Solved papers for JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

done JEE Main Paper (Held On 10 April 2016) Total Questions - 2

• question_answer1) A conducting metal circular -wire loop of radius r is placed perpendicular to a magnetic field which varies with time as \[B={{B}_{0e}}^{-t/\tau }\], where B0 and \[\tau \]are constants, at t = 0 . If the resistance of the loop is R then the heat generated in the loop after a long time \[(t-\infty )\] is:   JEE Main Online Paper (Held On 10 April 2016)
  A)
  \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}R}{\tau }\]                     
  done clear
  B)
  \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2\tau R}\]
  done clear
  C)
  \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{\tau R}\]                     
  done clear
  D)
  \[\frac{{{\pi }^{2}}{{r}^{4}}B_{0}^{2}}{2\tau R}\]
  done clear
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• question_answer2) Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed 'v' in a uniform magnetic field B going into the plane of the paper (see figure) . If charge densities \[{{\sigma }^{1}}\]and \[{{\sigma }^{2}}\] are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects)   JEE Main Online Paper (Held On 10 April 2016)
  A)
               \[{{\sigma }_{1}}=\frac{-{{\varepsilon }_{0}}vB}{2},\,{{\sigma }_{2}}=\frac{-2{{\varepsilon }_{0}}vB}{2}\]
  done clear
  B)
               \[{{\sigma }_{1}}={{\sigma }_{2}}={{\varepsilon }_{0}}vB\]              
  done clear
  C)
  \[{{\sigma }_{1}}=\frac{{{\varepsilon }_{0}}vB}{2},\,{{\sigma }_{2}}=\frac{-{{\varepsilon }_{0}}vB}{2}\]
  done clear
  D)
               \[{{\sigma }_{1}}={{\varepsilon }_{0}}vB,\,{{\sigma }_{2}}=-{{\varepsilon }_{0}}vB\]
  done clear
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