Question

The figure gives the electric potential V(x) along a copper wire carrying uniform current, from a point of higher potential Vs = 11.0 μV at x = 0 to a point of zero potential at xs = 2.80 m. The wire has a radius of 2.30 mm, and copper has a resistivity of 1.69 × 10-8 Ω·m. What is the current in the wire?

Answer 1

potential diffrence between the two points, delta_V = 11.0 - 0

= 11.0 micro V

length of the wire between the two points, L = 2.80 m

cross sectional area of the wire, A = pi*r^2

= pi*(2.3*10^-3)^2

= 1.66*10^-5 m^2

resistance of the segment of the wire, R = rho*L/A

= 1.69*10^-8*2.8/(1.66*10^-5)

= 0.00285 ohm

now use Ohm's law

delta_V = I*R

==> I = delta_V/R

= 11.0*10^-6/0.00285

= 3.86*10^-3 A or 3.86 mA <<<<<<<<<<---------Answer