Consider a coaxial cable with 2Amps in the center conductor coming out of the page, and 1Amp in the outer conductor going into the page. The center conductor has a radius of 1mm, the outer conductor's inner radius is 2mm, and the outer conductor's outter radius is 3mm. What is the magnitude of the magnetic field at point P which is 5mm from the cable's axis?
What was wrong with the answer I gave you earlier?
http://www.jiskha.com/display.cgi?id=1223804401 which is the same as i gave you ..
is the radius of the cable or of the axis to the point p?
To find the magnitude of the magnetic field at point P, we can use Ampere's law. Ampere's law states that the line integral of the magnetic field around a closed loop is equal to the product of the current passing through the loop and the permeability of free space.
1. First, let's consider a circular loop with radius r drawn around the cable's axis, passing through point P.
2. The loop encloses both the inner and outer conductors of the coaxial cable. Therefore, the net current passing through the loop is the difference between the current in the center conductor and the current in the outer conductor. In this case, it would be 2A - 1A = 1A.
3. According to Ampere's law, the line integral of the magnetic field (B) along the loop is equal to the product of the current (I) and the permeability of free space (μ₀).
∮ B · dl = μ₀ · I
Since the magnetic field is constant along the loop's path and parallel to the direction of the line element (dl), we can simplify the equation to:
B ∮ dl = μ₀ · I
4. The line integral ∮ dl represents the circumference of the circular loop. For a loop with radius r, its circumference is 2πr.
B × 2πr = μ₀ · I
5. Solving for B, we have:
B = (μ₀ · I) / (2πr)
Substitute the given values:
I = 1A, r = 5mm = 0.005m, and μ₀ = 4π × 10⁻⁷ T·m/A
B = (4π × 10⁻⁷ T·m/A × 1A) / (2π × 0.005m)
B = (2 × 10⁻⁷ T·m) / (0.01m)
B = 2 × 10⁻⁵ T
Therefore, the magnitude of the magnetic field at point P, 5mm from the cable's axis, is 2 × 10⁻⁵ Tesla.