JEE Main & Advanced Physics Magnetism JEE PYQ-Magnetism

  • question_answer
    A magnetic needle of magnetic moment \[6.7\times {{10}^{-2}}\,A{{m}^{2}}\]and moment of inertia \[7.5\times {{10}^{-6}}\,kg\,{{m}^{2}}\]is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is: [JEE Main 2017]

    A) 6.98 s

    B) 8.76 s

    C) 6.65 s

    D) 8.89 s

    Correct Answer: C

    Solution :

    [c] \[T=2\pi \sqrt{\frac{I}{MB}}\]
                \[I=7.5\times {{10}^{-6}}kg-{{m}^{2}}\]
    \[M=6.7\times {{10}^{-2}}A{{m}^{2}}\]
    By substituting value in the formula
    \[T=.665\,\sec \]
    for 10 oscillation, time taken will be
    Time = 10 T = 6.65 sec
    Answer option 3


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