<< Chapter < Page Chapter >> Page >
Figure shows a conducting rod that is pushed to the right through the perpendicular magnetic field at constant velocity. The resulting change in the magnetic flux induces a current in the circuit.
A conducting rod is pushed to the right at constant velocity. The resulting change in the magnetic flux induces a current in the circuit.

From an energy perspective, F a produces power F a v , and the resistor dissipates power I 2 R . Since the rod is moving at constant velocity, the applied force F a must balance the magnetic force F m = I l B on the rod when it is carrying the induced current I . Thus the power produced is

F a v = I l B v = B l v R · l B v = l 2 B 2 v 2 R .

The power dissipated is

P = I 2 R = ( B l v R ) 2 R = l 2 B 2 v 2 R .

In satisfying the principle of energy conservation, the produced and dissipated powers are equal.

This principle can be seen in the operation of a rail gun. A rail gun is an electromagnetic projectile launcher that uses an apparatus similar to [link] and is shown in schematic form in [link] . The conducting rod is replaced with a projectile or weapon to be fired. So far, we’ve only heard about how motion causes an emf. In a rail gun, the optimal shutting off/ramping down of a magnetic field decreases the flux in between the rails, causing a current to flow in the rod (armature) that holds the projectile. This current through the armature experiences a magnetic force and is propelled forward. Rail guns, however, are not used widely in the military due to the high cost of production and high currents: Nearly one million amps is required to produce enough energy for a rail gun to be an effective weapon.

Figure shows a schematic drawing of the rail gun. An armature is placed between two rails of opposite charge. Magnetic field is generated by currents in rails and interacts with the current in armature, generating the force.
Current through two rails drives a conductive projectile forward by the magnetic force created.

We can calculate a motionally induced emf    with Faraday’s law even when an actual closed circuit is not present . We simply imagine an enclosed area whose boundary includes the moving conductor, calculate Φ m , and then find the emf from Faraday’s law. For example, we can let the moving rod of [link] be one side of the imaginary rectangular area represented by the dashed lines. The area of the rectangle is lx , so the magnetic flux through it is Φ m = B l x . Differentiating this equation, we obtain

d Φ m d t = B l d x d t = B l v ,

which is identical to the potential difference between the ends of the rod that we determined earlier.

Figure shows a conducting rod of the length l that is pushed to the right for the distance x through the perpendicular magnetic field at constant velocity.
With the imaginary rectangle shown, we can use Faraday’s law to calculate the induced emf in the moving rod.

Motional emfs in Earth’s weak magnetic field are not ordinarily very large, or we would notice voltage along metal rods, such as a screwdriver, during ordinary motions. For example, a simple calculation of the motional emf of a 1.0-m rod moving at 3.0 m/s perpendicular to the Earth’s field gives

emf = B v = ( 5.0 × 10 −5 T ) ( 1.0 m ) ( 3.0 m/s ) = 150 μ V .

This small value is consistent with experience. There is a spectacular exception, however. In 1992 and 1996, attempts were made with the space shuttle to create large motional emfs. The tethered satellite was to be let out on a 20-km length of wire, as shown in [link] , to create a 5-kV emf by moving at orbital speed through Earth’s field. This emf could be used to convert some of the shuttle’s kinetic and potential energy into electrical energy if a complete circuit could be made. To complete the circuit, the stationary ionosphere was to supply a return path through which current could flow. (The ionosphere is the rarefied and partially ionized atmosphere at orbital altitudes. It conducts because of the ionization. The ionosphere serves the same function as the stationary rails and connecting resistor in [link] , without which there would not be a complete circuit.) Drag on the current in the cable due to the magnetic force F = I B sin θ does the work that reduces the shuttle’s kinetic and potential energy, and allows it to be converted into electrical energy. Both tests were unsuccessful. In the first, the cable hung up and could only be extended a couple of hundred meters; in the second, the cable broke when almost fully extended. [link] indicates feasibility in principle.

Practice Key Terms 1

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 2' conversation and receive update notifications?

Ask