2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    A long horizontal rigidly supported wire carries a current is = 96 A. Directly above it and parallel to it at a distance, another wire of 0.144 N weight per metre carrying a current ib = 24 A, in a direction opposite to that of ia. If the upper wire is to float in air due to magnetic repulsion, then its distance (in mm) from the lower wire is:

    A)  9 . 6                                      

    B)  4 . 8

    C)  3 . 2                                      

    D)  1 . 6

    Correct Answer: C

    Solution :

                     \[\frac{F}{l}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{i}_{a}}{{i}_{b}}}{r}\] \[\therefore \]  \[0.144=\frac{{{10}^{-7}}\times 2\times 96\times 24}{r}\] \[\therefore \]  \[r=\frac{{{10}^{-7}}\times 2\times 96\times 24}{0.144}\]                 \[=3.2\times {{10}^{-3}}m\]                 \[=3.2\,mm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner