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Punjab Medical Punjab - MET Solved Paper-2010

  • question_answer
    A wire in the form of a circular loop of radius\[r\]lies with its plane normal to a magnetic field\[B.\] If the wire is pulled to take a square shape in the same plane in time\[t\], the emf induced in the loop is given by

    A) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{10} \right)\]  

    B) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{8} \right)\]

    C) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{6} \right)\]     

    D) \[\frac{\pi B{{r}^{2}}}{t}\left( 1-\frac{\pi }{4} \right)\]

    Correct Answer: D

    Solution :

    Induced \[emf(e)\] \[=\frac{magnetic\,\,field\times change\,\,in\,\,area}{time}=\frac{B\Delta A}{t}\] Since, the circumference of the circular loop\[=2\pi r\], The side of the square loop\[=\frac{2\pi r}{4}=\frac{\pi r}{2}\] Therefore,                 \[\Delta A=\pi {{r}^{2}}-{{\left( \frac{\pi \,\,r}{2} \right)}^{2}}=\pi {{r}^{2}}\left( 1-\frac{\pi }{4} \right)\] \[\therefore \]  \[e=\frac{B(\pi {{r}^{2}})}{t}\left( 1-\frac{\pi }{4} \right)\]


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