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The resistivity of gold is 2.44 x 10⁻⁸ ohms.m at room temperature. A gold wire that is 1.9 mm in diameter and 48 cm long carries a current of 800 mA. What is the electric field in the wire?

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Final answer:

To calculate the electric field in a gold wire, we first determine the resistance using the wire's dimensions and resistivity. With the resistance and current value, we apply Ohm's Law to find the electric field.

Step-by-step explanation:

The question asks to calculate the electric field in a gold wire with given dimensions and current. To find the electric field, we first need to calculate the resistance of the wire, which can be done using the formula R = ρ × (L / A), where ρ is the resistivity, L is the length, and A is the cross-sectional area. Since we're given the diameter of the wire, we can find the area of the cross-section using A = π × (d/2)^2, where d is the diameter. After finding resistance, we can compute the electric field using Ohm's Law, where E = I × R, with I being the current.

First, we convert the diameter from millimeters to meters (d = 1.9 mm = 0.0019 m) and calculate the area: A = π × (0.0019/2)^2 m^2. Next, using the resistivity of gold (ρ = 2.44 x 10^-8 Ω.m) and the length of the wire (L = 48 cm = 0.48 m), we calculate the resistance, R.

With the resistance and the current (I = 800 mA = 0.8 A), we can calculate the electric field, E.

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