A) 2B
B) 4B
C) B/2
D) B
Correct Answer: D
Solution :
Key Idea: According to Amperes law, the line integral\[\oint{\overrightarrow{B}}.\overrightarrow{dl}\]of the resultant magnetic field along a closed, plane curve is equal to\[{{\mu }_{0}}\]times the total current crossing the area bounded by the closed curve. Using Ampere's law \[\oint{\overrightarrow{B}}.d\overrightarrow{l}={{\mu }_{0}}({{i}_{net}})\] ?.. (i) In our case, \[{{i}_{net}}=1\](number of turns inside the area) \[\times \](current through each turn)\[=(nl)i\] where n is number of turns per unit length. Then, Eq. (i) can be written as, \[Bl=({{\mu }_{0}})(nli)\] \[Bl={{\mu }_{0}}ni\] \[B\propto ni\] \[\therefore \] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{n}_{1}}{{i}_{1}}}{{{n}_{2}}{{i}_{2}}}\] Here, \[{{n}_{1}}=n,{{n}_{2}}=\frac{n}{2},{{i}_{1}}=i,{{i}_{2}}=2i,{{B}_{1}}=B\] Hence, \[\frac{B}{{{B}_{2}}}=\frac{n}{n/2}\times \frac{i}{2i}=1\] or \[{{B}_{2}}=B\]
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