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NEET Sample Paper NEET Sample Test Paper-44

  • question_answer
    A bar magnet is placed in the position of stable equilibrium in a uniform magnetic field of induction B. If is rotated through an angle\[180{}^\circ \], then the work is:

    A) MB                  

    B) 2MB  

    C) \[\frac{MB}{2}\]                       

    D) Zero

    Correct Answer: B

    Solution :

    Work done in rotating the dipole \[{{\operatorname{W}}_{{{\theta }_{1}}}}- {{W}_{{{\theta }_{2}}}}= MB \left[ cos{{\theta }_{1}}-cos{{\theta }_{2}} \right]\] Initially dipole is in stable position \[9=0{}^\circ \] \[{{\operatorname{U}}_{1}}=-MB\,\,cos0{}^\circ =-MB\] \[{{\operatorname{U}}_{2}}=-MB\,cos\,\,180{}^\circ  =MB\] \[{{\operatorname{W}}_{ext}} \left( We rotate it \right) = {{U}_{2}}-{{U}_{1}} = 2MB\]


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