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, B, C, D

Explanation

Suppose at `t=0,y=0 and t=t, y=y`
(a) Total magnetic flux=`vec(B).vec(A)`
where `vec(A)=a^(2)(hat k)and vec(B)=(B_(0)y)/(a)(hat k)`
`:. (phi)=(B_(0)y)/(a).(a^2)=B_(0)ya`
Net emf. `e=-(d phi)/(dt)=-b_(0)a(dy)/(dt)=-B_(0)a(dy)/(dt)=-B_(0)av(t)`
As total resistance =R
`:. |i|=(|e|)/R=(B_(0)av(t))/(R)`

(b) Each side of the cube will experience a forve as shown (since a current carrying segment in a magnetic field experience a force).
`vec(F_1)=i(vec(l)xx vec(B))=i(-a(hat i)xx(B_(0)y)/(a)(hat k)=B_(0)y(hat(i)xx hat (j),`
`vec(F_3)i(+a(hat i)xx(B_(0)(y+a))/(a)(hat k))=iB_(0)(y+a)(hat j)`
(c) Total net force =`mg (hat j)+vec(F)=[mg-(B_(0)^(2)a^(2)v(t))/(R)](hat j)`
`:. m(dv)/(dt)=mg-(B_(0)^(2)a^(2)v(t))/(R)`
Integrating it we get, `int_(0)^(v)(dv)/(g-(B_(0)^(2)a^(2)v(t))/(mR))=int_(0)^(t) dt`
`(log[(g-(B_(0)^(2)a^(2)v(t))/(mR))]_(0)^((v)t))/((B_(0)^(2)a^(2))/(mR))=t`
or log [(g-(B_(0)^(2)a^(2)v(t))/(mR))/(g)]=-(B_(0)^(2)a^(2))/(mR)`
or `1-(B_(0)^(2)a^(2)v(t))/(mgR))=e^(-(B_(0)^(2)a^(2)t)//(mR)`
`oe, 1-=e^(-(B_(0)^(2)a^(2)t)//(mR)`=(B_(0)^(2)a^(2))/(mgR)v(t),`
`:. v(t)=(mgR)/(B_(0)^(2)a^(2))[1-e^(-(B_(0)^(2)a^(2)t)//(mR)]`
When terminal velocity is attained, v(t) does not depend on t
`:. v(t) =(mgR)/(B_(0)^(2)a^(2))`