A) 2B
B) B
C) \[\frac{B}{2}\]
D) 3B
Correct Answer: B
Solution :
Magnetic field at midpoint M in first case is \[B={{B}_{PQ}}-{{B}_{RS}}\] (\[\because \]\[{{B}_{PQ}}\]and \[{{B}_{RS}}\]are in opposite directions) \[=\frac{4{{\mu }_{0}}}{4\pi d}-\frac{2{{\mu }_{0}}}{4\pi d}\] \[=\frac{2{{\mu }_{0}}}{4\pi d}\] When the current 2 A is switched off, the net magnetic field at M is due to current 1A \[B=\frac{{{\mu }_{0}}\times 2\times 1}{4\pi d}=B\]
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