A) \[1.4\times {{10}^{-4}}N\] towards the right
B) \[1.4\times {{10}^{-4}}N\] towards the left
C) \[2.6\times {{10}^{-4}}N\] to the right
D) \[2.6\times {{10}^{-4}}N\] to the left
Correct Answer: A
Solution :
Force on wire Q due to wire P is \[{{F}_{P}}={{10}^{-7}}\times \frac{2\times 30\times 10}{0.1}\times 0.1\]\[=6\times {{10}^{-5}}N\](Towards left) Force on wire Q due to wire R is \[{{F}_{R}}={{10}^{-7}}\times \frac{2\times 20\times 10}{0.02}\times 0.1=20\times {{10}^{-5}}N\] (Towards right) Hence \[{{F}_{net}}={{F}_{R}}-{{F}_{P}}=14\times {{10}^{-5}}N=1.4\times {{10}^{-4}}N\] (Towards right)
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