Page 300 - Engineering Electromagnetics, 8th Edition

282 ENGINEERING ELECTROMAGNETICS

(keeping the interior of the positive side of the surface on our left as usual), the
contribution E L across the voltmeter must be −Bνd, showing that the electric
field intensity in the instrument is directed from terminal 2 to terminal 1. For an up-
scale reading, the positive terminal of the voltmeter should therefore be terminal 2.
The direction of the resultant small current flow may be confirmed by noting that
the enclosed flux is reduced by a clockwise current in accordance with Lenz’s law.
The voltmeter terminal 2 is again seen to be the positive terminal.
Let us now consider this example using the concept of motional emf. The force
on a charge Q moving at a velocity v in a magnetic field B is

F = Qv × B

or
F = v × B (10)
Q
The sliding conducting bar is composed of positive and negative charges, and each
experiences this force. The force per unit charge, as given by (10), is called the
motional electric field intensity E m ,

E m = v × B (11)

If the moving conductor were lifted off the rails, this electric field intensity would force
electrons to one end of the bar (the far end) until the static field due to these charges
just balanced the field induced by the motion of the bar. The resultant tangential
electric field intensity would then be zero along the length of the bar.
The motional emf produced by the moving conductor is then

(v × B) · dL (12)
emf = E m · dL =
where the last integral may have a nonzero value only along that portion of the path
which is in motion, or along which v has some nonzero value. Evaluating the right
side of (12), we obtain
0
(v × B) · dL = νBdx =−Bνd
d
as before. This is the total emf, since B is not a function of time.
In the case of a conductor moving in a uniform constant magnetic field, we may
therefore ascribe a motional electric field intensity E m = v × B to every portion of
the moving conductor and evaluate the resultant emf by

(v × B) · dL (13)
emf = E · dL = E m · dL =
If the magnetic flux density is also changing with time, then we must include
both contributions, the transformer emf (5) and the motional emf (12),
∂B

emf = E · dL =− · dS + (v × B) · dL (14)
S ∂t