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JCECE Medical JCECE Medical 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 testa) 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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