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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.
If a conductor (purple bar) moves
with speed v in a magnetic field,
the electrons in the bar experience
a force
FM = qv  B = - ev  B
B 




v
–



+










“up”
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
ε= E
The electric field exerts a
“downward” force on the electrons:
FE = qE = - eE
An equilibrium condition is reached,
where the magnetic and electric
forces are equal in magnitude and
opposite in direction.
ε
evB = eE = e
 ε =B v
“up”
B 




v
–


E

+










ℓ
Homework Hint!
When the moving conductor is
“tilted” relative to its direction of
motion (when v is not perpendicular to the conductor), you must
use the “effective length” of the
conductor.*
V  E   E
 E
B   v

  

+


–
 ℓ
effective


cos
effective
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.