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Download PWE 19-3: Magnetic Levitation
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Transcript
Example 19-3 Magnetic Levitation You set up a uniform horizontal magnetic field that points from south to north and has magnitude 2.00 * 1022 T. (This is about 400 times stronger than Earth’s magnetic field, but easily achievable with common magnets.) You want to place a straight copper wire of diameter 0.812 mm in this field, then run enough current through the wire so that the magnetic force will make the wire “float” in midair. This is called magnetic levitation. What minimum current is required to make this happen? The density of copper is 8.96 * 103 kg>m3. Set Up In order to make the wire “float,” there must be an upward magnetic force on the wire that just balances the downward gravitational force. We know from Equation 19-5 that to maximize the magnetic force, the current direction should s. be perpendicular to the magnetic field B The right-hand rule then shows that the current should flow from west to east so that the magnetic force is directed upward. We’re not given the mass of the wire, but we can express the mass (and hence the gravitational force on the wire) in terms of its density. We’re also not given the length of the wire; as we’ll see, this will cancel out of the calculation. Solve Write expressions for the two forces (magnetic and gravitational) that act on the wire. Magnetic force on a current-carrying wire: F = i/B sin u magnetic force Definition of density: m r = V (11-1) Volume of a cylindrical wire of cross-sectional area A and length /: B i (19-5) up West East South i North gravitational force down V = A/ Assume the wire has length /. Since the current flows in a direction perpendicular to the magnetic field, the angle u in Equation 19-5 is 90°. So the magnitude of the upward magnetic force is F = i/B sin 90 = i/B The magnitude of the downward gravitational force on the wire of mass m is w = mg From Equation 11-1, the mass equals the density of copper multiplied by the volume of the wire: m = rV = rA/ The cross-sectional area A of the wire is that of a circle of radius r: A = pr 2 so m = rA/ = r1pr 2 2/ So the magnitude of the gravitational force is w = mg = r1pr 2 2/g If the wire is floating in equilibrium, the upward magnetic force must just balance the downward gravitational force. Use this to solve for the required current i. In equilibrium the net vertical force on the wire is zero: F 2 w = 0, so F = w and i/B = r1pr 2 2/g The length / of the wire cancels out of this equation (both the magnetic force and the gravitational force are proportional to /): iB = r(pr2)g Solve for the current i: i = = r1pr 2 2g B 18.96 * 103 kg>m3 2 1p2 10.406 * 10-3 m2 2 19.80 m>s 2 2 = 2.27 Reflect 2.00 * 10-2 T kg T s 2 = 2.27 A [Check on units: We know from Section 19-3 that 1 T = 1 N> 1A # m), and we also know that 1 N = 1 kg # m>s 2. Therefore 1 T = 1 kg> 1A # s 2 2, and so 1 kg> 1T # s 2 2 = 1 A.] A current of 2.27 A is relatively small, so this experiment in magnetic levitation is not too difficult to perform. Note that the required current i is inversely proportional to the magnitude B of the magnetic field. You can see that if you tried to make a wire “float” using Earth’s magnetic field, which is about 1>400 as strong as the field used here, you would need to use an immense current of 400 * 2.27 A = 909 A. That’s not practical because a current of that magnitude would cause the wire in this example to melt! (Recall from Equation 18-24 in Section 18-6 that the power into a resistor with resistance R that carries current i is P = i2R. The wire in this example has a small cross-sectional area, so its resistance R will be fairly large. The power delivered to the wire by a 909-A current will quickly increase its temperature to above the melting point of copper, 1085°C.) So you needn’t worry about any of your electrical devices floating in midair when you turn on the current. A practical application of magnetic levitation is train design. By using magnetic forces to make a train float just above the track, the rolling friction between the wheels and the track can be completely eliminated and very high speeds achieved. (Magnetic forces are also used to propel the train forward.) A train of this type in commercial operation in Shanghai, China reaches a top speed of 431 km>h (268 mi>h). Such train lines require special magnets and wires capable of sustaining very high currents.