JEE Main & Advanced JEE Main Online Paper (Held On 23 April 2013)

  • question_answer
    A particle of charge \[16\times {{10}^{-16}}\] C moving with velocity  \[10{{\operatorname{ms}}^{-1}}\] along \[x-\] axis enters enters a region where magnetic field of induction \[\overset{\to }{\mathop{\operatorname{B}}}\,\] is along the \[y\]-axis and an electric field of magnitude \[{{10}^{4}}{{\operatorname{V}}^{-1}}\] is along the negative z-axis. If the charged particle continues moving along \[x-axis\], the magnitude of magnitude  of \[\overset{\to }{\mathop{\operatorname{B}}}\,\]is :     [JEE Main Online Paper ( Held On 23  April 2013 )

    A)                  \[16\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]              

    B)                                         \[2\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]

    C)                                         \[1\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\] 

    D)                                         \[4\times {{10}^{3}}\operatorname{Wb}{{\operatorname{m}}^{-2}}\]

    Correct Answer: C

    Solution :

                      Since particle is moving un deflected So,          \[{{q}_{E}}={{q}_{v}}B\] \[\Rightarrow \]\[B=\frac{E}{V}=\frac{{{10}^{4}}}{10}={{10}^{3}}wb/{{m}^{2}}\]


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