Figure 28-50 shows a wood cylinder of mass m = 0.211 kg and length L = 0.0772 m,

Question

Figure 28-50 shows a wood cylinder of mass m = 0.211 kg and length L = 0.0772 m, with N = 15.0 turns of wire wrapped around it longitudinally, so that the plane of the wire coil contains the long central axis of the cylinder. The cylinder is released on a plane inclined at an angle ? to the horizontal, with the plane of the coil parallel to the incline plane. If there is a vertical uniform magnetic field of magnitude 0.627 T, what is the least current i (in A) through the coil that keeps the cylinder from rolling down the plane?

its chapter 28 problem 53 in the fundamentals of physics 8e book.

i'm just not getting how they found A=2rL and not 2prL.. everything else i get.

Explanation / Answer

* The wire is wrapped around the wood cylinder such a way that the plane of the wire coil contains the long central axis. * Hence, it is clear that the wire is wrapped around the cylinder in the form of a rectangle. Here, length of the rectangle = Length of the cylinder = L Breadth of the rectangle = Diameter of the wood cylinder = 2 r Area of the rectangular coil = Length * breadth = L * 2r = 2rL

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