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 ONLINE 11-04-2014]

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 :

[a] \[\begin{matrix}    {{I}_{1}}=30A & I=10A & {{I}_{2}}=20A  \\ \end{matrix}\]

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