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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
      A magnet of total magnetic moment \[{{10}^{-}}^{2}\widehat{i}\,A-{{m}^{2}}\] is placed in a time varying magnetic field, \[B\widehat{i}\left( cos\text{ }\omega t \right)\] where \[B=1\] Tesla and\[\omega =0.125\text{ }rad/s\]. The work done for reversing the direction of the magnetic moment at \[t=1\] second, is- [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 0.014 J                         

    B) 0.028 J

    C) 0.01 J              

    D)   0.007 J

    Correct Answer: A

    Solution :

    \[M\,=\,{{10}^{-2}}\widehat{i}\] \[B=1\,cos\,\left( 0.125t \right)\widehat{i}\] \[at\text{ }t=1\text{ }sec\] \[B=1\,\,cos\left( 0.125 \right)\text{ }\widehat{i}\] \[=\,\,\cos \left( \frac{125}{1000} \right)\widehat{i}\,=\,\cos \,\left( \frac{1}{8} \right)\widehat{i}\] = 1 \[{{W}_{ext}}=\left[ -MB\text{ }cos\text{ }180{}^\circ -\left( -MB\text{ }cos\text{ }0{}^\circ  \right) \right]\] \[=\text{ }MB+MB=2MB\] \[=\text{ }2\times {{10}^{-}}^{2}\times cos\left( \frac{1}{8} \right)=0.019\]


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