻⁵ m². Find the length of the wire. Use 1.7 × 10⁻⁸ Ω⋅m for the resistivity of copper.

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### Video Transcript

A copper wire with a resistance of 12.8 milliohms has a cross-sectional area of 1.15 times 10 to the negative fifth meter squared. Find the length of the wire. Use 1.7 times 10 to the negative eighth ohms times meters for the resistivity of copper.

Let’s begin by drawing a diagram of our copper wire. In our diagram, we’ve labeled our copper wire with a resistivity 𝜌 of 1.7 times 10 to the negative eighth ohm meters, a cross-sectional area 𝐴 of 1.15 times 10 to the negative fifth meter squared, a resistance 𝑅 of 12.8 milliohms, and we are solving for the length of the wire 𝐿.

Before we do any calculations, we need to recall an equation that relates the resistivity, cross-sectional area, length, and resistance of our wire to each other. That equation is 𝑅, the resistance of the wire, is equal to 𝜌, the resistivity of the wire, times 𝐿, the length of the wire, divided by 𝐴, the cross-sectional area of the wire. The problem asked us to solve for the length of the wire. Therefore, we must rearrange our formula so that we are solving for 𝐿.

To do this, we have to multiply both sides of the equation by 𝐴 over 𝜌. This will cancel out both the 𝐴 and the 𝜌 on the right side of the equation, leaving us with the relationship 𝐴 times 𝑅 divided by 𝜌 is equal to 𝐿. We can now substitute in our values for our variables. For our area, we use 1.15 times 10 to the negative fifth meter squared. The resistance is 12.8 milliohms. And the resistivity is 1.7 times 10 to the negative eighth ohms meters.

We need to be careful with our units. Our resistance is given to us in milliohms, but it needs to be converted into ohms. We need to remember that milli- is the prefix for 10 to the negative third. Therefore, we can replace 12.8 milliohms with 12.8 times 10 to the negative third ohms. When we multiply out our fraction, we get a length of the wire as 8.66 meters. The resistivity in our problem was given to two significant figures. Therefore, we must report back our length to two significant figures. 8.66 meters rounds to 8.7 meters. The length of the wire is 8.7 meters.