6. The rectangular loop shown below is pivoted about the side ab on the z-axis. When the loop is inposition 1, it lies entirely in the yz-plane. When the loop is in position 2, it has been rotated by 30°from position 1 toward the xz-plane. In position 1, the loop carries a current of 8.2 A in thedirection indicated by the purple arrows. The flow of current through the loop does not changewhen it is moved to position 2. The loop is immersed in a uniform magnetic field B = (0.48 T)ĵ.yZ6.00 cm8.00cmPosition 130.0⁰Position 2(a) When the loop is in position 1,i.ii.(b) When the loop is in position 2,i.ii.(c) What is the change in magnetic potential energy when the loop is moved from position 1 toposition 2?What is the magnetic dipole moment of the loop?What is the magnetic torque on the loop?xWhat is the magnetic dipole moment of the loop?What is the magnetic torque on the loop?

Solution: (a). (i). The magnetic dipole moment at position is given by, m=ia x^ =8.2A×0.06×0.08 x^ =0.039 Am2 x^ The right-hand thumb rule gives the direction of magnetic dipole moment. Curl the fingers of the right hand in the direction of the current flow, then the thumb will indicate the direction of the magnetic field.(ii). The cross-product of the magnetic dipole moment and the magnetic field gives the magnetic torque on the loop. τ=m×B =0.039×0.48sin90° z^ =0.039×0.48 z^ =0.02 Nm z^b). (i). The magnetic dipole moment at position 2 is the same as at position 1. m=0.039 Am2 But the direction of the dipole moment is not the same. In this case, the dipole moment makes an angle of 30 degrees with the x-axis. (ii). The magnetic torque on the loop at position 2 is, τ=m×B =0.039×0.48sin30° z^ =0.039×0.48×0.5 z^ =9.3×10-3 Nm z^(c). The change in magnetic potential energy is given by the following, ∆U=-m.Bf-m.Bi =-mBcos30° J =-0.039×0.48×0.86 J =-0.017 J

Answered: (a) When the loop is in position 1, i.…