NEET Sample Paper NEET Sample Test Paper-19

  • question_answer In a uniform magnetic field of induction B, a wire in the form of a semi-circle of radius r rotates about the diameter of the circle with angular frequency\[\omega .\]The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R, the mean power generated per period of rotation is:

    A) \[\frac{{{(B\pi r\omega )}^{2}}}{2R}\]                  

    B) \[\frac{B\pi {{r}^{2}}\omega }{2R}\]

    C) \[\frac{B\pi {{r}^{2}}\omega }{8R}\]                    

    D) \[\frac{{{(B\pi r{{\omega }^{2}})}^{2}}}{8R}\]   

    Correct Answer: B

    Solution :

    \[\Phi =B\frac{\pi {{r}^{2}}}{2}\cos \omega t\] \[\therefore \]\[{{e}_{ind}}=-\frac{d\Phi }{dt}=\frac{1}{2}B\pi {{r}^{2}}\omega \sin \omega t\] \[\therefore \]\[P=\frac{e_{ind}^{2}}{2}=\frac{{{B}^{2}}{{\pi }^{2}}{{\omega }^{2}}{{\sin }^{2}}\omega t}{4R}\] Now \[<{{\sin }^{2}}\omega t>=\frac{1}{2}\]\[\therefore \]\[<P>=\frac{{{(B\pi {{r}^{2}}\omega )}^{2}}}{8R}\] Hence, the correction option is [b].

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