A) \[{{A}_{2}}={{\mu }_{0}}{{A}_{1}}\]
B) \[{{A}_{2}}={{A}_{1}}\]
C) \[{{A}_{2}}=\frac{{{A}_{1}}}{{{\mu }_{0}}}\]
D) \[{{A}_{2}}={{\mu }_{0}}^{2}{{A}_{1}}\]
Correct Answer: A
Solution :
We know \[\vec{B}={{\mu }_{0}}(\vec{H}+\vec{l})\] \[\Rightarrow \]\[dB={{\mu }_{0}}dH+{{\mu }_{0}}dl\] \[\oint{\vec{H}.d\vec{B}={{\mu }_{0}}}\oint{\vec{H}.d\vec{H}+{{\mu }_{0}}}\oint{\vec{H}.d\vec{l}}\] \[\Rightarrow \]\[\oint{\vec{H}.d\vec{B}={{\mu }_{0}}}\oint{Hdl}\]\[\Rightarrow \]\[{{A}_{2}}={{\mu }_{0}}{{A}_{1}}\]
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