Physics · Electromagnetic Induction · NEET
| What changes | The conductor moves; field B is steady | The conductor is still; field B changes in time |
| Cause of EMF | Area A enclosed changes → flux changes | B changes → flux changes |
| Force on charges | Magnetic (Lorentz) force F = qv×B | Induced electric force F = qE |
| Shortcut formula | ε = Bvl (rod), ε = ½BωR² (rotating rod) | ε = A·(dB/dt) |
| General law (both) | ε = -dΦ/dt | ε = -dΦ/dt |
The MAGNETIC FIELD B is steady, but the FLUX Φ = B·A through the circuit still changes because the moving rod changes the enclosed area A. Faraday's law is about flux, not about B alone. So a steady B plus a growing/shrinking area gives a changing Φ, and that is the EMF. Formula: ε = Bvl.
In motional EMF the wire moves, so its free charges move with velocity v and feel the MAGNETIC (Lorentz) force F = qv×B. This magnetic force separates + and - charges along the rod. In changing-B EMF the wire is at rest, so v = 0 and there is no magnetic force on the charges; instead the changing B creates an INDUCED ELECTRIC FIELD E, and that electric force F = qE drives the charges. Different force, same result: an EMF.
They are the same law. ε = -dΦ/dt is the general Faraday law. For a rod of length l sliding at speed v in field B, the area swept per second is l·v, so dΦ/dt = B·l·v, which gives ε = Bvl. So Bvl is just Faraday's law worked out for the moving-rod case. You do not need two rules.
You can always use ε = -dΦ/dt; it never fails. Knowing the TYPE just helps you pick the fastest shortcut: if a conductor moves, use ε = Bvl (or ½BωR² for a rotating rod); if the field changes in time over a fixed area, use ε = A·(dB/dt). Same answer, less work.
Because a time-changing magnetic field sets up an induced electric field E in the surrounding space (this E can exist even where B is zero). That induced E field does work on charges around the whole loop. This is the deep reason NEET sometimes asks about EMF in a loop that is not itself in the field region.
A cycle wheel of radius 0.5 m is rotated with constant angular velocity of 10 rad/s in a region of magnetic field of 0.1 T which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is:
A rectangular wire loop of sides 8 cm and 3 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the plane of the loop. The EMF developed across the cut, if the velocity of the loop is 2 cm/s in a direction normal to the shorter side of the loop, will be:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Motional EMF is one way to produce an induced EMF — the one where the conductor moves through a steady field. The other way is a changing field acting on a stationary conductor. Both are 'induced EMF' and both obey Faraday's law ε = -dΦ/dt.
For a straight rod moving perpendicular to a field: ε = Bvl (B = field, v = speed, l = length of rod cutting field lines). For a rod rotating about one end: ε = ½ B ω R².
For a fixed loop of area A with B changing in time: ε = A · (dB/dt) (taking B perpendicular to the loop). More generally ε = -dΦ/dt with Φ = BA cosθ.
No — ε = -dΦ/dt always works. But spotting the type saves time: 'wire moves' means use Bvl; 'field changes' means use A·dB/dt. Read whether it is the conductor or the field that is changing.
The magnetic force is F = qv×B. In the changing-B case the wire is at rest, so its charges have v = 0 and qv×B = 0. The push comes instead from the induced electric field created by the changing B.
