Question

Two solenoids A and B are placed near to each other coaxially. Number of turns of A is $400$ and number of turns of B is $700$. A current of $3.5$A in coil produced an average flux of $300\mu$Wb through each turn of A and a flux of $90\mu Wb$ through each turn of B. Find induced emf (in millivolt) in B when the current in A increases at the rate of $0.5A/s$ ?

Answer 1

Total flux through $B$= $Mx$ current in solenoted A
$Ns{Q}_B=\frac{M}{A}$
$M=\frac{700\times 90\times {10}^{-6}}{3.5}=1.8\times {10}^{-2}$ or 18mH
$N_A{Q}_A=L_A{l}_A$
$L_A=\frac{450\times 300\times {10}^{-6}}{3.5}=342mH$
$EMF=-M\frac{dl}{dt}=-18/{10}^{-3}\times 0.5=-9mV$