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 |
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