question_answer

A coaxial cable consists of a thin inner conductor fixed along the axis of a hollow outer conductor. The two conductors carry equal currents in opposites directions. Let \[{{B}_{1}}\]and \[{{B}_{2}}\] be the magnetic fields in the region between the conductors and outside the conductor, respectively Then,     

A) \[{{B}_{1}}\ne 0,\,{{B}_{2}}\ne 0\]

B) \[{{B}_{1}}={{B}_{2}}=0\]

C) \[{{B}_{1}}\ne 0,{{B}_{2}}=0\]

D) \[{{B}_{1}}=0,{{B}_{2}}\ne 0\]

Correct Answer: C

Solution :

[c] Apply Ampere's circular law to the coaxial circular loops \[{{L}_{1}}\] and\[{{L}_{2}}\].            The magnetic field is \[{{B}_{1}}\] at all points on \[{{L}_{1}}\]and \[{{B}_{2}}\]at all points on \[{{L}_{2}}\]. \[\sum I\ne 0\]for\[{{L}_{1}}\]and 0 for \[{{L}_{2}}\]Hence, \[{{B}_{1}}\ne 0\] but \[{{B}_{2}}=0\] \[\left[ As\,\oint{\overset{\to }{\mathop{B}}\,.d\,\overset{\to }{\mathop{i}}\,={{\mu }_{0}}\sum \,I} \right]\]