JEE Main & Advanced Physics Magnetism Question Bank Magnetic Equipments

  • question_answer A magnet oscillating in a horizontal plane has a time period of 2 second at a place where the angle of dip is 30o and 3 seconds at another place where the angle of dip is 60o. The ratio of resultant magnetic fields at the two places is [Pb. PET 2001]

    A)            \[\frac{4\sqrt{3}}{7}\]            

    B)            \[\frac{4}{9\sqrt{3}}\]

    C)            \[\frac{9}{4\sqrt{3}}\]            

    D)            \[\frac{9}{\sqrt{3}}\]

    Correct Answer: C

    Solution :

               \[T\propto \frac{1}{\sqrt{{{B}_{H}}}}=\frac{1}{\sqrt{B\cos \varphi }}\]\[\Rightarrow \frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{B}_{2}}\cos {{\varphi }_{2}}}{{{B}_{2}}\cos {{\varphi }_{1}}}}\] \[\Rightarrow \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{T_{2}^{2}}{T_{1}^{2}}\times \frac{\cos {{\varphi }_{2}}}{\cos {{\varphi }_{1}}}={{\left( \frac{3}{2} \right)}^{2}}\times \frac{\cos 60}{\cos 30}\]\[\Rightarrow \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{9}{4\sqrt{3}}\]

adversite


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