A conducting frame abcd is kept in a vertical plane. A conducting rod of of mass m and length `l` can slide smoothly on it remaining always horizontal. The resistance of the loop is negligible and inductance is constant having value `L`. The rod is left from rest and allowed to fall under gravity and inductor has no initial current. A magnetic field of constant magnitude Bis present throughout the loop pointing inwards. Determine

(a) position of the rod as a function of time assuming initial position of the rod to be `x=0` and vertically downward as the positive x-axis.
(b) the maximum current in the circuit.
(c) maximum velocity of the rod

Answer

B

Explanation

a. Suppose `v` be the velocity of rod `ef` whenit has fallen a distane d. Then,
`V_(fe)=V_(cb)` or `Bvl=(di//dt)`
or (`Bdx//dt)l=L(di//dt)` or `Bl(dx)=L(dt)`
Integrating we get `Li=Blx`
or `i=((Bl)/L)x`………….(i)
Now magnetic force opposite to displacement x will be
`F=F_m=ilB=((B^2l^2)/L)x`
A constant downward force is `mg`.

so, this is similar situation like spring clock system in vertical position. in wich a force `F=kx` acts upwards and a constant force `mg` acts downwards.
hence, the wire will execute SHM, where
`k =(B^2l^2)/L`
Ampltude will be at `F_m=mg`
or `((B^2l^2)/L)A=mgimplies A=(mgL)/(B^2l^2)`
At `t=0`, rod is in its extreme position.
Therefore if we write the equation from mean position we will write
`X=-Acosomegat`
But `x=X+A=A-Acosomegat=A(1-cosomegat)`
where, `omega=sqrt(k/m)=sqrt((B^2l^2)/(mL))`
b. From Eqn i
`i_(max)=((Bl)/L) x_(max)`
Here `x_(max)=2A=(2mgL)/(B^2l^2)`
`:. i_(max)=((Bl)/L)((2mgL)/(B^2l^2))=(2mg)/(Bl)`
c. Maximum velocity
`v_0=omegaA =(sqrt((B^2l^2)/(mL)))((mgL)/(B^2l^2))=gsqrt((mL)/(B^2l^2))=(gsqrt(mL))/(Bl)`