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 \[{{B}_{2}}=0\] \[\left[ As\,\oint{\overset{\to }{\mathop{B}}\,.d\,\overset{\to }{\mathop{i}}\,={{\mu }_{0}}\sum \,I} \right]\]You need to login to perform this action.
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