KVPY Sample Paper KVPY Stream-SX Model Paper-5

  • question_answer
    Two mutually perpendicular infinitely long lines of charge having charge per unit length as \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] are located in air at a distance "a" from each other. The force of interaction between them is -

    A) \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{4{{\in }_{0}}\pi a}\]                      

    B) \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{\pi a{{\in }_{0}}}\]

    C) \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2{{\in }_{0}}\pi a}\]          

    D) \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{3\pi a{{\in }_{0}}}\]

    Correct Answer: C

    Solution :

    Consider an element of thickness dy at distance y
    Charge on element \[=dq={{\lambda }_{1}}dy\]
    E at the position of element = E
    Force on element \[=dF=dqE\]
    \[E=\frac{{{\lambda }_{2}}}{2\pi {{\in }_{0}}r}=\frac{{{\lambda }_{2}}}{2\pi {{\in }_{0}}\sqrt{{{a}^{2}}+{{y}^{2}}}}\]           \[dF=\frac{{{\lambda }_{2}}\,\,\times \,\,{{\lambda }_{1}}}{2\pi {{\in }_{0}}\sqrt{{{a}^{2}}+{{y}^{2}}}}.dy\]
    Taking component of dF in x and y direction By symmetry y component get cancel out \[d{{F}_{x}}=dF\cos \theta \]
    Where \[\cos \theta =\frac{a}{\sqrt{{{a}^{2}}+{{y}^{2}}}}\] and \[{{F}_{x}}=\int\limits_{-\infty }^{+\infty }{d{{F}_{x}}}\]
    \[{{F}_{net}}\] is in direction of a-axis \[\therefore {{F}_{net}}={{F}_{x}}\]
    \[{{F}_{x}}=\int\limits_{-\infty }^{+\infty }{\frac{{{\lambda }_{1}}{{\lambda }_{2}}a}{2\pi {{\in }_{0}}({{a}^{2}}+{{y}^{2}})}dy}\]\[\Rightarrow \frac{{{\lambda }_{1}}{{\lambda }_{2}}}{2\pi {{\in }_{0}}a}\]

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