VMMC VMMC Medical Solved Paper-2003

  • question_answer
    Two thin long parallel wires separated by a distance b are carrying a current i A each. The magnitude of the force per unit length will be:

    A) \[\frac{{{\mu }_{2}}{{i}^{2}}}{{{b}^{2}}}\]             

    B)                        \[\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi b}\]                   

    C)         \[\frac{{{\mu }_{0}}i}{2\pi {{b}^{2}}}\]                  

    D)        \[\frac{{{\mu }_{0}}i}{2\pi b}\]

    Correct Answer: B

    Solution :

    Magnetic induction due to long wire at a  distance b is \[B=\frac{{{\mu }_{0}}{{i}_{1}}}{2\pi b}\] and the force experienced by a conductor per unit length will be \[F=B{{i}_{2}}\] \[=\frac{{{\mu }_{0}}{{i}_{1}}{{i}_{2}}}{2\pi b}\] \[=\frac{{{\mu }_{0}}{{i}^{2}}}{2\pi b}\]                 \[[\because \,{{i}_{1}}={{i}_{2}}=i]\]


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