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JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    Two long conductors, separated by a distance d. carry currents\[{{I}_{1}}\]and\[{{I}_{2}}\]in the same direction. They exert a force F on each other. Now, the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d. The new value of the force between them is

    A) \[-2F\]

    B)                                        \[F/3\] 

    C)        \[-2F/3\]          

    D)        \[-F/3\]

    Correct Answer: C

    Solution :

    Force acting between two current carrying m conductors     \[F=\frac{{{\mu }_{0}}}{2\pi }\frac{{{l}_{1}}{{l}_{2}}}{d}I\]                                             ?..(i) where, \[d=\]distance between the conductors \[l=\]length of each conductor     Again,   \[F'=\frac{{{\mu }_{0}}}{2\pi }\frac{(-2{{l}_{1}})({{l}_{2}})}{(3d)}l\]                 \[F'=\frac{{{\mu }_{0}}}{2\pi }\frac{(-2{{l}_{1}})({{l}_{2}})}{(3d)}l\]                            ?.(ii) Thus, from Eqs. (i) and (ii), we get  \[\frac{F'}{F}=-\frac{2}{3}\] \[\Rightarrow \]               \[F'=-\frac{2}{3}F\]


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