JEE Main & Advanced Physics Magnetism Question Bank Magnetic Equipments

  • question_answer A magnetic needle is made to vibrate in uniform field H, then its time period is T. If it vibrates in the field of intensity 4H, its time period will be [MP Board 1988; MP PMT 1992; MH CET (Med.) 1999]

    A)            \[2T\]                                      

    B)            \[T/2\]

    C)            \[2/T\]                                    

    D)            \[T\]

    Correct Answer: B

    Solution :

               \[T=2\pi \sqrt{\frac{I}{M{{B}_{H}}}}\Rightarrow \frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{({{B}_{H}})}_{2}}}{{{({{B}_{H}})}_{1}}}}\] \[\Rightarrow {{T}_{2}}=T\sqrt{\frac{{{(BH)}_{1}}}{{{(BH)}_{2}}}}=\frac{T}{2}\ \ \ \left( \because {{({{B}_{H}})}_{2}}=4{{({{B}_{H}})}_{1}} \right)\]


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