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Haryana PMT Haryana PMT Solved Paper-2000

  • question_answer
    Currents of 10 A and 2 A are passed through two parallel wires A and B respectively in opposite directions. If the wire is uniformly long and the length of wire B is 2m. The force on conductor B at separation 10 cm from A will be :

    A)  \[40\times {{10}^{-7}}N\]                           

    B)  8\[\pi \] \[\times {{10}^{-7}}N\]

    C)  \[8\times {{10}^{-5}}N\]                             

    D)  \[4\times {{10}^{-5}}N\]

    Correct Answer: C

    Solution :

                    Here,  \[{{i}_{1}}=10A,\,\,\,{{i}_{2}}=2A\] \[r=10cm=0.1m,\,\,l=2m\] Using the relation                 \[F=\frac{{{\mu }_{0}}{{i}_{1}}{{i}_{2}}l}{2\pi R}\] or            \[F=\frac{4\pi \times {{10}^{-7}}\times 10\times 2\times 2}{2\pi \times 0.10}\] So,          \[F=8\times {{10}^{-5}}N\]


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