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.

Answer

A

Explanation

`L=pimu_(0)Nnb^(2)=pimu_(0)(nl)nb^(2)=mu_(0)n^(2)pib^(2)l`
`I=V/R(1-e^(-Rt//L))`…(i)
Determine the induced current in the ring The flux associated with ring is `phi_(rho)=Bpia^(2)=mu_(0)nlpia^(2)`
Here `B`=Magnetic field due to solenoid `=mu_(0)nl`
`therefore` Induced `emf` in the ring is `epsilon'=(dphi_(r))/(dt)=mu_(0)nra^(2)(dl)/(dt)`
`therefore` The current in the ring is `l'=(epsilon')/ r=(mu_(0)npia^(2))/r(di)/(dt)`
From equ.(i) we get `l=V/R(1-e^(-Rt//L))`
`(di)/(dt)=V/Le^(-Rt//L)`
`l'=(mu_(0)npia^(2))/rV/Le^(-Rt//L)`
Determine magnetic force on an element of length `dL` of the ring The magnetic force on the considered element is `dF=l'dLB`
`therefore` Force per unit length is `F_(0)=(dF)/(dL)=I'B`
`therefore` On putting the value we get , `F_(0)=(mu_(0)npia^(2)V)/(Lr)e^(-Rt//L)(mu_(0)nl)`
`=(mu_(0)^(2)n^(2)pia^(2)V)/(Lr)e^(-Rt//L)(1-e^(-Rt//L))`
For maximum value of `F_(0), (dF_(0))/(dt)=0`
After solving, we `F_(0max)=((dF)/(dL))_(max)=(mu_(0)^(2)a^(2)V^(2))/(4rRlb^(2))`