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A conductor of mass m and length l is sl...

A conductor of mass `m` and length `l` is sliding smoothly an two vertical conducting rails `AB` and `CD` as shown in figure. The top ends of two conducting rails are joined by a capacitor of capacitance `C`. The conductor is relased from rest when it is very close to `AC` i.e., `y = 0`. A uniform magnetic field `B_(0)` perpendicular to plane of figure existing. Neglect the resistance of rails and connecting wires. Take acceleration due to gravity to be `g`.

Based on above information answer the following questions:
Mark the correct statement about the motion of conductor

A

It is falling down with constant acceleration `g`

B

It is falling down with constant acceleration but not equal to `g`

C

It is falling down with increasing acceleration

D

It is falling down with decreasing acceleration

Text Solution

Verified by Experts

The correct Answer is:
B
Let us say that in time `t` the conductor falls down by `y` and acquires a velocity `v`, then at this instant induced emf is
`e = B_(0) vl` [with `Q` at higher potential]

Charge on the capacitor at this instant is,
`q = Ce = B_(0) Cl xx v`
the current in the circuit at this instant is,
`I = (d q)/(dt) = B_(0) Cl(dv)/(dt) = B_(0) Cl xx a`
where a is the acceleration of the conductor at this instant.
Writing Newton's `2^(nd)` law equation for conductor,
`mg - IB_(0)l = m a`
` rArr a = (m g)/(m + CB_(0)^(2)l^(2))`,
`v = (m g t)/(m + CB_(0)^(2)l^(2))` and `y = (m g t^(2))/(2(m + CB_(0)^(2) l^(2)))`
After solving above equations, we could have `q` as a function of `y`
as `I=B_(0)Clxxa`
As a is constant, `I` is also constant.
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Knowledge Check

  • A conductor of mass m and length l is sliding smoothly an two vertical conducting rails AB and CD as shown in figure. The top ends of two conducting rails are joined by a capacitor of capacitance C . The conductor is relased from rest when it is very close to AC i.e., y = 0 . A uniform magnetic field B_(0) perpendicular to plane of figure existing. Neglect the resistance of rails and connecting wires. Take acceleration due to gravity to be g . Based on above information answer the following questions: Current in the circuit is

    A
    constant
    B
    increasing with time
    C
    decreasing with time
    D
    First increase, then reaches a maximum value, and then starts decreasing to attain a constant value finally

  • A conductor of mass m and length l is sliding smoothly an two vertical conducting rails AB and CD as shown in figure. The top ends of two conducting rails are joined by a capacitor of capacitance C . The conductor is relased from rest when it is very close to AC i.e., y = 0 . A uniform magnetic field B_(0) perpendicular to plane of figure existing. Neglect the resistance of rails and connecting wires. Take acceleration due to gravity to be g . Based on above information answer the following questions: Charge on the capacitor as a function of distance travelled y is given by

    A
    `CB_(0)lsqrt((2m gy)/(m+CB_(0)^(2)l^(2)))`
    B
    `CB_(0)lsqrt((m+CB_(0)^(2)l^(2))/(2mgy))`
    C
    `CB_(0)l e^((mgy)/(m+CB_(0)^(2)l^(2)))`
    D
    `CB_(0)lsqrt((2mgy)/(m+CB_(0)^(2)l^(2)) ) xx e^((mgy)/(m+CB_(0)^(2)l^(2)))`

  • A conducting rod MN of mass m and length 'l' is placed on parallel smooth conducting rails connected to an uncharged capacitor of capacitance C and a battery of emf epsilon as shown. A uniform magnetic field B is existing perpendicular to the plane of the rails. The steady state velocity acquired by the conducting rod MN after closing switch S is (neglect the resistance of the parallel rails and the conducting rod)

    A
    `(2CBl epsilon)/((m+CB^(2)l^(2)))`
    B
    `(CBl epsilon)/((m+CB^(2)l^(2)))`
    C
    `(CBl epsilon)/(2(m+CB^(2)l^(2)))`
    D
    `(CBl epsilon)/(4(m+CB^(2)l^(2)))`

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