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Figure given in the question is a cross-...

Figure given in the question is a cross-sectional view of a coaxial cable. The centre conductor. surrounded by a rubber layer, which is surrounded by an outer conductor, which is surround 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.

Text Solution

Verified by Experts

Applying Ampere's circular circuital law.
`B_A=mu_0/(2pi) i_("in")/r`

`=((2xx10^-7)(1))/10^-3`
`=2xx10^-4T`
This due to `(*)` current of `1A`. Hence, magnetic lines are circular and anti-clockwise. Hence, magnetic field is upwards.
`B_b=mu_0/(2pi) i_("in")/r`
`=((2xx10^-7)(3-1))/(3xx10^-3)`
`=1.33xx10^-4T`
This is due to net `(ox)` current. Hence, magnetic lines are clockwise.
So, magnetic fied at `B` is downwards.
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