Let’s look in detail at each of these four ways of using flux change or motion to produce an emf. Method 3…

Example 3 of motional emf: moving conductor in B field. Motional emf is the emf induced in a conductor moving in a magnetic field. “up” v B B If a conductor (purple bar) moves with speed v in a magnetic field, the electrons in the bar experience a force       –                               ℓ                         + The force on the electrons is “up,” so the “top” end of the bar acquires a net – charge and the “bottom” end of the bar acquires a net + charge. The charges in the bar are “separated.” This is a simplified version of reality, but it gives you the right “feel.”

The separated charges in the bar produce an electric field pointing “up” the bar. The emf across the length of the bar is The separated charges in the bar produce an electric field pointing “up” the bar. The emf across the length of the bar is “up” The electric field exerts a “downward” force on the electrons: v B B       –                           E     ℓ An equilibrium condition is reached, where the magnetic and electric forces are equal in magnitude and opposite in direction.                         +

Homework Hint! When the moving conductor is “tilted” relative to its direction of motion (when is not perpen-dicular to the conductor), you must use the “effective length” of the conductor.* v B B             –                          ℓeffective             +             ℓeffective = ℓ sin  if you use the if you define  as the angle relative to the horizontal Caution: you do not have to use this technique for homework problems 29.25 or 29.28 (if assigned this semester). Be sure to understand why! *This is because the magnetic force is , and not directed along the conductor. Let’s not worry about showing this.