2. Questions Bank
  3. JEE Main & Advanced
  4. Physics
  5. Electro Magnetic Induction

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

  • question_answer
    Shown in the figure is a circular loop of radius r and resistance R. A variable magnetic field of induction \[B={{B}_{0}}{{e}^{-t}}\]is established inside the coil. If the key (K) is closed, the electrical power developed right after closing the switch is equal to.                  

    A) \[\frac{B_{0}^{2}\pi {{r}^{2}}}{R}\]

    B) \[\frac{{{B}_{0}}10{{r}^{3}}}{R}\]

    C) \[\frac{B_{0}^{2}{{\pi }^{2}}{{r}^{4}}R}{5}\]

    D) \[\frac{B_{0}^{2}{{\pi }^{2}}{{r}^{4}}}{R}\]

    Correct Answer: D

    Solution :

    [d] \[e=-\frac{d\phi }{dt}--A\frac{dB}{dt}=-\pi {{r}^{2}}\frac{d}{dt}({{B}_{0}}{{e}^{-t}})=\pi {{r}^{2}}{{B}_{0}}{{e}^{-t}}\] At\[t=0\],\[e=\pi {{r}^{2}}{{B}_{0}}\], \[P=\frac{{{e}^{2}}}{R}=\frac{{{(\pi {{r}^{2}}{{B}_{0}})}^{2}}}{R}=\frac{{{\pi }^{2}}{{r}^{4}}{{B}_{0}}^{2}}{R}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner