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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    A particle of mass m and charge q is in an electric and magnetic field given by \[\vec{E}=2\hat{i}+3\hat{j};\vec{B}=4\hat{j}+6\hat{k}.\] The charged particle is shifted from the origin to the point \[P(x=1;y=1)\]along a straight path. The magnitude of the total work done is [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) (0.15)q                         

    B) (0.35)q

    C) (2.5)q               

    D)   5q

    Correct Answer: D

    Solution :

    Work done on moving charge in a magnetic field is zero.                           For electric field, \[W=q\Delta V=q(\vec{E}.d\vec{r})\] \[=q\left[ (2\hat{i}+3\hat{j}).(\hat{i}+\hat{j}) \right]=5q\]


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