JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity\[\omega ,\]the maximum e.m.f. induced in the coil will be [JEE Main 16-4-2018]

    A)  \[nBA\omega \]                      

    B) \[\frac{3}{2}nBA\omega \]

    C) \[3nBA\omega \]                     

    D) \[\frac{1}{2}nBA\omega \]  

    Correct Answer: A

    Solution :

    Given, a coil of N turns an area A, rotated with angular speed\[\omega \] in a uniform magnetic field B. The flux linking to the coil, sin\[\phi =NBA\sin (\omega t)\] Therefore, Induced \[E=d(NBA\sin (\omega t)/dt)\] E= NBA Emax= NBA


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