VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    Find the value of magnetic field between plates of capacitor at a distance 1 m from centre, where electric field varies by \[{{10}^{10}}V/m\]per second.

    A)  \[5.56\times {{10}^{-8}}T\]

    B)  \[5.56\times {{10}^{-3}}T\]

    C)  \[\text{5}\text{.56 }\!\!\mu\!\!\text{ T }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{ }\]

    D)  \[5.55\text{ }T\]  

    Correct Answer: A

    Solution :

    Magnetic field, \[B=\frac{{{\mu }_{0}}{{\varepsilon }_{0}}r}{2}\frac{dE}{dt}=\frac{1}{9\times {{10}^{16}}\times 2}\times {{10}^{10}}\] \[=5.56\times {{10}^{-8}}T\]


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