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12th Class Physics Magnetism Question Bank Case Based (MCQs) - Moving Charges and Magnetism

  • question_answer
    An electron having momentum \[2.4\times {{10}^{-23}}kg\text{ }m/s\]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) 1mm   

    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 \[r=\frac{mv\bot }{eB}=\frac{mv\,\sin \,\theta }{eB}=\frac{(2.4\times {{10}^{-23}}kg\,m/s)\times sin\,30{}^\circ }{(16\times {{10}^{-19}}C)\times 0.15\,T}\]\[=5\times {{10}^{-4}}m=0.5\times {{10}^{-3}}m=0.5\,mm\]


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