A long coaxial cable consists of a solid inner cylindrical conductor of radius a and a thick outer cylindrical conductor with inner radius b and outer radius c . The inner cylinder carries a current with a uniform current density J₀ . If the magnetic field at a radial distance r > c from the central axis is observed to be zero, what is the magnitude of the uniform current density in the outer conductor?

Step-by-step solution

By Ampere's circuital law, the magnetic field outside the coaxial cable ( r > c ) is given by B d l = ₀ I_ net . Since B = 0 for r > c , the net enclosed current must be zero. This means the total current in the inner conductor must be equal in magnitude and opposite in direction to the total current in the outer conductor. The current in the inner conductor is I_ in = J₀ ( a^2) . Let the uniform current density in the outer conductor be J_ out . The cross-sectional area of the outer conductor is c^2 - b^2 = (c^2 -