In a nickel conductor 72 cm long and a cross-sectional area of 0.8 mm squared, a current of 200 mA

In a nickel conductor 72 cm long and a cross-sectional area of 0.8 mm squared, a current of 200 mA was established. What is the voltage at its ends?

The electrical resistance R in a nickel conductor with a length of L = 72 cm = 0.72 m and a cross-sectional area S = 0.8 mm ^ 2 is determined by the formula R = (ρ ∙ L): S, where ρ = 0.1 Ohm ∙ mm ^ 2 / m – resistivity of nickel, we determine it according to reference tables. From the condition of the problem, it is known that a current with a force of I = 200 mA = 0.2 A. was established in the conductor. To determine what the voltage at its ends is, we will use Ohm’s law for a section of the circuit:

I = U: R, then U = I ∙ R or U = (I ∙ ρ ∙ L): S.

We get:

U = (0.2 A ∙ 0.1 Ohm ∙ mm ^ 2 / m ∙ 0.72 m): 0.8 mm ^ 2;

U = 0.018 V; U = 18 mV.

Answer: the voltage at the ends of the nickel conductor was 18 mV.