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CMC Medical CMC-Medical Ludhiana 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 6, The magnitude of B (in tesia) 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 \[F=Bil\text{ sin }\theta =Bil\text{ sin }90{}^\circ =Bil\] 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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