A thin wire ring of radius a and resitance `r` is located inside a long solenoid is equal to `l`, its cross-sectional radius , to `b`. At a certain moment the solenoid was connected to a source of a constant voltage `V`. The total resistance of the circuit is equal to `R`. Assuming the the radial force acting per unit length of the ring.

Explanation

The solenoid has an inductance,
`L = mu_(0) pi^(2) pi b^(2) l`,
where `n` = number of turns of the solenoid per unit length. When the solenoid is connected to the source an e.m.f. is set up, which, because of the inductance and resistance, rises slowly, according to the equacation,
`RI + L'I = V`
This has the knows solution,
`I = (V)/(R) (1 - e^(-1R//L))`.
Corresponding to this current, an e.m.f is induced in the ring. Its magnetic field `B = mu_(0) n I` in the solenoid produces a force per unit length, `(dF)/(dl) = B i = mu_(0)^(2) n^(2) pi a^(2) II//r`
`= (mu_(0)^(2) pi a^(2) V^(2))/(r) ((n^(2))/(R L)) e^(-1 R//L) (1 - e^(-1 R//L))`,
acting on each segment of the ring. This force is zero initially and zero for large `t`. Its maximum value is for some finite `t`. The maximum value of
`e^(-1 R//L) (1 - e^(-1 R//L)) = (1)/(4) - ((1)/(2) - e^(-1 R//L))^(2)` is `(1)/(4)`,
So, `(dF_(max))/(dl) = (mu_(0)^(2) pi a^(2) V^(2))/(r) (n^(2))/(4 R L) = (mu_(0) a^(2) V^(2))/(4 r R l b^(2))`