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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    A straight wire of mass 200 g and length \[\text{1}.\text{5 m}\] 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

    A) \[0.\text{65}\]                                  

    B) \[0.\text{55}\]   

    C) \[0.\text{75}\]                                  

    D) \[0.\text{45}\]

    Correct Answer: A

    Solution :

     Given, mass of the wire = 200 g = 02 kg Length of the wire = 15 m Current \[i=2A\] Magnitic field B =?                    The force acting on the current carrying wire in uniform magnetic field \[F=Bil\sin \theta \] \[F=Bil\]                               \[(\because \theta ={{90}^{\circ }})\] Weight of the wire w = mg = 0.2 \[\times \] 9.8N In the position of suspension \[Bil=mg\] \[B=\frac{mg}{il}=\frac{0.2\times 9.8}{2\times 1.5}=0.65T\]


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