Question

Figure given in the question is a cross sectional view of a coaxial cable.The centre conductor is surrounded by a rubber layer,which is surrounded by an outer conductor,which is surrounded by another rubber layer.The current in the inner conductor is 1.0 A out of the page,and the current in the outer conductor is 3.0 A into the page.Determine the magnitude and direction of the magnetic field at points a and b.

Answer 1

Applying Ampere's circuit at law:
$B_A=\frac{{\mu }_o}{2\pi }\frac{i_{in}}{r}$
$=\frac{(2\times {10}^{-7})\left(1\right)}{{10}^{-3}}=2\times {10}^{-4}T$
This is due to $(\cdot )$ current of 1 A.Hence magnetic lines are circular and anticlockwise.Hence magnetic field is upwards.
$B_b=\frac{{\mu }_o}{2\pi }\frac{i_{in}}{r}$
$=\frac{(2\times {10}^{-7})(3-1)}{3\times {10}^{-3}}=1.33\times {10}^{-4}T$
This is due to net $⨂$ current. Hence magnetic lines are clockwise.
So, magnetic field at B is downwards.