JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Mock Test - Electromagnetic Induction

  • question_answer
    A conducting ring of radius r is rolling without slipping with a constant angular velocity \[\omega \](figure). If the magnetic field strength is B and is directed into the page then the emf induced across PQ is

    A) \[B\omega {{r}^{2}}\] 

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

    C) \[4B\omega {{r}^{2}}\]

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

    Correct Answer: A

    Solution :

    [a] \[\left| E \right|=\]Magnitude of induced emf \[=\frac{B{{\ell }^{2}}}{2}\omega ,\ell =\sqrt{2}r\] \[E=\frac{B(2{{r}^{2}})\omega }{2}=B\omega {{r}^{2}}\]


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