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Question
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero
(a) outside the cable
(b) inside the inner conductor
(c) inside the outer conductor
(d) in between the tow conductors.
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Solution
(a) outside the cable
(b) inside the inner conductor
According to Ampere's law, in a coaxial, straight cable carrying currents i in the inner conductor and -i (equally in the opposite direction) in the outside conductor.
Inside the inner conductor
\[\oint \vec{B} . d \vec{l} = \mu_o i_{\text{ inside }} \]
\[\oint \vec{B} . d \vec{l} = 0\]
\[ \Rightarrow B . l = 0\]
\[ \Rightarrow B = 0\]
In between the 2 conductors
\[\oint \vec{B} . d \vec{l} = \mu_o i\]
\[ \Rightarrow B = \frac{\mu_o i}{2\pi r}\]
Outside the outer conductor
\[\oint \vec{B} . d \vec{l} = \mu_o (i - i)\]
\[ \Rightarrow B = 0\]
Therefore, the magnetic field is zero outside the cable.
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