JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    Currents of a 10 ampere and 2 ampere are passed through two parallel thin wires A and B respectively in opposite directions. Wire A is infinitely long and the length of the wire B is 2 m. The force acting on the conductor B, which is situated at 10 cm distance from A will be   JEE Main Online Paper (Held On 26-May-2012)  

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

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

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

    D)                        \[4\pi \times {{10}^{-7}}N\]

    Correct Answer: A

    Solution :

                    Force acting on conductor B due to conductor A is given by relation\[F=\frac{{{\mu }_{0}}{{I}_{1}}{{I}_{2}}l}{2\pi r}\]\[l\]-length of conductor Br-distance between two conductors \[\therefore \]\[F=\frac{4\pi \times {{10}^{-7}}\times 10\times 2\times 2}{2\times \pi \times 0.1}=8\times {{10}^{-5}}N\]


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