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The magnetic flux density is constant, and the rod is being moved inside the field.

And if an emf is induced, how does the flux linkage change in this rod? The cross sectional area of this rod remains constant, so there shouldn't be a change in flux linkage. I feel like I'm getting something wrong here, but I just can't understand it.

Also, I would appreciate if the answers could be given in language a 12th grader can understand, I've looked at some of the other answers on here and I can't understand the maths of it all. I'm clear on integration, but we haven't done vector fields or even vector calculus yet, so I'd appreciate if those could be kept out (or simplified)

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Your problem statement is not entirely clear, but if the rod is perpendicular to the field, and is moving sideways across the field, then each free electron in the rod will be subject to a force (evB) pushing it toward one end of the rod. They will move until the separation of charge produces an electrostatic force opposing the magnetic force (eE = evB). E will be constant and the potential difference between the ends of the rod will be El. To observe or use this “emf” the ends of the rod wold have to be connected to external conductors so that it becomes part of a circuit. The flux through this circuit changes as the rod moves. One option would be for the rod to be rolling or sliding along parallel rails.

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  • $\begingroup$ thanks for the answer, but how would the flux through the circuit change? wouldn't the entire circuit be moving through the field, and so the cross sectional area of the circuit should remain the same. Are we assuming, that only the rod moves through the field, while the rest of the circuit remains stationary, thereby increasing the cross sectional area and hence the amount of flux through the circuit? $\endgroup$ – higuys Oct 5 '19 at 14:30
  • $\begingroup$ We are assuming that only the rod moves. $\endgroup$ – R.W. Bird Oct 7 '19 at 20:05

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