Magnetic Field Produced by Straight Currents Video Lessons

Concept

# Problem: A coaxial cable consists of a solid inner conductor of radius R1 surrounded by a concentric cylindrical tube of inner radius R2 and outer radius R3 (the figure). The conductors carry equal and opposite currents Io distributed uniformly across their cross sections (Figure 1).Let Io = 1.50 A, R1 = 1.00 cm, R2 = 2.00 cm, and R3 = 2.50 cm. Graph B from R = 0 to R = 3.00 cm.

###### FREE Expert Solution

In this problem, we are dealing with the magnetic field produced by straight currents. We'll use amperes law to determine the magnetic field around a current.

Ampere's law:

$\overline{){\mathbf{\oint }}{\mathbf{B}}{\mathbf{·}}{\mathbf{dl}}{\mathbf{=}}{{\mathbf{\mu }}}_{{\mathbf{0}}}{{\mathbf{I}}}_{\mathbf{e}\mathbf{n}\mathbf{c}\mathbf{l}\mathbf{o}\mathbf{s}\mathbf{e}\mathbf{d}}}$

B•∫dl = μ0Ienc

91% (34 ratings)
###### Problem Details

A coaxial cable consists of a solid inner conductor of radius R1 surrounded by a concentric cylindrical tube of inner radius R2 and outer radius R3 (the figure). The conductors carry equal and opposite currents Io distributed uniformly across their cross sections (Figure 1).

Let Io = 1.50 A, R1 = 1.00 cm, R2 = 2.00 cm, and R3 = 2.50 cm. Graph B from R = 0 to R = 3.00 cm.