A magnetic field `B = B_(0) (y//a)(hat)k` is into the paper in the +z direction, `B_(0)` and a are positive constants. A square loop EFGH of side a, mass m and resistance R, in x-y plane, starts falling under the influence of gravity see figure. Note the direction of x and y axis in figure.

Find
(a) the induced current in the loop and indicate its direction.
(b) the total Lorentz force acting on the loop and indicate its direction, and
(c) an expression for the speed of the loop, v(t) and its terminal value.

Answer

`(a) i=(B_(0)av)/R` in anticlockwise direction,
`v=` velocity at time `t`
(b) `F_(nett)=B_(0)^(2)a^(2)V//R`,
(c ) `V=(mgR)/(B_(0)^(2)a^(2))(1-e(B_(0)^(2)a^(2)t)/(mR))`

Explanation

`dphi=B.ady=(B_(0)y)/a ady`
`phi=int_(Y)^(Y+a)B_(0)ydy=B_(0)[y^(2)/2]_(Y)^(Y+a) =B_(0)/2[(Y+a)^(2)-Y^(2)]`
`(dphi)/(dt)=B_(0)/2[2(Y+a)(dY)/(dt)-2y(dY)/(dt)]=B_(0)a(dY)/(dt)`
`e=-B_(0)av implies i=(B_(0)av)/R`

main flux increase so induce flux `o.` & direction of current anti clockwire.
(a) Motional emf if `EF=(B_(0)Ya)/a=B_(0)Yv`
Motional emf in `HG=(B_(0)(Y+a)av)/R`
`e=e_(2)-e_(1)=B_(0)av implies I=(B_(0)av)/R`

(B) Force on `F_(EG)=IaB=(B_(0)av)/R a (B_(0)Y)/a=(B_(0)^(2)+avY)/Rdarr`
`F_(GH)=IaB=(B_(0)av)/R a (B_(0)(Y+a))/auarr`
`=(B_(0)^(2)aV(Y+a))/R`
`F_(net)=F_(GH)-F_(EF)=(B_(0)^(2)a^(2)v)/Ruarr`

(C ) Expresion for terminal speed
`m(dv)/(dt)=mg-(B_(0)^(2)a^(2)v)/Rimplies int_(0)^(v)(dv)/(g-(B_(0)^(2)a^(2)v)/(mR))=int_(0)^(t)dt`
`-(mR)/(B_(0)^(2)a^(2))[log(g-(B_(0)^(2)a^(2)v)/(mR))]_(0)^(v)=[t]_(0)^(t)`
`t=-(mR)/(B_(0)^(2)a^(2))[log (g-(B_(0)^(2)a^(2)v)/(mR))/g]`
`ln' (g-B_(0)^(2)a^(2)v//mR)/g=-(B_(0)^(2)a^(2)t)/(mR)`
ln `(1-(B_(0)^(2)a^(2)v)/(mgR))=-(B_(0)^(2)a^(2)t)/(mR)`
`1-(B_(0)^(2)a^(2)v)/(mgR)=e^((B_(0)^(2)a^(2)t)/(mR)) implies (B_(0)^(2)a^(2)v)/(mgR) =1-e^((B_(0)^(2)a^(2)t)/(mR))`
`v=(mgR)/(B_(0)^(2)a^(2))(1-e^(-B_(0)^(2)I^(2)t)/(mR)) implies v_(t)=(mgR)/(B_(0)^(2)a^(2))`