In a coaxial cable,the solid inner conductor carries a current of I.The shield,which also carries a current of I but in the opposite direction,has an inner radius of a and an outer radius of b.The current density is equally distributed over each conductor.What is the magnitude of at a radial distance of 1.05 mm from the center of the coaxial cable if a = 1 mm,b = 1.1 mm and I = 100 mA?

Question

Quizplus · Accepted Answer

The magnetic field at a radial distance r from the center of a coaxial cable with current I in the inner conductor and opposite current -I in the shield is given by:
B = (μ0/2π) x Iln(b/a) / r
where μ0 is the permeability of free space and ln(b/a) is the natural logarithm of the ratio of the outer and inner radii of the shield.

Substituting the given values, we get:
B = (4π×10^-7/2π) x 100×10^-3 ln(1.1/1) / 1.05×10^-3
B = 10 μT

Thus, the magnetic field magnitude at a radial distance of 1.05 mm from the center of the coaxial cable is 10 µT. Therefore, the correct answer is option C.

In a Coaxial Cable,the Solid Inner Conductor Carries a Current