A straight copper wire that is 1 milimeter in diameter carries a current of 20 miliamps. Whas the magnitude of the largest magnetic field created by this wire in Tesla? pls help :(

To calculate the magnitude of the largest magnetic field created by a straight wire, we can use Ampere's Law.

Ampere's Law states that the magnetic field around a closed loop is directly proportional to the current passing through the loop, and inversely proportional to the distance from the wire.

The formula for calculating the magnetic field (B) is:

B = (μ₀ * I) / (2π * r)

Where:
- B represents the magnetic field
- μ₀ represents the permeability of free space (μ₀ ≈ 4π × 10⁻⁷ T·m/A)
- I represents the current passing through the wire
- r represents the distance from the wire

In this case, we have a copper wire with a diameter of 1 millimeter, which means the radius (r) is 0.5 millimeters or 0.0005 meters. The current (I) is 20 milliamps or 0.02 amps.

Plugging these values into the formula, we get:

B = (4π × 10⁻⁷ T·m/A * 0.02 A) / (2π * 0.0005 m)
B = (4π × 10⁻⁷ * 0.02) / (2π * 0.0005)
B = (8π × 10⁻⁸) / π * 0.0005
B = (8 × 10⁻⁸) / 0.0005
B ≈ 1.6 × 10⁻⁴ T

Therefore, the magnitude of the largest magnetic field created by this wire is approximately 1.6 × 10⁻⁴ Tesla.