# Double solenoid problem

$B =mu*N*I/L$ magnetic induction in smaller solenoid.

$U(emf) = -d(Phi)/d t$

$Phi$ is the flux of smaller solenoid in larger solenoid.

$Phi = B*S$

$Delta(phi) = mu*N*I/L *(pi*R^2)$

$Delta(t) =87 ms =87*10^{-3} sec$

$Delta(phi) = – 4*pi*10^{-7} *240*2.20 / (0.01) *pi*(0.034)^2 =2.41*10^{-4} Wb$

$U(emf) = -2.41*10^-4/87*10^{-3} = – 2.77*10^{-3} V$