A rectangular coil of dimensions 4.80 cm & 32.50 cm consists of 22 turns of wire and carries a current of 18.0 mA. A 0.280-T magnetic field is applied parallel to the plane of the coil. Calculate a) the magnitude of the magnetic dipole moment of the coil, b) the magnitude of the torque acting on the loop. Consider the loop of wire in Figure 28.25a. Imagine it is pivoted along side , which is parallel to the z axis and fastened so that side : remains fixed and the rest of the loop hangs vertically in the gravitational field of the Earth but can rotate around side (Fig. 28.25b). The mass of the loop is 50.0 g, and the sides are of lengths a = 0.200 m and 6 = 0.100 m. The loop carries a current of 3.50 A and is immersed in a vertical uniform magnetic field of magnitude 0.010 0 T in the positive y direction (Fig. 28.25c). What angle does the plane of the loop make with the vertical? Figure 28.25 (a) The dimensions of a rectangular current loop. (b) Edge view of the loop sighting down sides and . (c) An edge view of the loop in (b) rotated through an angle with respect to the horizontal when it is placed in a magnetic field. (2) A The loop hangs vertically and is pivoted so that it can rotate around side . x The magnetic torque causes the loop to rotate in a clockwise direction around side 4, whereas the gravitational torque is in the opposite direction. g sin cos e y B Solution Conceptualize In the edge view of Figure 28.25b, notice that the magnetic moment of the loop is to the left. Therefore, when the loop is in the magnetic field, the magnetic torque on the loop causes it to rotate in a clockwise direction around side , which we choose as the rotation axis. Imagine the loop making this clockwise rotation so that the plane of the loop is at some angle 0 to the vertical as in Figure 28.25c. The gravitational force on the loop exerts a torque that would cause a rotation in the counterclockwise direction if the magnetic field were turned off. Categorize At some angle of the loop, the two torques described in the Conceptualize step are equal in magnitude and the loop is at rest. We therefore model the loop as a rigid object in equilibrium. B = = B sin (90 0) = IAB cos 0 k = IabB cos 0 k Analyze Evaluate the magnetic torque on the loop B about side from Equation 28.18: Evaluate the gravitational torque on the loop about side 4, noting that the gravitational force can be modeled to act at the center of the loop: From the rigid body in equilibrium model, add the torques and set the net torque equal to zero: Solve for 0: B b sin 7 = 7 mg = mg-sin k =-IabB cos 0 + mg 7=-1 b b sin 0 k = 0 2IaB mg IabB cos 0=mg_sin 0 tan 0 = Substitute numerical values: 0 = tan-1 21aB mg 0 = tan-1 [2(3.50 A)(0.200 m) (0.010 0 T) (0.050 0 kg) (9.80 m/s) 1.64 Finalize The angle is relatively small, so the loop still hangs almost vertically. If the current I or the magnetic field B is increased, however, the angle increases as the magnetic torque becomes stronger. 7 = 7 B (28.18)