question_answer

Three straight parallel current carrying conductors are shown in the figure. The force experienced by the middle conductor of length 25 cm is:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

A) \[3\times {{10}^{-4}}N\] toward right

B) \[6\times {{10}^{-4}}N\] toward right

C)  \[9\times {{10}^{-4}}N\]toward right

D) Zero

Correct Answer: A

Solution :

                \[\begin{matrix}    {{I}_{1}}=30A & I=10A & {{I}_{2}}=20A  \\ \end{matrix}\] Also given; length of wire Q = 25 cm = 0.25 m Force on wire Q due to wire R \[{{F}_{QR}}={{10}^{-7}}\times \frac{2\times 20\times 10}{0.05}\times 0.25\] \[=20\times {{10}^{-5}}N\](Towards left) Force on wire Q due to wire P \[{{F}_{QP}}={{10}^{-7}}\times \frac{2\times 30\times 10}{0.03}\times 0.25\] \[=3\times {{10}^{-4}}N\](Towards right) Hence, \[{{F}_{net}}={{F}_{QP}}-{{F}_{QR}}\] \[=50\times {{10}^{-5}}N-20\times {{20}^{-5}}N\] \[=3\times {{10}^{-4}}N\]towards right