JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron \[=1.6\,\times \,{{10}^{-19}}C\]) [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) \[1.6\times {{10}^{-19}}kg\]                 

    B) \[1.6\times {{10}^{-27}}\text{ }kg\]

    C) \[9.1\times {{10}^{-}}^{31}kg\]   

    D)               \[2.0\times {{10}^{-}}^{24}kg\]

    Correct Answer: D

    Solution :

    \[R\,\,=\,\,\frac{mV}{qB}\] \[\left( qVB \right)\,\,=\,\,qE\] \[V\,\,=\,\,\frac{E}{B}\] \[R\,\,=\,\frac{mE}{q{{B}^{2}}}\] \[m\,\,=\,\frac{qR{{B}^{2}}}{E}\,\,=\,\,\frac{1.6\,\times {{10}^{-19}}\,\times 0.5\times {{10}^{-2}}\,\times \,0.25}{100}\] \[=\text{ }2\times {{10}^{-}}^{24}kg\]


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