Final answer:
Using Ampère's Law, the maximum current a wire can carry to produce a magnetic field no larger than Earth's (0.50×10⁻⁴T) at 12 cm distance is calculated to be about 3.0 A when rounded to two significant figures.
Step-by-step explanation:
The question is asking for the maximum current that a wire can have while only producing a magnetic field equivalent to that of Earth's at a 12 cm distance from the wire. To find this current, we can apply the formula for the magnetic field produced by a long straight current-carrying wire, which is given by Ampère's Law:
B = μ_0 * I / (2πr)
Where B is the magnetic field strength, μ_0 is the permeability of free space (4π x 10⁻⁷ T·m/A), I is the current in the wire, and r is the distance from the wire. Given that the maximum magnetic field strength B is 0.50 x 10⁻⁴ T and the distance r is 12 cm (0.12 m), we can re-arrange the formula to solve for I, the maximum current.
I = B * (2πr) / μ_0
Plugging the values into the equation, we can calculate the maximum current:
I = (0.50 × 10⁻⁴ T) * (2π * 0.12 m) / (4π × 10⁻⁷ T·m/A)
After doing the math, we find that the maximum current the wire can carry is approximately 3.0 A to two significant figures, respecting the given precision of the Earth's field.