A long hollow cylindrical thick-walled conductor (inner radius 1.50cm, outer radius 4.00cm) carries a current of 24.0A distributed uniformly across its cross section. A long thin wire runs down the center of the cylinder and carries a current of 32.0A in the opposite direction. What is the magnitude of the magnetic field 3.00cm from the axis?

B = 13.47 * 10⁻⁵ T

Explanation:

The magnetic field will be the difference between the magnetic field due to the current passing through the center of the cylinder and the one due to the current carried by the hollow cylindrical thick-walled conductor.

Magnetic field due to the current passing through a conductor can be given by the equation:  

Let the magnetic field due to the current passing through the center of the cylinder be B₁

B₁ = 0.0002133 T = 21.33 * 10⁻⁵ T

Let the magnetic field due to the current carried by the hollow cylindrical thick-walled conductor be B₂

B = B₁ - B₂

B = (21.33 * 10⁻⁵) - (7.86 * 10⁻⁵)

B = 13.47 * 10⁻⁵ T