derive the expression for "torque on rectanglular coil placed in a uniform magnetic field"
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
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 âÂ»
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 âÂ«
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 â sin 9. Likewise, the moment arm of Fj about 0 is also â 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 Â©j = IlbB sin 9
= IAB sin 9
where A = Ibis the area of the loop. This result shows that the torque has the maximum value IAB when the field is parallel to the plane of the loop (9 = 90Â°) and is zero when the field is perpendicular to the plane of the loop (9 = 0). As we see in Fig. (5.24) the loop tends to rotate to smaller values of 9 (that is, such that the normal to the plane of the loop rotates 1
A convenient vector expression for the torque is the following cross product relationship:
T = /AxB ...(/0
where A , a vector perpendicular to the plane of the loop, has a magnitude equal
to the area of the loop. The sense of A is determined by the right-hand rule as described in Fig. (5.25). By wrapping the four fingers of the right hand in the direction of the current, the thumb points in the direction of A. The product IA is defined as the magnetic moment m of the loop.