Magnetic fields: clearing up some confusion

difference between magnetic field and B-field? What 's the difference between those two?

why the direction of magnetic field of a loop is a circle and the professor use the word" B field" to refer to the direction at a certain point which is tegant to the circle?

Magnetic field and "B-field" are the same. Since we usually use "B" to denote magnetic field, we sometimes refer to the magnetic field as the "B-field". Consider it physics slang. As confusing to the uninitiated as OMG, LOL. ;)


Now, for you second question, if we consider the magnetic field around a current carrying wire, we know that the magnetic field forms a closed loop around the wire. But the magnetic field is a vector, right? So those vectors are tangential to these magnetic field lines and represent the vector field. Those lines just make it a little easier to visualize (without too many crazy arrows all over the place, because I drew 3 and that doesn't look too bad, but to truly represent it we need one for every point along that circle as finely as we are interested in).

to clarify the meaning of "perpendicular distance" when talking about B-fields.

perpendicular distance is the distance from the wire to the point in the magnetic field we are interested in (the angle between this line and the wire does not have to be 90).

Please correct if needed.

The perpendicular distance does indeed need to be perpendicular. I think you are still referring to the magnetic field from a straight segment of wire. In this case, the magnetic field produced is uniform along the wire, and varies radially away from it (~1/r). Since the only variation is radially, you only need this radial distance (which is always perpendicular to the wire).

Please let me know if you are concerned about a different situation.

Force of B-Field of one current-carrying wire on another.

Example: one horizontal wire (I1) has a current flowing to the right. Another, perpendicular to it, is above it with a current flowing upwards. The direction of the force of I1 is downward, by RHR. But, the direction of the force acting on I2 is to the right, because, though the direction of B1 is taken into consideration, the direction of the force is based on the direction of I2 and not I1. Why?

Ahh! But you have already taken into account the direction of I₁ when determining the direction of the magnetic field (B₁) produced by it. You then consider the force exerted by this magnetic field on the second wire, so you must then consider the direction of I₂.

I hope that clears up some confusion concerning magnetic fields.

Question: clarification of lecture notes on magnetic fields

I would like to have clarified a number of points regarding the Magnetism slides.

1) Regarding B and decrease of B as distance from I increases.

The relationship between B and distance from I (r) is B ~ 1/r (re: Capa 8 q 1a). However, in the class slides (slide 1 p. 4) it says

" delta B ~ 1/r2 (r squared)"

What r is this? I notice this is not the perpendicular r, but in the following delta B equation, sin theta is multiplied, and thus becomes the perpendicular r. Confusion?

2) Slide 2 p. 5, for a circular clockwise I, P1 within the circle points into the page, while P2 outside the circle points out of the page. Is it that B's outside a circular I are always in the opposite direction as that in the I?

3) slide 4 p. 3. The current loops in the diagram show a B opposite to RHR. Is it because the loops are electron orbitals, and electrons produce B's opposite to RHR?

4) similar question as previous. slide 1 p. 6. why is the B opposite to the RHR?

Thank you.

Regarding your first question, while there is an apparent inconsistency here, both are correct. The problem is one expression is for "B" and one is for "ΔB". To get B~1/r you have to integrate "ΔB" over the particular geometry appropriate to the problem. This is why you get different expressions for the B-field generated by current in a straight wire, a loop of wire, etc.

The "r" being referred to for "ΔB" is the distance from the point on the wire generating the field when integrating. Since all parts of the wire contribute to the magnetic fields in all parts of space you have to consider "off-axis" contributions as well as those perpendicular. Once the expression is integrated, for example in the case of a straight current carrying wire, the r will refer to the perpendicular distance away since the field will be symmetric and constant along the length of the wire anyhow. Otherwise, it might be important!

For your second question, I assume you are using a right hand rule where you curl your fingers in the direction of the current and your thumb tells you the direction of the magnetic field inside the loop. You can also point your thumb in the direction of the current and curl your fingers around to determine the direction of the magnetic field around the wire. If you do this you will quickly see that your fingers point up on the outside of the loop and down on the inside of the loop no matter what part of the loop you look at. And yes, the field should point in the opposite direction on the outside of the coil compared to the inside. This is because magnetic field lines must always close (the field lines would look like rings around the loop). It might help you to take a look at The "Right-hand Rules" and Magnetic fields.

For the last two questions, so far as I can tell, the directions are as given by the RHR. Pg. 3 slide 1, I get B pointing right, and for slide 4, I get B pointing left. It's not so easy to see which way the loops are supposed to be coming out of the page... look for where the breaks in lines are, indicating that that part of the line is passing behind another line.

Hope that helps!

Topic open: Magnetostatics

The following topic is now open for questions: Magnetic fields.

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Not related to electromagnetism, but some good clean Physics Phun

Professor of Physics at MIT Peter Fisher helps Conan O'Brien with the ring spin record...





I just had to share when I saw this.

Edit: Why do you think too much vaseline would be a bad thing for the ring spin? Discuss...

found via physicsknits blog

Topic open: DC Circuits

The following topic is now open for questions: Circuits. Anything about resistors, capacitors, currents, batteries, etc.

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Question: Electric PE could be 0? This is very distrurbing...

I know we have down tons of questions where we have to find where V is zero (or similar like it..)

I never made the connection until now that V =0 means the electric potential energy at that point is 0 too! This is deeply disturbing.

Does this mean that if say a charge move between two other charges.. suddenly, middle of the way, it losts all of its potential energy? It is certain possible for V to be zero, which would imply electric PE be 0 as a consequence.

I am so used to the potential energy in a gravitational sense, where potential energy is 0 only when you hit the bottom.. This is like jumping off the a building then middle of the way, finding all of your potential energy all gone of a sudden. Where did it go?

Disturbing indeed.

Rest assured, physics is not broken, but there are two interesting points to be made here.

Firstly, while the analogy to gravitational energy is a good one, it is incomplete due to the fact that there is no negative mass, but there are indeed negative charges. This means that the potential surface can dip below zero such that a local minimum in potential energy may very well be "beyond zero". Really what the universe seeks is minima, not "zero", especially since we can arbitrarily choose where zero is (think about gravitational energy and choosing the top of a cliff to be at zero PE... one still loses PE if one falls to the bottom).

Secondly, what matters in terms of motion is force, not PE, and that is related to the slope of the potential surface, not the "value" of it.

Let's look at a potential map:



I started this off with some small rectangular regions at set voltage (these could be metal plates held at a potential for example) and then calculated the potential in between. We can see that there are some "hills" (white, or lighter yellow) and some "valleys" (darker shades) and a gradient in between representing a smooth potential surface (ignore the pixellation, I didn't calculate this on a very fine mesh).

In particular, take a look at the region between the highest positive potential and the lowest negative potential. We can see that there is an equipotential line running between them at V=0. But, if we think of this like an elevation map that V=0 crossing is on the middle of a hill! A positive charge (we would invert the map for a negative charge) moving across this potential surface would act like a marble rolling on an elevation map. If we put it on top of the white hill of +2V it would roll down into the -5V valley, still experiencing a force due to the electric field as it passed through 0V.

So, "zero potential" is a somewhat arbitrary thing, and because of negative charges it comes up naturally more often for electrostatic potential that you might be used to from gravitation. Also, the force, the thing which drives motion, is related to the slope (really a directional kind of slope which senses the steepest change).

The very useful thing about electrostatic "potential" is that it gives a convenient way to map out the landscape that a charge will "see" without reference to that charge, just as electric field maps out the force a charge would experience without reference to that charge, and the two are related by slope. Such a map can give an intuitive feel for what will happen when a charge is introduced even for complex arrangements of charges or plates.

I hope that helps. This is far from a complete explanation, so feel free to stimulate further discussion in the comments here!

Topic open: Electric fields, potentials and capacitors

The following topic is now open for discussion: Electric fields, potentials, capacitors and parallel plates

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Electric fields and electrostatic potential

I've had an interesting discussion regarding Electric field and Electric potential and thought I would repost it here so it wasn't buried in the comments of an old post...

In relation to the post Question: E=V/d or V=-Ed? (to minus or not to minus), Peter asked:

Hey, I was just reading Cutnell & Johnson Physics 7th edition (Competitor textbook to Giancoli) and on pg 585 it had some comments about this. I am not sure if it is applicable?

They first derived the -form of the equation. So they said
W = Fd =qEd
But W = -PE

qEd = -PE

qEd/q = -PE/q

Ed = -V

V = -Ed

Ok so far so good.

then they had some comments about the form of the equation where V = Ed (no negative sign)

"When applied strictly to a parallel plate capacitor, however, this expression is often used in a slight different form. In figure 19.16, the metal plates of the capacitor are marked A( higher potential) and B (lower potential). Traditionally, in discussions of such a capacitor, the potential difference between the plates is referred to by using the symbol V to denote the amount by which the higher potential exceeds the lower potential. V = Va-Vb.

Thus,

E = -V/d = - (Vb-Va)/d = (Va-Vb)/d = V/d.

Would this be a valid explanation as well since in this course we are mostly dealing with parallel plate capacitors? Thanks.

Hi Peter,

Excellent idea to check out another text book, sometimes a different perspective is all you need...

However, I disagree with Cutnell's approach here. They are essentially taking advantage of two negatives which cancel, and I think it is confusing. It is true that, as a sort of shorthand, the potential difference between the plates is simply given as the amount by which the higher potential exceeds the lower. The problem I see with using this "no minus" version of the equation is that it does not represent the true relationship between E and V (ie. that the electric field points from the high potential to the low potential). This takes a relationship and turns it into an equation, which I dislike. Equations are limited in scope and easy to misapply, relationships can give you better understanding which is a much more solid basis.

I'd rather see you sketch a diagram showing the high and low potential and where the positive and negative charges are separated on the two plates and applying the "V" given as the magnitude of the potential difference. It will be much harder to go wrong with this picture in front of you, and the whole "-" issue more or less disappears.

I hope that helps.

Gee you are totally right. I posted this early this morning and now I already have a different idea, which you have already hinted here but I want to make sure I got this down 100%.

This just came to me.



Is this way of thinking correct?

Thanks!

Which is exactly right. In fact, the "full" form (using vector calculus) is given by:



which essentially says to add up (integrate) the components of the electric field parallel to the path taken (the dl). If the electric field is constant and the path is straight, then everything reduces to what we have above.

Conversely, the electric field is proportional to a sort of directional slope of the potential (called the gradient) such that the greatest forces will be felt by charges in the "steepest" regions of electrostatic potential. More on that later perhaps...

Topic open: Electric Potential and Potential Energy

The following topic is now open for discussion: Electric Potential and Potential Energy.

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Topic open: Forces on charges and Electric fields

The following topic is now open for discussion: Forces on charges and Electric fields.

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New Example! Chapter 16 #15

Now posted on the main WebTA site:

Chapter 16, #15

Before heading over to check it out, I suggest you try it yourself and then take a look.

Welcome to the physics 102 webTA blog: winter '08

Welcome back to a new term!

I will be continuing as the webTA for physics 102 for the 3rd time now. As with every other year, I hope that I can continue to make this online resource more and more useful. If you have any thoughts or suggestions on how to improve the online materials, I am always open to suggestions. You can contact me through the blog, discussions or by email (webta101 at physics.mcgill.ca) with your ideas.

This year the webTA persona will take on three aspects:

1. the static site with examples and explanations
   
2. this blog, where answers to your most common questions will be answered in some detail
   
3. the WebCT discussion forum, to provide a more interactive environment for providing answers
   

For the sake of my email inbox, I will ask you to refrain from emailing me questions. Instead, I suggest you either post your question to the discussion boards (where you may find a friend will answer before I get a chance!) or post your question in one of the "topic open" posts as a comment on the blog. There are a few groundrules for asking questions that I suggest you check before asking. This has worked out for two years so far, so let's try to stick with it. Also, while the WebCT discussion board will always leave your name, anything posted to the blog will remain anonymous. You can post questions here anonymously and when I post answers I will never include your name. I hope this encourages you to ask any question you want even if you might feel it is "stupid". Often those solve the most trouble and end up being the most worthwhile.

One great thing about you being the 3rd year to have access to this is that there is a considerable compilation of previous students' question now on the blog. I'm hoping to move a few of these examples over to the static site, but for now if you are having a problem I suggest you check backposts of the blog for possible answers. I've tried to label all the posts with their relevant topics and a list is available in the sidebar under "post topics". You can also use the search bar at the top and hit "search blog" to search within the archives. You may not find exactly what you are looking for, but it might give you a starting point, and a better place to start asking questions.

A couple more things: I may (likely) not be able to answer every single question (the webTA is a single person, me, and not a collection of TA's and I only have so many hours to work on this). My strategy in the past has been to post about the most common problems (eg: 2 people ask about the same problem, and someone else asks about a related concept). This means that you can't solely rely on the webTA for help. Which brings me to my other point: I hope that you will see this as an additional resource and not a replacement for the regular tutorials. It is often easier to sit down face-to-face with a real TA to work out what you are having trouble with. Make use of whatever resources you need.

Good luck, and I look forward to a good semester working with you!

Sarah :)

Good Luck and signing off!

Your friendly neighbourhood webTA here wishing you good luck on your exam tomorrow! Wise advice from Douglas Adams and the Hitchiker's Guide to the Galaxy:

DON'T PANIC!

Relax, try to stay clear-headed, and don't stay up all night studying, get some sleep.

Good luck to you all.

Question: AC circuits and resonance

This question relates to Lab 6 but also to the lecture material (it was in the lab that I figured out that my understanding of the material was totally wrong).

It makes sense that in an A/C circuit the capacitor charges when there is a voltage and then discharges through the inductor when the voltage goes away. Also, it seems that the capacitor may or may not reach it's max charge depending of the frequency of the A/C generator (is that correct?). But why does the charging/discharging create a pattern with deteriorating voltage? It seems like the charge in the capacitor should just go up and down to the same extream values over and over. And how is this related to the resonance of the circuit at all? For some reason the frequency of these oscillations is the same as the resonance frequency, but it seems like this shouldn't be the case at all, I would have thought they would be very different values.

Let's see if I can help...

Your first comment is regarding the decay in the oscillations in the LC circuit. It's been awhile since I was a lab demonstrator, so I don't recall the setup exactly for this experiment, however, it sounds like there is a resistance in the circuit (whether there is a resistor there or not... it could be the fact that you have a "real" inductor which consists of a length of wire and will have some resistance). In that case, you have a damped harmonic oscillator system. The inductor and capacitor act as an oscillator (where the capacitor acts like a mass, and the inductor provides a restoring force like a spring), but the resistance provides a means for energy to leave the system, so the oscillations will decay. An analogous mechanical system would be a mass on a spring in molasses... for an entertaining visual. As for what you think it "should" do, you are correct, in that if you had a perfect capacitor and a perfect inductor with no resistances, the system would keep on oscillating forever.

As for resonance... this seems to confuse a lot of people every year. Here's a question and answer from a couple of years ago just to define what we are talking about:

Q: I am a little confused about frequency. What exactly is the resonant frequency and how and why does it effect the current compared to the frequency?

A: In an AC circuit, depending on the components in the circuit, the impedance (a sort of complex resistance) may depend on the frequency. Thus, if an oscillating voltage is applied, the amplitude of the current (peak amount of current that flows) may also depend on the frequency of the applied voltage. This is how the current depends on frequency.

In the case of an LRC circuit, there is a phenomenon called "resonance", whereby the amplitude response of the current has a maximum at the so-called "resonant frequency". Resonance is a common in oscillatory systems, where the amplitude response of the system, be it a mechanical system, or an electrical system, has a maximum at some frequency. At frequencies other than the resonance frequency the amplitude response of the system (the amplitude of the current in this case) will be smaller.

Now, in an electrical system, the resonance will be established by the capacitor and the inductor (the resistance provides the damping), and the values of these components will give the resonance frequency, which is the natural frequency at which the oscillating system operates at, and where it will have the greatest response if you drive it with a frequency. If you input an oscillating voltage to an LC circuit, the amplitude of the current should be very small until you get near the resonance frequency. However... I'm not sure if this is what you did?

If you input a square wave at a much much lower frequency than the resonance, then it will be like turning on and off the voltage, and you would see oscillations at the resonance frequency, which would decay until the next cycle when they would start again. Perhaps this is what you did in the lab.

I hope this is somewhat helpful. Without knowing what you did in lab, I can't be sure where the confusion lies. Please do comment (click the # comment link below... or email if that doesn't work) to follow up!

Question: Induced EMF coil-solenoid; How do areas play a role?

If a coil is placed around a solenoid that has a fluctuating current, how do the radii of the coil and solenoid (so basically, the areas) affect the induced EMF in the coil? Thanks.

Thanks for your question.

The EMF induced in the outer coil is a result of the magnetic field produced in the solenoid.

First let's consider the magnetic field produced by the solenoid, which is given by:

B=μnl

where n is the number of turns/unit length. But wait! There's no mention of area here... in fact as long as the approximation that l>>r holds (and that you are not near the ends), the magnetic field generated by a solenoid is independent of the cross-sectional area.

So let's consider the EMF induced in the coil surrounding the solenoid. This is given by:

EMF=-ΔΦB/Δt=-A_{perpendicular}ΔB/Δt

So the cross-sectional area of the coil does enter into the induced EMF.

Hope that is helpful. For a related question check out: Question: Solenoid in a coil