BVP Medical BVP Medical Solved Paper-2001

  • question_answer
    The magnetic field at P on the axis of a solenoid having 100 turns/metre and carrying a current of 5 A is :

    A)  \[500\sqrt{2}{{\mu }_{0}}\]                       

    B) \[300\sqrt{2}{{\mu }_{0}}\]

    C) \[200\sqrt{2}{{\mu }_{0}}\]                        

    D)  \[250\sqrt{2}{{\mu }_{0}}\]

    Correct Answer: D

    Solution :

                                                Here, \[n=100\]turns/m, \[i=5\]amp, \[\theta ={{45}^{o}}\] Using the relation, \[B=\frac{{{\mu }_{0}}ni}{2}\]           \[(\cos 45+\cos 45)\] or \[B=\frac{{{\mu }_{0}}\times 5\times 100}{2}\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right)\] or \[B=250\sqrt{2}{{\mu }_{0}}\]


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