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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove Alfven's theorem: In a perfectly conducting fluid (say, a gas of free electrons), the magnetic flux through any closed loop moving with the fluid is constant in time. (The magnetic field lines are, as it were, "frozen" into the fluid.)

(a) Use Ohm's law, in the form of Eq. 7.2, together with Faraday's law, to prove that if σ=and is J finite, then

Bt=×(v×B)

(b) Let S be the surface bounded by the loop (P)at time t , and S'a surface bounded by the loop in its new position (P')at time t+dt (see Fig. 7.58). The change in flux is

=S'B(t+dt)da-SB(t)da

Use ·B=0to show that

S'B(t+dt)da+RB(t+dt)da=SB(t+dt)da

(Where R is the "ribbon" joining P and P' ), and hence that

=dtSBt·da-RB(t+dt)da

(For infinitesimal dt ). Use the method of Sect. 7.1.3 to rewrite the second integral as

dtP(B×v)·dI

And invoke Stokes' theorem to conclude that

dt=S(Bt-×v×B)·da

Together with the result in (a), this proves the theorem.

Suppose

E(r,t)=14πε0qr2θ(rυt)r^; B(r,t)=0

(The theta function is defined in Prob. 1.46b). Show that these fields satisfy all of Maxwell's equations, and determine ρ and J. Describe the physical situation that gives rise to these fields.

A familiar demonstration of superconductivity (Prob. 7.44) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. Treat the magnet as a perfect dipole , m a height z above the origin (and constrained to point in the z direction), and pretend that the superconductor occupies the entire half-space below the xy plane. Because of the Meissner effect, B = 0 for Z0, and since B is divergenceless, the normal ( z) component is continuous, so Bz=0just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at - z , as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region z>0.

(a) Which way should the image dipole point (+ z or -z)?

(b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to Mg (where M is the mass of the magnet) to determine the height h at which the magnet will "float." [Hint: Refer to Prob. 6.3.]

(c) The induced current on the surface of the superconductor ( xy the plane) can be determined from the boundary condition on the tangential component of B (Eq. 5.76): B=μ0(K×z^). Using the field you get from the image configuration, show that

K=-3mrh2π(r2+h2)52ϕ^

where r is the distance from the origin.

Refer to Prob. 7.11 (and use the result of Prob. 5.42): How long does is take a falling circular ring (radius a, mass m, resistance R) to cross the bottom of the magnetic field B, at its (changing) terminal velocity?

A square loop is cut out of a thick sheet of aluminum. It is then placed so that the top portion is in a uniform magnetic field , B and is allowed to fall under gravity (Fig. 7 .20). (In the diagram, shading indicates the field region; points into the page.) If the magnetic field is 1 T (a pretty standard laboratory field), find the terminal velocity of the loop (in m/s ). Find the velocity of the loop as a function of time. How long does it take (in seconds) to reach, say, 90% of the terminal velocity? What would happen if you cut a tiny slit in the ring, breaking the circuit? [Note: The dimensions of the loop cancel out; determine the actual numbers, in the units indicated.]

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free