Editing Content :: Admin Scraper

Chapter

Content type

Position

Value

Moving Charges and Magnetism We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. This is illustrated in Fig. 4.18(a). The field exerts no force on the two arms AD and BC of the loop. It is perpendicular to the arm AB of the loop and exerts a force F o1 it which is directed into the plane of the loop. Its magnitude is, F1= I b B Similarly, it exerts a force2F on the arm CD and F 2 is directed out of the plane of the paper. F2= I b B = F1 Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces F and F . 1 2 Figure 4.18(b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anticlockwise. This torque is (in magnitude), a a τ = F1 + F2 2 2 = IbB a + IbB a = I( ) B 2 2 FIGURE 4.18 (a) A rectangular = I A B (4.20) current-carrying coil in uniform where A = ab is the area of the rectangle. magnetic field. The magnetic moment We next consider the case when the plane of the loop, m points downwards. The torque τ is is not along the magnetic field, but makes an angle with along the axis and tends to rotate the it. We take the angle between the field and the normal to coil anticlockwise. (b) The couple acting on the coil. the coil to be angleθ(The previous case corresponds to θ =π/2). Figure 4.19 illustrates this general case. The forces on the arms BC and DA are equal, opposite, and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms AB and CD are F and1F . The2 too are equal and opposite, with magnitude, F1= F 2 I b B But they are not collinear! This results in a couple as before. The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.19(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is, a a τ = F12 sin θ+ F2 2 sinθ = I ab B sin θ = I A B sin θ (4.21) 125 Reprint 2026-27