In Fig. a three-side frame is pivoted at AC and hangs vertically. Its sides are each of the same length and have a linear density of `0.10 kg m^-1`. A current of 10.0 A is sent through the frame, which is in a uniform magnetic field of 10 mT directed upward. Through what angle will the frame be deflected?

Explanation

Refracting to the figure in question, we note that the magnetic forces on the slanting sides are parallel to AC and therefore, do not produce any torque about the axis AC. The magnetic force on the horizontal side of the frame is
`F_(mag)=iLxxB...(i)`
This force has magnitude `F_(mag)=iLBsin90^@=iLB` and points horizontally to the right on this plane of this figure, perpendicular to L and B. The magnetic torque about AC has magnitude
`T_m=iLB(Lcos theta)=iL^2Bcos theta....(ii)`
and tends to increase the angle `theta`. The gravitational torque about
the axis AC is
`T_g=-[(lambdaL)g(Lsin theta)+2(lambdaL)]`
`g(1/2L sin theta).....(iii)`
where the minus sign indicates that gravity tends to turn the frame in the direction of decreasing `theta`. Let `lambda` denote the mass per unit length of the wire. In equilibrium, the torque evaluted in equations (ii) and (iii) must cancel, which implies that `tan theta=(Bi)//(2lambdag)`.

Subtituting the data, we obtain `theta=tan^-1((Bi)/(2lambdag))`.