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AFMC AFMC Solved Paper-2008

  • 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 tesia) is (assume g = 9.8ms-2)

    A) 2                                             

    B) 1.5         

    C) 0.55                                       

    D) 0.65

    Correct Answer: D

    Solution :

    Magnetic   force   on   straight   wire \[F=Bil\text{ }sin\text{ }\theta =\text{ }Bil\text{ }sin\text{ }{{90}^{o}}=\text{ }Bil\] 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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