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.

Explanation

(a) Let v = velocity of the loop at any time t.
`implies ` The induced emf between E & F.
`= e_(1)= B_(1) av =(B_(0)y)/(a) av = B_(0)yv`
The induced emf between H and G
`= e_(1)= B_(2) av= (B_(0)(y+a))/(a) av =B_(0)(y+a)`
`implies` The net emf `= e=e_(z) -e_(1)= B_(0)av`
`implies` The induced current =i` = -(e )/(R )=(B_(0)av)/(R )` anticlockwise because
`e_(z) gt e_(1)`

(b ) The Lorentz force on EF & HG are given as
`F_(1)=(B_(0)Y)/(a) i a = B_(0) yi & F_(2) = (B_(0)(Y+a))/(a) ia = B_(0)(y+a)i`
`implies F =F_(2)-F_(1)=B_(0)` a i along negative -Y axis
`implies F =-B_(0) ai haty`
`implies F = -(B_(0)^(2)a^(2)v)/(R ) `
( c) `F_("net") = mg -(B_(0)^(2)a^(2)v)/(R) haty`
`implies m""(dv)/(dt)= mg -(B_(0)^(2)a^(2)v)/(R) `
`implies int_(0)^(v)(dv)/((mg-(B_(0)^(2)a^(2)v)/(R)))=int_(0)^(1)(dt)/(m) `
Evaluating the integration and rearranging the terms we get
`v= (mg R)/(B_(0)^(2)a^(2)) [1-e^(-((B_(0)^(2)a^(2)1)/(mR)))]`
When `t rarr oo , v rarr V_("terminal ")`
`implies V_("terminal") = (mgR)/(B_(0)^(2)a^(2))`.