JCECE Medical JCECE Medical Solved Paper-2011

  • question_answer
    A current of 1 A flows in a circular area of wire which subtends an angle of \[\left( \frac{3\pi }{2} \right)\text{rad}\]at its centre, whose radius is R. The magnetic induction B at the centre is

    A) \[\frac{{{\mu }_{0}}I}{R}\]

    B)  \[\frac{{{\mu }_{0}}I}{2R}\]

    C)  \[\frac{2{{\mu }_{0}}I}{R}\]

    D)  \[\frac{3{{\mu }_{0}}I}{8R}\]

    Correct Answer: D

    Solution :

     For a loop, magnetic induction at center \[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\pi I}{R}\] When loop subtends angle \[\theta \] at center, then \[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{\theta I}{R}\] In the given problem \[\theta =\frac{3\pi }{2}\] \[\therefore \] \[B=\frac{{{\mu }_{0}}}{4\pi }\times \frac{3\pi }{2}\times \frac{I}{R}\] \[=\frac{3{{\mu }_{0}}I}{8R}\]


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