A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to both the rod and its velocity. If the length of the rod is doubled, what happens to the electric field in the rod?
Since it is a conducting metal, there is no net E field. Negative charge accumulates at one end of the rod, while the rod moves, but there is also a magnetic force on electrons that opposes the field created by displaced electron charge.
If a closed circuit were formed, so that electrons were free to move in a current loop, the EMF (E-field times length) would double if the length of the rod doubled. The E-field would stay the same. I believe the value would be B * (V/c), but am not sure.
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To determine what happens to the electric field in the rod when its length is doubled, we need to apply the principle of electromagnetic induction. According to Faraday's law of electromagnetic induction, whenever a conductor (such as the metal rod) moves relative to a magnetic field, an electric field is induced in the conductor. This induced electric field is responsible for the flow of electric current within the conductor.
The magnitude of the induced electric field (E) depends on three factors: the velocity of the conductor (v), the magnetic field strength (B), and the length of the conductor (L). Mathematically, E is given by the equation:
E = vBL
From the equation, we can observe that the induced electric field (E) is directly proportional to the length of the conductor (L). If the length of the rod is doubled (2L), then the induced electric field will also double in magnitude.
Therefore, when the length of the rod is doubled, the electric field in the rod also doubles.