A) 2B
B) 4B
C) \[B/2\]
D) B
Correct Answer: D
Solution :
[d] 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) |
× (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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