A rectangular loop PQRS made from a uniform wire has length a, width b and mass m. It is free to rotate about the arm PQ, which remains hinged along a horizontal line taken as the y-axis (see figure). Take the vertically upward direction as the z-axis. A uniform magnetic field `vec(B) = (3 hat(i) + 4 hat(k)) B_(0)` exists in the region. The loop is held in the x-y plane and a current I is passed through it. The loop is now released and is found to stay in the horizontal position in equilibrium

(a) What is the direction of the current I in PQ?
(b) Find the magnetic force on the arm RS.
(c) Find the expression for I in terms of `B_(0)`, a, b and m

Answer

`B_(0)=(momega)/(omegat-sin omegat)`

Explanation

The block moves with speed `upsilon = omega a` in the `x-y` plane.
The velocity of the block, when the framce has totated through an angle `theta=omegat`,

`vec(v)=omegaa(- sin theta hat(i)cos thetahat(j))+upsilon_(z)hat(k)`
Force on the block,
`F=qvec(v)xxvec(B)=[omegaa(-sin thetahat(i)+cos thetahat(j))+upsilon_(z)hat(k)]xxB_(0)hat(j)`
Force in `z-` direction `=omega a B_(0)(-hat(k))sin theta`
Acceleration of block `, a_(z)=(omegaa B_(0)sin theta)/(m)(-hat(k))`
or `" "(dupsilon_(z))/(dt)=-(omegaaB_(0))/(m)sin omegathat(k)`
or `upsilon_(z)=-(omega a B_(0))/(m)int_(0)^(t)sin omega t dt=-(aB_(0))/(m)(1-cos omegat)`
`(dz)/(dt)=-(aB_(0))/(m)(1-cos omegat)`
or `" "dz=-(aB_(0))/(m)int_(0)^(1)(1-cos omegat)dt`
`z=-(aB_(0))/(m)[t-(sin omegat)/(omega)]_(0)^(t)=-(aB_(0))/(m)[t-(sin omega t)/(omega)]`
When the block reaches `B,z=-a`.
or `-a=-(aB_(0))/(m)(t-(sin omegat)/(omega))`or`B_(0)=(m omega)/(omegat-sin omegat)`