A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12 mm from its center

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

    (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  

The current density is given by the total area and the total current:

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

 (2)

You replace the last result of equation (2) into the equation (1):

Finally. you replace the values of all parameters:

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT