A rectangular coil of wire 22.0 cm by 35.0 cm and carrying a current of 1.90 A i
ID: 1902808 • Letter: A
Question
A rectangular coil of wire 22.0 cm by 35.0 cm and carrying a current of 1.90 A is oriented with the plane of its loop perpendicular to a uniform 1.60 T magnetic field, as shown in the figure 1.
1. The coil is now rotated through a 30.0 angle about the axis shown, the left side coming out of the plane and the right side going into the plane.Calculate the net force that the magnetic field exerts on the coil. (Hint: In order to help visualize this three-dimensional problem, make a careful drawing of the coil as viewed along its axis of rotation.) The answer to this part is 0 N.
2.Calculate the torque that the magnetic field exerts on the coil in part (C).
Explanation / Answer
F = i(L) x B = 1.4(2*0.22)(1.5)(sin 90) = 0.92 Newtons, I assumed the short side of the loop was perpendicular to the magnetic field. Half the force is on each 22 cm wire. If the long side is perpendicular the answer is 1.4(2*0.35)(1.5)(sin 90) = 1.47 Newtons. The torque in the first case is (1/2)(F)(D) = (0.5)(0.92)(0.35)(sin 90) = 0.161 newton meters. And if the long side is perpendicular to the magnetic field, T = (0.5)(1.47)(0.22)(sin 90) = 0.161 newton meters. They are equal as they should be. Torque = I(A) x B where I is the current in the loop, A the area of the loop, and B the magnetic field density. x is the cross product, sine of the angle between the loop perpendicular and the magnetic field. if you rotate the loop 30 degrees, sine 30 is 1/2 so the torque drops by 50%. Torque = (1/2)F x D (1/2) (0.92)(0.35) (sin 30)= 0.08 mewton meters