3
$\begingroup$

wire cutting field lines

The magnitude of an induced E.M.F, according to Faraday's law, is proportional to the rate of change of magnetic flux linkage.

As the field is uniform, the field strength, or magnetic flux density, is identical throughout; with this lack of a change in flux linkage, why is a voltage still induced?

Is it due to the fact that although the magnitude stays constant, the actual linkage is changing, due to a change in location? What would this change in flux linkage be if the magnitude of magnetic flux density is the same?..

|cite|improve this question
$\endgroup$
  • $\begingroup$ A conductor moving in a magnetic field gets polarized. This creates an E.M.F which can drive a current when the circuit is completed. $\endgroup$ – Yashas Feb 13 '17 at 11:34
  • $\begingroup$ I'd very much appreciate some elaboration. $\endgroup$ – Tobi Feb 13 '17 at 11:35
  • $\begingroup$ @Yashas Samaga Why is a conductor polarised, what does this mean? Could you apply this to my actual question regarding a wire, flux linkage and density. $\endgroup$ – Tobi Feb 13 '17 at 11:40
  • $\begingroup$ Can you make your question more specific using a diagram? How is the wire traversing in the magnetic field? $\endgroup$ – Yashas Feb 13 '17 at 11:43
  • $\begingroup$ Do you know about Lorentz force? $\endgroup$ – Lapmid Feb 13 '17 at 14:21
1
$\begingroup$

When a charged particle is moving in a magnetic field, a force known as the Lorentz force acts on the particle. The magnitude and the direction of the force are given by the following equation:

$$\vec{F} = q(\vec{E} + \vec{v}\times \vec{B})$$

Consider a piece of conducting wire. The conducting wire consists of freely moving electrons which experience a force in the presence of a magnetic field. However, there won't be any force acting on the wire as a whole as the electrons are moving in random directions and the net force adds up to zero.

When you move the wire in a particular direction, the vector sum of the direction of motion of the electrons in the conductor adds up to give a finite value. Hence, when you sum the forces up vectorially, you get a net force.

This causes the electrons to move in a particular direction which in turn results in a separation of charge as shown in the diagram.

enter image description here

This establishes an electric field along the wire which prevents recombination of the charges. Hence, there exists an E.M.F along the wire.

When you complete the circuit by joining wires at the ends of the rod, the electric field which was preventing the charges from recombining drives a current in the circuit.

|cite|improve this answer
$\endgroup$

protected by ACuriousMind♦ Nov 25 '17 at 19:46

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.