Physics-
General
Easy

Question

In the arrangement shown in the figure, uniform magnetic field B is upward to the plane of paper. Connecter AB is smooth and conducting, having mass m and length l. Initially spring has extension X0. Spring is non-conducting (neglect induction in spring). Connector is released at t = 0. Maximum charge on the capacitor is

  1. X subscript 0 end subscript square root of fraction numerator K C over denominator 1 plus fraction numerator B to the power of 2 end exponent l to the power of 2 end exponent C over denominator m end fraction end fraction end root    
  2. X subscript 0 end subscript square root of fraction numerator K C over denominator 1 plus fraction numerator m over denominator B to the power of 2 end exponent l to the power of 2 end exponent C end fraction end fraction end root    
  3. square root of fraction numerator K C over denominator 2 plus fraction numerator B to the power of 2 end exponent l to the power of 2 end exponent C over denominator m end fraction end fraction end root    
  4. square root of fraction numerator 4 K C over denominator 1 plus fraction numerator B to the power of 2 end exponent l to the power of 2 end exponent C over denominator 2 m end fraction end fraction end root    

The correct answer is: X subscript 0 end subscript square root of fraction numerator K C over denominator 1 plus fraction numerator m over denominator B to the power of 2 end exponent l to the power of 2 end exponent C end fraction end fraction end root


    At any time
    K x minus I l B equals m a (a)
    fraction numerator q over denominator C end fraction equals B v l
    I equals B l C a (b)
    From (a) and (b)
    k x equals a left parenthesis B to the power of 2 end exponent l to the power of 2 end exponent C plus m right parenthesis
    a equals open parentheses fraction numerator K over denominator B to the power of 2 end exponent l to the power of 2 end exponent C plus m end fraction close parentheses x (c)
    Since motion of connector will be opposite to the displacement from equilibrium. Therefore equation (c) represents S.H.M.
    omega equals square root of fraction numerator K over denominator B to the power of 2 end exponent l to the power of 2 end exponent C plus m end fraction end root
    V 0 subscript m a x end subscript
    \q m a x subscript m a x end subscript
    q 0 subscript m a x end subscript
    X subscript 0 end subscript square root of fraction numerator K C over denominator 1 plus fraction numerator m over denominator B to the power of 2 end exponent l to the power of 2 end exponent C end fraction end fraction end root

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