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

  • question_answer
    There are three wire MO, NO and PQ, wires MO and NO are fixed and perpendicular to each other. Wire PQ moves with a constant velocity v as shown in the figure and resistance per unit length of each wire is\[\lambda \]and magnetic field exists   perpendicular and inside the paper then. Which of the following is wrong?

    A) current in loop is anticlockwise

    B) magnitude of current in the loop is\[\frac{Bv}{\lambda (\sqrt{2}+1)}\]

    C) current in the loop is independent of time.

    D) magnitude of current decreases as time increases.

    Correct Answer: D

    Solution :

    [d]  \[\phi =BA=\frac{B}{2}\frac{x}{\sqrt{2}}\frac{x}{\sqrt{2}}\] \[\varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}Bv\] \[\varepsilon =\frac{d\phi }{dt}=\frac{2x}{4}\frac{dx}{dt}B=\frac{x}{2}B\frac{d(2y)}{dt}\] \[i=\frac{xBv}{2\lambda (x+\sqrt{2}x)}=\frac{Bv}{\lambda (1+\sqrt{2})}\]


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