Motional emf
Suppose a straight conducting rod of length L moves through a uniform B field with its
velocity perpendicular to B as shown below.
X
X
X
X
X
X
X
X
G
G G
Then the conduction electrons feel a force given by F = ev × B and they migrate along the wire
until they quickly reach an equilibrium with
eE = evB or E = vB.
The electric field is related to the potential difference across the wire as ∆V = EL, so that
∆V = EL = vBL
so that as long as the rod moves through the magnetic filed, there is a potential difference across
the rod.
If the rod is part of a closed circuit loop so that the charges can flow around a loop, then
the magnetic flux through the closed loop is changing, since the area A changes in
G G
Φ B = ∫∫ B ⋅ dA
and we have that the induced emf is given by
d [ BA]
dΦB
dA
dx
ε =−
=−
= −B
== BL
= − BLv
dt
dt
dt
dt
using the labeling in the figure.
X
X
X
R
X
L
X
X
X
X
x
Then the induced current in the loop is given by I = ε/R = BLv/R. As a result of the induced
G
G G
current there will be a magnetic force on the rod to the left according to F = IL × B . An external
force equal to this and directed to the right is needed to maintain the constant velocity motion of
the rod. Furthermore, the input energy from this external force is dissipated in the resistor from
the induced current and P = I2R. This can be seen from calculating I2R.
B 2 L2 v 2  ε 2 
Pdissipated = I 2 R =
 =  = [ ILB ] v=Fext v=Pinput
R  R
Thus we have mechanical energy being converted to electrical energy and then to thermal
energy.