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

  • question_answer
                An electron having momentum \[2.4\times {{10}^{-23}}kg\text{ }m{{s}^{-1}}\] enters a region of uniform magnetic field of 0.15 T. The field vector makes an angle of \[30{}^\circ \] with the initial velocity vector of the electron. The radius of the helical path of the electron in the field shall be:

    A) 2 mm                           

    B) 1 mm    

    C) \[\frac{\sqrt{3}}{2}mm\]                       

    D) 0.5 mm

    Correct Answer: D

    Solution :

    (d) 0.5 mm. The radius of the helical path of the electron in the uniform magnetic field is             \[r=\frac{m{{v}_{\bot }}}{eB}=\frac{mv\,\sin \theta }{eB}\] \[=\frac{(2.4\times {{10}^{-23}}kg\,m{{s}^{-1}})\times sin\,30{}^\circ }{(16\times {{10}^{-19}}C)\times (0.15\,T)}\]             \[=5\times {{10}^{-4}}m=0.5\times {{10}^{-3}}m=0.5\,mm.\]


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