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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2006

  • 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\[g\text{ }=\text{ }9.8\text{ }m{{s}^{-2}}\])

    A)  2

    B)  1.5

    C)  0.55

    D)  0.65

    Correct Answer: D

    Solution :

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


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