Induced Electric Fields and Motional EMF
A time-varying magnetic field produces a circulating electric field, even without a conductor. A conductor moving through a static magnetic field can instead develop motional emf through magnetic forces on its charges. Force and flux-change calculations agree for a rod sliding on rails.
What a learner can do afterwards
- Computes the EMF of a rod moving on rails in two ways and shows they agree
- Describes the induced electric field around a region of changing flux, with no conductor needed
- Explains eddy current braking as induction plus Lenz's law
1 · Read
A rod of length l slides on rails at speed v, perpendicular to the rod. Uniform B is normal to the rail plane. Here v cross B points along the rod by the right-hand rule, perpendicular to v and B, with magnitude v B. Magnetic force q(v cross B) separates charges. Dividing force by charge gives v B; integrating this constant along length l means multiplying by l, giving emf B l v. Loop area l x gives flux B l x, whose rate of change is B l v.
Use SI units and multiply B, l and v for the stated perpendicular geometry. A 0.50 m rod at 2.0 m/s in a 0.40 T field has emf magnitude 0.40 times 0.50 times 2.0 = 0.40 V. The formula gives a magnitude; use v cross B or Lenz's law to determine polarity.
Circulation, integral E dot dl, sums each field component along the path times the small segment length. For a fixed closed path it equals -dPhi/dt. Use the right-hand rule for path direction and surface normal; add signed flux-change rates before applying the minus sign. Changing B induces E without a wire. Electrostatic circulation is zero. Circular field lines need circular symmetry. A moving loop's emf also includes integral (u cross B) dot dl.
In the perpendicular rod-and-rails arrangement, emf magnitude is proportional to speed. Doubling v with B and l fixed doubles B l v. For other orientations, integrate (v cross B) dot dl along the rod. Motion perpendicular to B alone does not guarantee B l v: the rod must also align with v cross B.
A conducting sheet entering or leaving a magnetic-field region has changing flux through internal current paths. Eddy currents then produce magnetic forces opposing the motion. Resistance converts mechanical energy to heat. A sheet translating wholly inside a uniform, steady field need not develop continuing eddy-current drag. Slots or insulated laminations interrupt large current paths.
For a closed rod-and-rails circuit with total resistance R, negligible inductance and no other source, I = B l v / R in the perpendicular geometry. First calculate the emf, then divide by R. A 0.30 m rod at 4.0 m/s in 0.50 T gives 0.60 V; with R = 3.0 ohms, the current is 0.20 A.
In uniform B, a rod aligned with v cross B and moving perpendicular to B has emf B l v. Changing B induces electric-field circulation. Eddy-current drag converts motion to heat.
Original lesson, questions and flashcards authored for this learning dataset.
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