question_answer

What is the net force on the square coil  [DCE 2000; RPMT 2000]

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.