Question

A spool of copper wire 200 m long and with a diameter of 0.360 mm is at 20.0°C. For copper, the resistivity is 1.70 ✕ 10⁻⁸Ω · m and the temperature coefficient of resistivity is 3.90 ✕ 10⁻³ (°C)⁻¹. What is the magnitude of the electric field (in V/m) in the wire if it carries a current of 0.450 A?

Answer 1

To find the magnitude of the electric field in the wire, we can use Ohm's Law and the formula for resistance. The magnitude of the electric field is 1.49 V/m.

Step-by-step explanation:

To find the magnitude of the electric field in the wire, we can use Ohm's Law and the formula for resistance. Ohm's Law states that the current through a conductor is equal to the voltage across the conductor divided by the resistance of the conductor. In this case, the current is given as 0.450 A and the resistance can be calculated using the formula R = (ρ * L) / A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire.

First, we need to find the cross-sectional area of the wire. The formula for the cross-sectional area of a circle is A = π * (d/2)^2, where d is the diameter of the wire. Plugging in the values, we get A = π * (0.360 mm / 2)^2 = 0.1017 mm².

Next, we can calculate the resistance using the formula R = (1.70 * 10^-8 Ω·m * 200 m) / (0.1017 mm² * 10^-6 mm²/m²) = 3.32 Ω.

Finally, we can use Ohm's Law to find the voltage across the conductor. V = I * R = 0.450 A * 3.32 Ω = 1.49 V.

Therefore, the magnitude of the electric field in the wire is 1.49 V/m.