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

  • question_answer
    A conductor of length \[l\] and mass m can slide without any friction along the two vertical conductors connected at the top through a capacitor (figure). A uniform magnetic field B is set \[\bot \]up 1 to the plane of paper. The acceleration of the  conductor

    A) Is constant        

    B) Increases

    C) Decreases          

    D) Cannot say

    Correct Answer: A

    Solution :

    [a] Let v is the velocity of conductor at any time, then induced emf:\[e=Blv\]                                    ...(i) Charge on capacitor: \[q=Ce=CBlv\] Current in circuit: \[I=\frac{dq}{dt}=CBl\frac{dv}{dt}\] For conductor: \[mg-IBl=\frac{mdv}{dt}\] \[\Rightarrow mg-C{{B}^{2}}{{t}^{2}}\frac{dv}{dt}=\frac{mdv}{dt}\Rightarrow \frac{dv}{dt}=\frac{mg}{m+C{{B}^{2}}{{t}^{2}}}\]This is the acceleration of conductor which is constant.


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