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JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    An electric charge +q moves with velocity \[\vec{v}=3\hat{i}+4\hat{j}+\hat{k},\]in an electromagnetic field given by: \[\vec{E}=3\hat{i}+\hat{j}+2\hat{k}\]and \[\vec{B}=\hat{i}+\hat{j}+3\hat{k}.\]The y - component of the force experienced by + q is :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A)  11q

    B)  5q

    C)  3q

    D)  2q

    Correct Answer: A

    Solution :

                                    \[\vec{F}=q[\vec{E}+\vec{v}\times \vec{B}]\]                 \[=q\left[ 3\hat{i}+\hat{j}+\hat{k}+3\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    3 & 4 & 1  \\    1 & 1 & -3  \\ \end{matrix} \right| \right]\]                 \[=q[3\hat{i}+\hat{j}+2\hat{k}+\hat{i})-12-1)-\hat{j}(-9-1)+\hat{k}(3-4)]\]                 \[=q[3\hat{i}+\hat{j}+2\hat{k}-13\hat{i}10-\hat{j}-\hat{k}]\]                 \[=q[-10\hat{i}+11\hat{j}+\hat{k}]\]                 \[={{F}_{y}}=11q\,\hat{j}.\]


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