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 F_l = 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

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi