question_answer

Two long parallel wires carry currents of equal magnitude but   in   opposite  directions.    These wires are suspended from rod PQ by four chords of same length L as shown in the figure. The mass per unit length of the wire is \[\lambda \]. Determine the value of\[\theta \]assuming it to be small.

A)  \[I\sqrt{\frac{{{\mu }_{0}}}{4\pi \lambda gL}}\]

B)  \[I\sqrt{\frac{4\pi \lambda gL}{{{\mu }_{0}}}}\]

C)  \[I\sqrt{\frac{{{\mu }_{0}}}{2\pi \lambda g}}\]  

D)  \[2\pi \lambda \sqrt{\frac{{{\mu }_{0}}}{g}}I\]

Correct Answer: A

Solution :

[a] : The force per unit length between current carrying parallel wires is    \[\frac{dF}{dL}=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}}{2\pi d}\] If two wires carry current in opposite directions the magnetic force is repulsive, due to which the parallel wires have moved out so that equilibrium is reached. Figure shows free body diagram of each wire. In equilibrium, \[\Sigma {{F}_{y}}=0,2T\cos \theta =(\lambda {{L}_{0}})g\]                  ...(i) \[\Sigma {{F}_{z}}=0,2T\sin \theta ={{F}_{B}}\]                                   ...(ii) Now dividing eqn. (ii) by eqn. (i) we get \[\tan \theta =\frac{{{F}_{B}}}{{{L}_{0}}\lambda g}\]where, the magnetic force, \[{{F}_{B}}=\left( \frac{dF}{dL} \right)\times {{L}_{0}}=\frac{{{\mu }_{0}}{{I}^{2}}}{4\pi \sin \theta }\frac{{{L}_{0}}}{L}\] For small \[\theta ,\tan \theta \simeq \sin \theta \simeq \theta \] \[\therefore \]\[\theta =I\sqrt{\frac{{{\mu }_{0}}}{4\pi \lambda gL}}\]