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.

Answer

A, B, C

Explanation

By newton's law `mg-ilB=m(dv)/(dt)`-(1)
By `KVL Blv=iR+q/c`..(2)
differentiate (2) w.r.t. time `Bl (dv)/(dt)=R(di)/(dt)+i/c`..(3)
Eliminate `(dv)/(dt)` by (1) & (3) `mg-ilB=m/(Bl)[R(dv)/(dt)+i/c]`
`rArr mg Bl-iB^(2)l^(2)=mR(di)/(dt)+(mi)/c`..(4)
`i` will be maximum when `(di)/(dt)=0`. Use this in (4)
`rArr mg Blc=i(B^(2)l^(2)c+m) rArr i_(max)=(mgBlc)/(m+B^(2)l^(2)c)`