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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    An electron of mass m and charge q is travelling with a speed v along a circular path of radius r at right angles to a uniform of magnetic field B. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of:

    A)  \[\frac{r}{4}\]                                  

    B)  \[\frac{r}{2}\]

    C)  \[2r\]                                   

    D)  \[4r\]

    Correct Answer: D

    Solution :

                     In a perpendicular magnetic field, Magnetic force = centripetal force i.e.,        \[Bqv=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \]               \[r=\frac{mv}{Br}\]   \[\Rightarrow \]\[r\propto {{v}^{2}}\] \[\therefore \]  \[\frac{{{r}_{2}}}{r}=\frac{v_{2}^{2}}{v_{1}^{2}}={{\left( \frac{2v}{v} \right)}^{2}}=4\] \[\Rightarrow \]               \[{{r}_{2}}=4r\]


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