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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A thin wire of length 0.2 m and mass\[5\,\mu g\]remains suspended in air between the pieces of a magnet. If the wire is carrying a current of 0.5 A, the strength of the magnetic field is: (take\[=10m/{{s}^{2}}\])

    A)  50 gauss       

    B)         5 gauss

    C)  0.5 gauss            

    D)  0.05 gauss

    E)  0.005 gauss

    Correct Answer: D

    Solution :

    Since, wire remains suspended, force due to magnetic field is balanced by the weight of wire \[\therefore \]  \[F=ilB=mg\] \[\Rightarrow \]               \[B=\frac{mg}{il}\] \[\therefore \]   \[B=\frac{5\times {{10}^{-3}}\times {{10}^{-3}}\times 10\times {{10}^{-2}}}{0.5\times 0.2}=0.05\,gauss\]


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