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12th Class Physics Magnetism Question Bank MCQ - Moving Charges and Magnetism

  • question_answer
    An electron of energy 1800 eV describes a circular path in magnetic field of flux density 0.4 T. The radius of path is \[(q=1.6\times {{10}^{-19}}C,\,{{m}_{e}}=9.1\times {{10}^{-31}}kg)\]

    A) \[2.58\times {{10}^{-4}}m\]        

    B) \[3.58\times {{10}^{-4}}m\]

    C) \[2.58\times {{10}^{-3}}m\]                 

    D) \[3.58\times {{10}^{-3}}m\]

    Correct Answer: B

    Solution :

    (b) \[3.58\times {{10}^{-4}}m\]             \[E=\frac{1}{2}m{{v}^{2}}\Rightarrow v=\sqrt{\frac{2E}{m}}(\because {{m}_{e}}=m)\]             \[r=\frac{mv}{Be}=\frac{m}{Be}\sqrt{\frac{2E}{m}}=\frac{\sqrt{2mE}}{Be}\]             \[r=\frac{\sqrt{2\times 1800\times 1.6\times {{10}^{-19}}\times 9.1\times {{10}^{-31}}}}{1.6\times {{10}^{-19}}\times 0.4}\]             \[=3.58\times {{10}^{-4}}m\]


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