1. May 11, 2008

### pmennen

The answer to this question (attached gif) is supposed to be "E".
I can get bL^2/R if I just compute the flux and differentiate to
compute the induced voltage and current, but how does one
figure out that it goes counter-clockwise. Also I'm at a loss to
explain why whatever effects the field has on the left side of the
loop wouldn't cancel out the effects on the right side of the loop.
(i.e. I was also tempted to answer "0"). Thanks in advance for any
insight. (Sorry if this is a double post, the 1st one seems lost?)

~Paul

#### Attached Files:

• ###### Q56.gif
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2. May 11, 2008

### Redbelly98

Staff Emeritus
The induced current produces a magnetic field in addition to the external field "B". This induced field will be opposite to the change in B (not necessarily opposite to B itself). What direction must the current flow to produce a field in the opposite direction of $$\frac{dB}{dt}$$ ?

You can look up Lenz's Law in your textbook for more discussion of this.

3. May 11, 2008

### pmennen

Lenz's law

Ok, yes I'm aware of Lenz's law and I can understand which direction the
induced current goes when you push a magnet into a loop of wire. But this
situation is a bit different. So when the magnetic field in this problem decreases,
it decreases both inside and outside of the loop. So the counter clockwise direction
indeed counters the change inside the loop, but it goes in the same direction of the
change outside the loop. So it didn't seem to me that Lenz's law tells me the

4. May 11, 2008

### Tedjn

What do you mean by inside and outside the loop? The magnetic field runs through the loop (at least through an open surface attached to the loop). If it helps, picture the magnetic field coming out of the end of a bar magnet.

5. May 12, 2008

### pmennen

I don't understand your question. The problem shows a "uniform" magnetic field shown by the dots on the page. Note that there are dots both inside the loop of wire and outside, and they are uniformly spaced. That is not the same as the field coming out of a bar magnet that is being pushed thru the loop. In that case the field would be string near the end of the magnet and would fall off in strength farther away from the magnet.

~Paul

6. May 12, 2008

### Tedjn

Faraday's law relates induced emf in a loop to the change in electric flux across an open surface attached to that loop. Magnetic field lines that do not cross that open surface--in other words ones that do not cross the loop--do not affect induced emf. Thus, they do not affect the current. I understand that your question is why magnetic field lines outside the loop do not seem to affect the current. Maybe there is a more sophisticated answer in advanced physics that I don't know that someone else can fill you in on, but empirical evidence supports Faraday's law the way it is.

Last edited: May 12, 2008
7. May 12, 2008

### pmennen

I wasn't really trying to question Faraday's law at this stage of my limited understanding. My problem was I hadn't read it carefully enough to understand what it was saying. Your answer however has cleared it up for me nicely. Many thanks!

~Paul

8. May 12, 2008

### misho

This can generally be shown with Gauss' Law (or Divergence Theorem), or in this 2-D case, with Green's Theorem, which implies that what happens along a closed contour is dependent on what happens in the area within the contour. If you are comfortable with vector calculus, take a look at the wikipedia entry, and it should at least become clear in a mathematical sense, if not intuitively.