Dear student,
consider a  rectangular  loop carrying a current/ in the  presence of a uniform  magnetic field in  the plane of the loop as in  Fig. [5.23 (a)]. The  forces on the sides of  length b are zero since these
wires are parallel to the field and hence dl x B=0 for these sides. On the other hand, the  magnitude of  the forces on the sides of the length I is given
by A
where F\ is the force  on the left side of the loop and Fi is the force on the  right side.
The direction jf F\ is out of the paper in —»
Fig.  (5.23) and that of Fi is into the paper. An  end  viev/ of the  direction of these forces, shown in Fig. 5.23 (b),   demonstrates that  the forces form a couple. Therefore, even though they   cancel each  other, they tend to rotate the loop. The torque due to this   couple  about the point 0 in Fig. [5.23 (6)] 

has a magnitude given  by
T = Fi- + F2-  =  IIbB . 2 2
where the moment arm about 0 is b/2 for both forces.This in fact, is the torque about any point. But the area of the loop  is given by A = ab. Hence, the torque can be expressed as
T=IAB | ... (0
Note that this result  is valid only when the  field B is in the plane of the loop. The sense pf the rotation   is clockwise when viewed from the bottom end, as indicated in Fig.  5.23 (b). If the   current were reversed, the forces would reveise their  directions and   the rotational tendency would be counterclockwise.

Now  consider a rectangular loop carrying a  current I in a  uniform  magnetic field. Suppose the field makes an angle 9  with the  normal to  the plane of the loop as in Fig. 5.24 (a). For  convenience,  we shall  assume that —«
the field B is  perpendicular to the sides  of length I. In this   case, the magnetic forces Fi and Fa on the  sides of length b cancel  each  other and produce no torque since they pass through a common  origin 0.  However, the forces F,and F2 acting on  the sides of length I form a couple and hence produce a   torque about any point 0. Referring to the end view shown  in Fig. 5.24 (b) we note that  the moment arm of  the force F\ about  the  point 0 is equal  to — sin 9. Likewise, the moment arm of Fj about 0 is also — sin 0. 2 2
Since Fl = F2 = / / B, the net torque about 0 has a magnitude given by
T = Fi - sin 0 + F2-  sin 0 .2 2
= I IB || sin flj + I IB sin ©j = IlbB sin 9
= IAB sin 9
 
















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