JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    A planar loop of wire rotates in a uniform magnetic field. Initially, at t = 0, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10 s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at [JEE MAIN Held on 07-01-2020 Evening]

    A) 2.5 s and 7.5 s

    B) 5.0 s and 7.5 s

    C) 2.5 s and 5.0 s

    D) 5.0 s and 10.0 s

    Correct Answer: C

    Solution :

    [c] \[~\phi (t)=\,AB\text{ }cos\text{ }\omega t\] \[E=\frac{-\,d\phi }{dt}\,=AB\,\omega \,\sin \,\omega \,t=AB\,\omega \,\sin \,\left( \frac{2\pi }{T}\cdot t \right)\] Induced emf, \[|\varepsilon |\]is maximum when \[\frac{2\pi \,t}{T}=\frac{\pi }{2},\frac{3\pi }{2}\] \[\Rightarrow \,\,t=\frac{T}{4}\] or \[\frac{3T}{4}\] i.e. 2.5 s or 7.5 s. For induced emf to be minimum i.e zero \[\frac{2\pi \,t}{T}=n\pi \Rightarrow \,t=n\frac{T}{2}\], \[\Rightarrow \]     Induced emf is zero at t = 5 s, 10 s

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