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MGIMS WARDHA MGIMS WARDHA Solved Paper-2006

  • question_answer
    A uniform magnetic field is at right angle to the direction of motion of proton. As a result, the proton describes a circular path of radius 2.5 cm. If the speed of proton is doubled then      the radius of the circular path will be :

    A) 0.5 cm                                  

    B) 2.5 cm

    C) 5.0 cm                                  

    D) 7.5 cm

    Correct Answer: C

    Solution :

    \[\frac{m{{v}^{2}}}{r}=evB\] \[r=\frac{mv}{eB}\] \[\frac{{{r}_{2}}}{{{r}_{1}}}=\frac{{{v}_{2}}}{{{v}_{1}}}\] \[\frac{{{r}_{2}}}{2.5}=\frac{2v}{v}\] \[\Rightarrow \]               \[{{r}_{2}}=5\,cm\]


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