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Figure shows a conducting rod of length ...

Figure shows a conducting rod of length `l = 10 cm`, resistance `R` and mass `m = 100 mg` moving vertically downward due to gravity. Other parts are kept fixed. Magnetic filed is `B = 1 T`. `MN and PQ` are vertical, smooth, the capacitor is `C = 10mF`. The rod is released from rest. Find the maximum current (in mA) in the circuit.

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The correct Answer is:
5
,
By Newton's law, `mg = ilB = m(dv)/(dt)` (i)
Using `KVL blv = iR + (q)/(C )`
Differenting (ii) w.r.t. time, we get
`Bl(dv)/(dt) = R(di)/(dt) + (i)/( c)`
Eliminating `(dv)/(dt)` from equations (i) and (iii), we get
`mg - ilB = (m)/(Bl)[R(di)/(dt) + (i)/( C)]`
`rarr` `mgBl - iB^(2)l^(2) = m(R(di)/(dt) + (mi)/(C ))`
I will be maximum when `(di)/(ct) = 0`. Use this in equation (iv)
`rarr` `mgBlC = i(B^(2)l^(2)C + m)`
`rarr` `i_(max) = (mgBlC)/(m + B^(2)l^(2)C)`
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