question_answer

Three long, straight parallel wires, carrying current, are arranged as shown in figure. The force experienced by a 25 cm length of wire C is.

A) \[{{10}^{-3}}N\]         

B) \[2.5\times {{10}^{-3}}N\]

C) \[zero\]

D) \[1.5\times {{10}^{-3}}N\]

Correct Answer: C

Solution :

[c] The magnetic field due to wire D at wire C is \[{{B}_{D}}=\left( \frac{{{\mu }_{0}}}{4\pi } \right)\frac{2I}{r}=\frac{{{10}^{-7}}\times 2\times 30}{0.03}=2\times {{10}^{-4}}T\]which is directed into the page. The magnetic field due to wire G at C is \[=\frac{{{10}^{-7}}\times 2\times 20}{0.02}=2\times {{10}^{-4}}T\]which is directed out of the page. Therefore, the field at the position of the wire C is \[B={{B}_{D}}-{{B}_{G}}=2\times {{10}^{-4}}-2\times {{10}^{-4}}=\]zero. \[\therefore \]The force on 25 cm of wire C \[F=BI\text{ }L\sin \theta =\] zero.