A) \[{{10}^{-4}}N\]
B) \[3.6\times {{10}^{-4}}N\]
C) \[1.8\times {{10}^{-4}}N\]
D) \[5\times {{10}^{-4}}N\]
Correct Answer: D
Solution :
Force on SR and PQ, are equal but opposite so their net will be zero. Force between two parallel conductors carrying currents \[{{I}_{1}}\] and \[{{I}_{2}}\]\[F=\frac{{{\mu }_{0}}}{2\pi }\frac{{{I}_{1}}{{I}_{2}}l}{r}\]where r = distance between two parallel conductors \[{{F}_{PS}}=\frac{{{10}^{-7}}\times 2\times 20\times 20\times 15\times {{10}^{-2}}}{2\times {{10}^{-2}}}\] \[=6\times {{10}^{-4}}N\] \[{{F}_{QR}}=\frac{{{10}^{-7}}\times 2\times 20\times 20\times 15\times {{10}^{-2}}}{2\times {{10}^{-2}}}\] \[=1\times {{10}^{-4}}N\] \[{{F}_{net}}={{F}_{PS}}-{{F}_{QR}}\] \[=6\times {{10}^{-4}}-1\times {{10}^{-4}}=5\times {{10}^{-4}}N\]
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