A insulated square frame `ABCD` of side a is able to rotate about one of its side taken as positive `z`-axis. A magnetic field `B` is present in the region given by `vecB=B_(0)hatj`. A small block of mass `m` and charge `q` movable along side `CB` is initially near `C`, when frame lies in `x-z` plane. Now, frame is rotated by constant angular velocity `omega` about `z`-axis. Whole system lies in gravity free space. If after time `t` block reaches point `B`, find `B_(0)` in terms of `t`.

Answer

`(a)mg.(a)/(2)(hat(j)) " "(b)IB_(0)b(3hat(k)-4hat(j))" "(c )I=(mg)/(6B_(0)b)`

Explanation

`(a)` Torque due to weight of coil
`vec(tau)=((a)/(2)hat(i))xx(-mghat(k))=mg.(a)/(2)(hat(j))`
For the equilibrium of loop, torque on it must be along negative `y-` axis . Let the magnetic moment of loop be `Mhat(k)`. As the loop lies in `xy-` plane, its magnetic moment vector ( from right `-` hand thumb rule ) either points up to down. Torque due to magnetic force,
`vec(tau)_(B)=vec(mu)xxvec(B)=mu hat(k)xx(3 hat(i)+4 hat(k))B_(0)=3 mu B_(0)hat(j)`
`(b)` Force acting on arm `RS=I(hat(l)xxvec(B))`
`=I[(-bhat(j))xx(3 hat(i)+4 hat(k))B_(0)]=IB_(0)b(3hat(k)-4hat(i))`
`(c )` In equilibrium , `vec( tau)_("gravity")+vec(tau)_(B)=0`
Hence `3(abI)B_(0)=(mga)/(2)` or `I=(mg)/(6B_(0)b)`