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JEE Main & Advanced Physics Electro Magnetic Induction Question Bank Self Evaluation Test - Electromagnetic Induction

  • question_answer
    A plane loop, shaped as two squares of sides a =1 m and b=0.4 m is introduced into a uniform magnetic field\[\bot \]to the plane of loop. The magnetic field varies as \[B={{10}^{-3}}\sin \](100t) T. The amplitude of the current induced in the loop if its resistance per unit length is\[r=5\,m{{\Omega }^{-1}}\]is

    A) 2 A

    B)   3 A

    C) 4 A

    D) 5 A   

    Correct Answer: B

    Solution :

    [b] \[\phi \] (flux linked) \[={{a}^{2}}B\,\cos \,{{0}^{o}}-{{b}^{2}}B\,\cos \,{{180}^{o}}\] \[E=-\frac{d\phi }{dt}=-({{a}^{2}}-{{b}^{2}})\frac{dB}{dt}\] \[=({{a}^{2}}-{{b}^{2}}){{B}_{0}}\,\omega \cos \,\omega t\]               where \[B={{B}_{0}}\], \[\sin \,\omega t\], \[{{B}_{0}}={{10}^{-3}}T\], \[\omega =100\] \[\therefore \,\,\,\,{{I}_{\max }}=({{a}^{2}}-{{b}^{2}})\frac{{{B}_{0}}\omega }{R}\] and \[R=(4a+4b)r=4(a+b)r\] \[\therefore \,\,\,\,{{I}_{\max }}=\frac{(a-b){{B}_{0}}\omega }{4r}=\frac{(1-0.4)\times {{10}^{-3}}\times 100}{4\times 5\times {{10}^{-3}}}=3A\]


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