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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A very small circular loop of radius a is initially (at\[t=0\]) coplanar and concentric with a much larger fixed circular loop of radius b. A constant current \[I\] flows in the larger loop. The smaller loop is rotated with a constant angular speed\[\omega \]about the common diameter. The emf induced in the smaller loop as a function of time t is

    A) \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega \cos \,(\omega t)\]

    B)  \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega \sin \,({{\omega }^{2}}{{t}^{2}})\]

    C)  \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega \sin \,(\omega t)\]

    D)  \[\frac{\pi {{a}^{2}}{{\mu }_{0}}l}{2b}\omega {{\sin }^{2}}\,(\omega t)\]

    Correct Answer: C

    Solution :

     We know that \[\tau =NBA\omega \sin \omega t\] Where \[N=\]number of loops \[=1\] \[B=\frac{{{\mu }_{0}}I}{2b}\]newton/amp-m \[A=\pi {{a}^{2}}metr{{e}^{2}}\] \[\therefore \] \[\tau =\frac{{{\mu }_{0}}I}{2b}(\pi {{a}^{2}})\omega sin\omega t\] \[=\frac{\pi {{a}^{2}}{{\mu }_{0}}I}{2b}.\omega \sin \omega t\]


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