Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    Rate of change of torque t with deflection \[\text{ }\!\!\theta\!\!\text{ }\] is maximum for a magnet suspended freely in a uniform magnetic field of induction B, when

    A) \[\text{ }\!\!\theta\!\!\text{ }\,\text{=}\,{{\text{0}}^{\text{o}}}\]                           

    B)  \[\text{ }\!\!\theta\!\!\text{ }\,\text{=}\,{{45}^{0}}\]

    C)  \[\text{ }\!\!\theta\!\!\text{ }\,\text{=}\,{{60}^{0}}\]                                   

    D)  \[\text{ }\!\!\theta\!\!\text{ }\,\text{=}\,{{90}^{0}}\]

    Correct Answer: A

    Solution :

    Torque acting on a bar magnet suspended - magnetic field of induction B is \[\tau =MB\,\sin \theta \] On differentiation we get \[\frac{d\tau }{d\theta }=MB\,\cos \theta \] \[\therefore \] Rate of change of torque \[\Rightarrow \]               \[\frac{dz}{d\theta }=MB\,\cos \theta \] Obviously, for maximum value of \[\frac{d\tau }{d\theta }\]                 \[\cos \,\theta =1\] or            \[\theta ={{0}^{o}}\]


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