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