Question

The figure below is a cross-sectional view of a coaxial cable. The center conductor is surrounded by a rubber layer, an outer conductor, and another rubber layer. In a particular application, the current in the inner conductor is 11 = 1.16 A out of the page and the current in the outer conductor is 12 = 3.20 A into the page. Assuming the distance d = 1.00 mm, answer the following. d (a) Determine the magnitude and direction of the magnetic field at point a. magnitude 232 direction downward (b) Determine the magnitude and direction of the magnetic field at point b. -136 X magnitude You can approach this problem by finding the field produced by current 11 and the field produced by 12 and then adding them vectorially. ut direction upward

Answer 1

Solution:

(a)

Ampere's law

$\oint \boldsymbol{\overrightarrow{B}}\cdot \boldsymbol{\overrightarrow{dl}}=\mu _{o}I_{enc}$

Β (2πά) = μΙ1

$B\left ( 2\pi \left ( 1\times 10^{-3}m \right ) \right )=\left ( 4\pi \times 10^{-7}\frac{T-m}{A} \right )\left (1.16A \right )$

$B=232\mu T$

Magnitude = $B=232\mu T$

Direction: upwards

(b)

Ampere's law

$\oint \boldsymbol{\overrightarrow{B}}\cdot \boldsymbol{\overrightarrow{dl}}=\mu _{o}I_{enc}$

B (27 (3d)) = u(11)
$B\left ( 2\pi \left (3d \right ) \right )=\mu\left ( I_{2} -I_{1} \right )$

$B\left ( 2\pi \left ( 3\times 10^{-3}m \right ) \right )=\left ( 4\pi \times 10^{-7}\frac{T-m}{A} \right )\left (3.20A-1.16A \right )$

$B=136\mu T$

Magnitude = $B=136\mu T$

Direction: downward