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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    A, B and C are parallel conductors of equal lengths carrying currents\[I,I\]and\[2I\]respectively. Distance between A and B is X. Distance between B and C is also\[x\].\[{{F}_{1}}\]is the force exerted by B on A.\[{{F}_{2}}\]is the force exerted by C on A. Choose the correct answer:

    A)  \[{{F}_{1}}=2{{F}_{2}}\]                               

    B)  \[{{F}_{2}}=2{{F}_{1}}\]                               

    C)  \[{{F}_{1}}={{F}_{2}}\]                 

    D)         \[{{F}_{1}}=-{{F}_{2}}\]

    E)  \[{{F}_{2}}=4{{F}_{1}}\]

    Correct Answer: D

    Solution :

    \[{{F}_{BA}}=\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi x}\](attractive force) and\[{{F}_{CA}}=\frac{{{\mu }_{0}}(I)(2I)}{2\pi (2x)}=\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi x}\]repulsive force) Hence, \[{{F}_{BA}}=-{{F}_{CA}}\Rightarrow {{F}_{1}}=-{{F}_{2}}\]


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