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epsilonjon
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[PLAIN]http://img826.imageshack.us/img826/2928/19002018.jpg [Broken]
Suppose I have a slidewire generator as in the image above.
If I move the sliding rod to the right I generate a current in the loop, directed anticlockwise. It takes work to move the rod, since the induced current is interacting with the magnetic field and experiencing a force to the left. The rate at which work is done by the applied force is equal to the rate of energy dissipation in the circuit, since you can't violate energy conservation.
Is it not the case here that the magnetic force is doing (negative) work on the current-carrying rod as it moves? I do not understand this, as the magnetic force on a moving charge is always perpendicular to its velocity, and so can never do work.
This caused me to go over the derivation of the [tex]\vec{F}=I\vec{L}\times\vec{B}[/tex] equation for the force on a current-carrying wire in a magnetic field. The few books I have all use the same sort of method, whereby they take the magnetic force on an electron at some instant then multiply it by the total number of electrons in the wire. Now I am thinking how can this be correct, as they're implying that the force on the electrons in the wire will always be in this same initial direction, contradicting what they've said earlier about moving charges in a magnetic field undergoing circular motion?
If someone could straighten this all out for me I would really appreciate it, as I don't want to keep reading before I understand this properly.
Many thanks!
Jon.
Suppose I have a slidewire generator as in the image above.
If I move the sliding rod to the right I generate a current in the loop, directed anticlockwise. It takes work to move the rod, since the induced current is interacting with the magnetic field and experiencing a force to the left. The rate at which work is done by the applied force is equal to the rate of energy dissipation in the circuit, since you can't violate energy conservation.
Is it not the case here that the magnetic force is doing (negative) work on the current-carrying rod as it moves? I do not understand this, as the magnetic force on a moving charge is always perpendicular to its velocity, and so can never do work.
This caused me to go over the derivation of the [tex]\vec{F}=I\vec{L}\times\vec{B}[/tex] equation for the force on a current-carrying wire in a magnetic field. The few books I have all use the same sort of method, whereby they take the magnetic force on an electron at some instant then multiply it by the total number of electrons in the wire. Now I am thinking how can this be correct, as they're implying that the force on the electrons in the wire will always be in this same initial direction, contradicting what they've said earlier about moving charges in a magnetic field undergoing circular motion?
If someone could straighten this all out for me I would really appreciate it, as I don't want to keep reading before I understand this properly.
Many thanks!
Jon.
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