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JCECE Engineering JCECE Engineering Solved Paper-2009

  • question_answer
    A straight wire of mass \[200\,\,g\] and length \[1.5\,\,m\] carries a current of\[2\,\,A\]. It is suspended in mid- air by a uniform horizontal magnetic field\[B\]. The magnitude of \[B\] (in tesla) is (assume\[g=9.8\,\,m{{s}^{-2}})\]

    A) \[2\]                                     

    B) \[1.5\]

    C) \[0.55\]                               

    D) \[0.65\]

    Correct Answer: D

    Solution :

    Magnetic force on straight wire\[F=Bil\sin \theta =Bil\sin {{90}^{o}}=Bil\] For equilibrium of wire in mid-air,                 \[F=mg\]                 \[Bil=mg\] \[\therefore \]  \[B=\frac{mg}{il}=\frac{200\times {{10}^{-3}}\times 9.8}{2\times 1.5}=0.65\,\,T\]


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