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 \[\theta ,\tan \theta \simeq \sin \theta \simeq \theta \] \[\therefore \]\[\theta =I\sqrt{\frac{{{\mu }_{0}}}{4\pi \lambda gL}}\]You need to login to perform this action.
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