10th Class Science Electricity and Circuits Question Bank Electricity

  • question_answer
    A proton of a mass m and charge +e is moving in magnetic field with energy 1 MeV. should be the energy of \[\alpha \]-particle (mass \[4m\] and charge\[+2e\]) so that it can revolve in the path of same radius?

    A)  \[1\,\,MeV\]      

    B)         \[4\,\,MeV\]

    C)  \[2\,\,MeV\]    

    D)         \[0.5\,\,MeV\]

    Correct Answer: A

    Solution :

                      \[r=\frac{mv}{qB}\] \[\Rightarrow \]               \[{{r}^{2}}=\frac{{{m}^{2}}{{v}^{2}}}{{{q}^{2}}{{B}^{2}}}\] In terms of energy, above equation can be written as                 \[{{r}^{2}}=\frac{2m\times (1/2m{{v}^{2}})}{{{q}^{2}}{{B}^{2}}}=\frac{2mE}{{{q}^{2}}{{B}^{2}}}\] Here \[r\] and \[B\]are same, hence         \[\frac{{{m}_{1}}{{E}_{1}}}{{{q}_{1}}^{2}}=\frac{{{m}_{2}}{{E}_{2}}}{{{q}_{2}}^{2}}\] \[\Rightarrow \]   \[\frac{m\times 1}{{{e}^{2}}}=\frac{3m\times {{E}_{2}}}{4{{e}^{2}}}\] \[\therefore \]        \[{{E}_{2}}=1MeV\]


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