# Why isn't current negative in this example?

• x86
In summary, the figure shows the electric potential along a copper wire carrying a uniform current from a point of higher potential at x = 0 to a point of zero potential at x = 3.20 m. The wire has a radius of 1.50 mm and copper has a resistivity of 1.69 × 10-8 Ω·m. Using the formula p(I/A)=V/L, the current in the wire can be calculated to be +1.176e-3, with all values considered as magnitudes. The current direction can be determined from the rule that current flows from higher to lower potential.

## Homework Statement

The figure gives the electric potential V(x) along a copper wire carrying uniform current, from a point of higher potential Vs = 9.00 μV at x = 0 to a point of zero potential at xs = 3.20 m. The wire has a radius of 1.50 mm, and copper has a resistivity of 1.69 × 10-8 Ω·m. What is the current in the wire?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c26/q24.jpg

## Homework Equations

p(I/A)=V/L

p = 1.69*10^-8 (ohm * meter) for copper

## The Attempt at a Solution

I solved it by just plugging values into the above equation give. It is clear that A>0 (cross section), L > 0 (length), p > 0 (given constant). And V < 0 because Vf = 9*10^-6 and Vi = 0.

The answer is +1.176e-3, but I got a negative value. As you can see, every value is positive except delta V, because Vf < Vi, so the current should be negative, should it not?

Probably the formula was intended to give only the magnitude of the current. So, all of the quantities in the formula should be considered as magnitudes (positive). You can always get the direction of the current from the rule that current in a conductor flows from higher to lower potential.