Main Answer:
a) The resistance of the copper wire is approximately 0.115 ohms.
b) The current flow through the wire, given a potential drop of 0.05 V, is around 0.435 A.
c) The current density in the wire is roughly 1.58 x 10^6 A/m².
d) The magnitude of the electric field across the ends of the wire is about 0.026 V/m.
Therefore, the correct answer is All of them.
Step-by-step explanation:
In part (a), the resistance of the copper wire is 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. The resistivity of copper is a fundamental property, and the cross-sectional area is determined from the diameter of the wire.
Moving on to part (b), Ohm's Law (V = IR) is applied to find the current flow through the wire. The potential drop (V) is given, and rearranging the formula to solve for current (I) provides the answer.
Part (c) involves calculating current density (J), which is the current per unit area. Current density is obtained by dividing the current by the cross-sectional area of the wire, providing an indication of how densely the electric current is flowing through the material.
Finally, in part (d), the magnitude of the electric field is determined using the formula E = V/L, where V is the potential drop and L is the length of the wire. This reflects the electric field strength across the wire.
These calculations help understand the electrical properties of the copper wire under the given conditions, shedding light on its resistance, current flow, current density, and electric field strength.
Therefore, the correct answer is All of them.