Faradays Law and Moving conductor


Faradays Law and Moving Circuits/Conductor.

In the previous two lectures, we’ve been investigating Faraday’s law, which is written as

As discussed in Section 9.3, however, there is an alternate form of (1) that is sometimes used when the circuit moves in space. This is especially true if B FONT FACE="Symbol">` is also a function of time. Consider the circuit shown below:

This loop is traveling with velocity v (relative to the frame defining B FONT FACE="Symbol">` ) in a time-varying B FONT FACE="Symbol">` field, B FONT FACE="Symbol">` ((t) ). The emf measured by a high-impedance voltmeter connected to this circuit is given in (1). Charges in the wire are induced to move under the influence of a force given by the Lorentz force equation
F FONT FACE="Symbol">`=q(E FONT FACE="Symbol">` +V FONT FACE="Symbol">`´B FONT FACE="Symbol">`)
Motional and Transformer EMF
However, for an observer traveling with the loop, there is no apparent velocity. Instead, this observer would observe a force `F ' due only to an electric field `E `F ' = q `E The two forces F ` and `F ' must be equal, so that we find that
`E= `E ++ v `´ `B
Now, let’s substitute (4) back into Faraday’s law (1). After considerable simplification we find that

This is an alternate form to Faraday’s law (1). Important points to note in (5):
is called the motional emf. Circuit defined by contour c is said to “cut” the `B field lines, whether or not `B is constant or changing with time. (Note that it’s assumed here the shape of the circuit does not change with time.)
is called the transformer emf. (Notice that the time derivative appears inside the integral.) There is no motion present in this term.
. The same emf would be measured by a voltmeter whether it is moving with the circuit as in (5) or stationary as in (1) since

Either of these two equations (1) and (5) yield exactly the same emf (this will be illustrated in the two examples to be shown next.) However, one form may be preferred over another depending on the problem.
Example N4.1: Determine the induced resistor voltage V ` in the rotating loop circuit shown below using the form of F `araday’s law given in (5). The loop is immersed in the field `B =aˆY `B0 [T].

The direction of c was arbitrarily chosen as shown. From (5)

The first term is zero because `B is not a function of time.
We will consider contributions to( (v ` ´ B).dl from each of the four line segments that comprise the contour c:

The cross product of unit vectors can be computed by drawing a top-down sketch of the rotating loop:



Next, draw the equivalent lumped-element circuit:

By KVL: V++ emf= 0 Þ V =- -emf =wLWB sin(wt) [V]
We can interpret the results of the line integral of ((v ` ´ `B)) dl for each of the four segments in the context of which segments actually “cut” the B field lines. Segments `BC and D `A are in motion, but since these segments do not “cut” the B field lines, no emf is generated by their motion through `B . Their motion is always parallel to the `B field lines.
Conversely, the vertical section CD ` is “cutting” the `B field
lines as it rotates. Consequently, an emf is generated in this segment as it rotates through `B . Note that for Æ = 0 and Æ = 180º, this segment does not cut the `B field lines so no emf is generated near these angles
Example N4.2: Repeat the previous example, but use the form of Faraday’s law in (1).

As indicated in the figure on page 3, d `s Consequently

Similar to the previous example problem, this inner product of unit vectors can be evaluated by making a quick sketch:


which is the same answer as in the previous example, as expected. Next, draw the equivalent lumped-element circuit:

By KVL: V++ emf= 0 Þ V =- -emf =wLWB sin(wt) [V]
Again, this is the same answer as in the previous example, as expected. Note that here in Example N4.2 we have solved for V without resorting to motional emf and/or transformer emf, in contrast to the previous Example N4.1 (where we used only motional emf). Instead, we have just applied Faraday’s law (1), which is always valid.
Lastly, notice that when the loop is in the xz plane, ym is maximum, but V y 0! Further, when the loop lies in the yz plane, ym=0 but V is maximum! Very strange, what is happening?
Recall that the emf is not equal to ym in Faraday’s law. Rather, it is equal to the negative time-rate-of-change of ym . When the loop is in the xz plane:

while when the loop is in the yz plane: