Motional Emf

In order to help answer this question, let us consider a simple circuit in which a conducting rod of length slides along a U-shaped conducting frame in the presence of a uniform magnetic field. This circuit is illustrated in Fig. 36. Suppose, for the sake of simplicity, that the magnetic field is directed perpendicular to the plane of the circuit. To be more exact, the magnetic field is directed into the page in the figure. Suppose, further, that we move the rod to the right with the constant velocity .

The magnetic flux linked by the circuit is simply the product of
the perpendicular magnetic field-strength, , and the area of the circuit,
, where determines the position of the sliding rod.
Thus,

(202) |

(203) |

Thus, the emf generated in the circuit by the moving rod is simply the product of the magnetic field-strength, the length of the rod, and the velocity of the rod. If the magnetic field is not perpendicular to the circuit, but instead subtends an angle with respect to the normal direction to the plane of the circuit, then it is easily demonstrated that the

(205) |

Since the magnetic flux linking the circuit *increases* in time, the
emf acts in the *negative* direction (*i.e.*, in the opposite sense to the
fingers of a right-hand, if the thumb points along the
direction of the magnetic field). The emf, , therefore, acts in the
*anti-clockwise* direction in the figure. If
is the total resistance of the circuit,
then this emf drives an anti-clockwise electric current of magnitude
around the circuit.

But, where does the emf come from? Let us again remind ourselves what an
emf is. When we say that an emf acts around the circuit
in the anti-clockwise direction, what we really mean is that a charge
which circulates once around the circuit in the anti-clockwise direction
acquires the energy . The only way in which the charge
can acquire this energy is if something does *work*
on it as it circulates.
Let us assume that the charge circulates very *slowly*. The magnetic
field exerts a negligibly small force on the charge when it is traversing the
non-moving part of the circuit (since the charge is moving very slowly).
However, when the charge is traversing the moving rod
it experiences an *upward* (in the figure) magnetic force of magnitude
(assuming that
). The net work done on the charge by this force as
it traverses the rod is

(206) |

But, if we think carefully, we can see that there is something
seriously wrong with the above explanation.
We seem to be saying that the charge acquires the energy
from the *magnetic field* as it moves around the circuit once in the
anti-clockwise direction. But, this is impossible, because a magnetic field
*cannot* do work on an electric charge.

Let us look at the problem from the point of view of a charge
traversing the moving rod. In the frame of reference of the rod,
the charge only moves very slowly, so the magnetic force on it is negligible. In fact, only an electric field can exert a significant
force on a slowly moving charge. In order to account for the motional emf generated
around the circuit, we need the charge to experience an upward force of
magnitude . The only way in which this is possible is if the charge
sees an upward pointing *electric field* of magnitude

(207) |

More generally, if a conductor moves in the laboratory frame
with velocity in the presence of a magnetic field then
a charge inside the conductor experiences a magnetic force
. In the frame of the conductor, in which the charge is
essentially stationary, the same force takes the form of an electric
force
, where is the electric field in
the frame of reference of the conductor. Thus,
if a conductor moves with velocity through a magnetic field
then the electric field which appears in the rest frame of the conductor
is given by

(208) |

We can now appreciate that Faraday's law is due to a combination of two apparently distinct effects. The first is the space-filling electric field generated by a changing magnetic field. The second is the electric field generated inside a conductor when it moves through a magnetic field. In reality, these effects are two aspects of the same basic phenomenon, which explains why no real distinction is made between them in Faraday's law.