What is the magnetic field outside a 2.5-mm-diameter copper wire carrying a 33-A current?
The magnetic field outside the wire, 2.5 mm from the surface, is approximately 2.64 × 10^(-4) T.
Understanding Ampere's Law
To determine the magnetic field outside the copper wire, we can use Ampere's law. According to Ampere's law, the magnetic field line integral around a closed loop is equal to the product of the current encompassed by the loop and the permeability of free space (μ₀).
Application of Ampere's Law
In this case, we can consider a circular loop with a radius of 2.5 mm, which is located outside the copper wire. The current passing through this loop is 33 A.
Applying Ampere's law, we have:
∮ B · dl = μ₀I
Determination of Magnetic Field
Since the magnetic field is uniform across the cross section of the wire and the loop is circular, the magnetic field is constant along the path of integration and perpendicular to the loop. Therefore, the dot product B · dl simplifies to B · 2πr, where r is the radius of the loop.
Substituting the values into the equation, we get:
B · 2πr = μ₀I
Solving for B, we have:
B = (μ₀I) / (2πr)
Numerical Calculation
Given that the current I is 33 A and the radius r is 2.5 mm (or 0.0025 m), and the permeability of free space μ₀ is approximately 4π × 10^(-7) T·m/A, we can substitute these values to calculate the magnetic field outside the wire:
B = (4π × 10^(-7) T·m/A * 33 A) / (2π * 0.0025 m)
Simplifying the equation, we get:
B ≈ 2.64 × 10^(-4) T
Therefore, the magnetic field outside the wire, 2.5 mm from the surface, is approximately 2.64 × 10^(-4) T.