Induced emf in a Conducting Rod Moving in a Uniform Magnetic Field:
In fig, the uniform magnetic field \(\vec B\) is represented by dots. Its direction is perpendicular to the plane of paper and is directed upwards. A conducting rod ab of length l is placed perpendicular to the magnetic field. Thus conducting rod is moved with velocity \(\vec V\) normally to both ‘ength l and magnetic field \(\vec B\).
The free electrons present in conducting rod also moves with velocity \(\vec V\) in magnetic field. Thus, magnetic force acts on these electrons whose value,
Here, q is the charge of electron. For a positive charge, the direction of \(\vec V\times \vec B\) is from b to a (from fig). Thus, for negative charge of electron, the direction of magnetic force in the conductor will be from a and b. Due to drift velocity of electrons, electrons are present in large amount on end b and it is lesser at end a. Due to lesser electrons at a, the end b is negatively charged and end a has more number of positive charge.
When both the ends of different charges of the rod come close, then an electrostatic force is developed between both the ends of the conducting rod. This process continues till the force on electrons due to electric field is not balanced by the force on electrons due to magnetic field.
If the developed electric field is E, then the force that acts on the electron of charge q,
i.e., \(\vec E\) is directed opposite to the direction of \(\vec V\times \vec B\) or is from end a to b in the conductor.
The magnitude of electric field, E = vB
Due to this electric field, induced emf or potential difference e is developed between the two ends of the conductor.
Thus, ε = Work done against the field in displacing the unit positive charge from one end to another = Force on unit positive charge × Displacement
ε = El
thus, ε = vBl …………. (3)
Here, l is directed from the negative terminal to the positive terminal.
