As a lecture demonstration a short cylindrical bar magnet is dropped down a vertical aluminum pipe of slightly larger diameter, about 2 meters long. It takes several seconds to emerge at the bottom, whereas an otherwise identical piece of unmagnetized iron makes the trip in a fraction of a second. Explain why the magnet falls more slowly.

Short Answer

Expert verified

The attraction force act between the magnet and ring of pipe. Therefore, the magnet falls slowly.

Step by step solution

01

Determine the currents in the falling magnet and ring of pipe.

Let’s consider the current is flowing into the falling magnet in the counter clockwise direction; then, according to the flaming’s rule, the magnetic field due to the magnet is in the upward direction.

Now assume the ring of pipe below the falling magnet, and the magnetic flux into the ring increases as the magnate approaches the pipe. Therefore, the induced current into the pipe ring is clockwise.

02

Determine the reason of falling the magnet slowly.

When the magnet falls above the pipe, the current in the magnet is in the counter clockwise direction, the induced current into the pipe is in a clockwise direction, and the opposite current repels each other.

When the magnet falls below the pipe, the current's directions become in the same direction, which means the attraction force act between them, and the current attracts each other. Therefore, the magnet falls more slowly.

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Most popular questions from this chapter

Question: The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (Fig. 7.46a). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w << a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

(a) Find the electric field between the plates, as a function of t.

(b) Find the displacement current through a circle of radius in the plane mid-way between the plates. Using this circle as your "Amperian loop," and the flat surface that spans it, find the magnetic field at a distance s from the axis.

Figure 7.46

(c) Repeat part (b), but this time uses the cylindrical surface in Fig. 7.46(b), which is open at the right end and extends to the left through the plate and terminates outside the capacitor. Notice that the displacement current through this surface is zero, and there are two contributions to Ienc.

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Problem 7.61 The magnetic field of an infinite straight wire carrying a steady current I can be obtained from the displacement current term in the Ampere/Maxwell law, as follows: Picture the current as consisting of a uniform line charge λmoving along the z axis at speed v (so that I=λv), with a tiny gap of length E , which reaches the origin at time t=0. In the next instant (up to t=E/v) there is no real current passing through a circular Amperian loop in the xy plane, but there is a displacement current, due to the "missing" charge in the gap.

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