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AMU Medical AMU Solved Paper-2013

  • question_answer
    Figure shows a conducting loop consisting of a half-circle of radius r= 0.20m and three straight sections. The half-circle lies in a uniform magnetic field B that is directed out of the page, the field magnitude is given by \[B=(4.0\,T/{{s}^{2}}){{t}^{2}}+(2.0\,T/s)t+3.0\,T\] An ideal battery with \[{{\varepsilon }_{0}}=2.0\,V\] is connected to the loop. The resistance of the loop is \[2.0\,\Omega \]. The current in the loop at \[t=10\,s\] will be close to                

    A)                  3.6 A                    

    B)                  1.6 A

    C)  6.2 A                    

    D)  4.2 A

    Correct Answer: C

    Solution :

                     Given that r = 0.20 m, t = 10s, \[R=2\,\Omega \]                 \[B=(4.0\,T/{{S}^{2}}){{t}^{2}}+(2.0\,T/S)\,t+3.0\,T\]                 \[\frac{dB}{dt}=8\,t+2\] From \[E=-\frac{d\phi }{dt}=-A\frac{d\phi }{dt}\]                 \[E=-\pi {{r}^{2}}[8\,t+2]\]                 \[=3.14\times 0.2\times 0.2\times 82\]                 = 10.30 Total      \[E=10.30+2\]                 \[=12.30\,V\]                 \[l=\frac{E}{R}\]                 \[=\frac{12.30}{2.0}=6.15\]                 \[\approx 6.2\,A\]


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