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Uttarakhand PMT Uttarakhand PMT Solved Paper-2010

  • question_answer
    An electron moving with velocity\[2\times {{10}^{7}}m/s\] describes a circle in a magnetic field of strength\[2\times {{10}^{-2}}T\].    If   \[\left( \frac{e}{m} \right)\] of   electron   is\[1.76\times {{10}^{11}}C/kg\]then the diameter of the circle is nearly

    A)  1.1 cm          

    B)  1.1 mm

    C)  1.1m         

    D)  11cm

    Correct Answer: A

    Solution :

     \[r=\frac{mv}{eB}=\frac{v}{\left( \frac{e}{m} \right)B}\] \[\therefore \] \[r=\frac{2\times {{10}^{7}}}{1.76\times {{10}^{11}}\times 2\times {{10}^{-2}}}\] \[\Rightarrow \] \[r=0.0056\,m\] The diameter of the circle \[D\approx 2r-0.011m\] \[D=1.1cm\]


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