A) \[25\times {{10}^{-7}}N\] moving towards wire
B) \[25\times {{10}^{-7}}N\] moving away from wire
C) \[35\times {{10}^{-7}}N\] moving towards wire
D) \[35\times {{10}^{-7}}N\] moving away from wire
Correct Answer: A
Solution :
Force on side BC and AD are equal but opposite so their net will be zero. But \[{{F}_{AB}}={{10}^{-7}}\times \frac{2\times 2\times 1}{2\times {{10}^{-2}}}\times 15\times {{10}^{-2}}=3\times {{10}^{-6}}N\] and \[{{F}_{CD}}={{10}^{-7}}\times \frac{2\times 2\times 1}{\left( 12\times {{10}^{-2}} \right)}\times 15\times {{10}^{-2}}\]\[=0.5\times {{10}^{-6}}N\] Þ \[\,{{F}_{net}}={{F}_{AB}}-{{F}_{CD}}\] \[=2.5\times {{10}^{-6}}N\] \[=25\times {{10}^{-7}}N\], towards the wire.You need to login to perform this action.
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