Question

A coaxial cable consists of an inner cylindrical conductor of radius 3 cm on the axis ofan outer hollow cylindrical conductor of inner radius 7 cm and outer radius of 10 cm.The inner conductor carries current 4.5 A in one direction, and the outer conductor carries current 8 A in the opposite direction. What is the value of the magnetic field atfollowing distances from the axis of the cable:

Answer 1

Hello your question lacks some information and that is the distance

At r = 0.150 m (outside the cable) note : (μ 0 = 4π × 10-7 T · m/A)

answer: 4.7 * 10^-6 T

Step-by-step explanation:

R1 = 0.03 m

R2 = 0.07 m

R3 = 0.1 m

Inner conductor carries current = 4.5A in one direction

outer conductor carries current = 8A in opposite direction

Determine the value of magnetic field at r = 0.150 m

net current = 8 - 4.5 = 3.5 A

∴ β ( value of magnetic field )

= ( 4π * 10^-7 * 3.5 ) / ( 2π * 0.150 )

= 4.4*10^-6 / 0.94

= 4.7 * 10^-6 T