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If the surface was defined so that it only covered one side of a dipole, would Gauss' Law then give a non-0 answer?

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(moved your question to lecture 12, since you ask about Gauss' law).

As long as you're not enclosing any current, the net magnetic flux is always zero.  Think about the very top end of the north pole of a magnetic dipole.  Here the field lines are all pointing radially outwards.  If you were to draw a circular loop around them, then B and dL would be perpendicular at every point so their dot product would be zero at every point, leading to an integral of zero.

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What does the integral look like if we integrate dL "over a loop"?
What limits are we using that we get 2piR from dL?

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With the example from the first part of lecture 7, is the electric field emanating from the hollow sphere with Q in it due to q(outer) the same as the electric field that would be produced by just Q if the hollow sphere were taken away?

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Excess Charge Expression on the Surface of a Conductor.

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Excess Charge Expression on the Surface of a Conductor.

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I'm not completely sold on the explanation given in the 'sea of charge video'. Namely in that I would have thought that maximally increasing the distance between charges would be satisfied by a maximal circle packing, demanding excess charge in the centre; or to use a thought experiment: a positively charged ring would repel excess positive charge to the centre. I would therefore assume that excess charge being expressed on the surface would be due to local movements motivated by the different charges in the medium moving inward to cancel the excess in the inside, leaving an excess on the outside? Or am I simply misunderstanding the model used?

Thank you in advance.

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Hi. In lecture 3, the restoring force from Hooke's Law is seen to be a result of an electric charge being displaced from an equilibrium point whilst in between two other charges.
However, when the limit of proportionality of a band is exceeded, the force is no longer proportional to the displacement from the equilibrium point. In terms of the charge configuration, what happens when the limit of proportionality of a rubber band is exceeded which leads to Hooke's Law no longer being applicable?
Thanks.

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Answer - Charge in a hollow spherical conductor

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Answer - Charge in a hollow spherical conductor

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The simple answer is yes  because Q represents all of the excess charge in the system.  Without the hollow sphere, Q is just a point charge.  When the hollow sphere is in place, the excess charge is now at the surface of the sphere and there is no net charge beneath the surface of the sphere.  Of course the complex answer to this question is “it depends where you are evaluating the field”.  At a large distance away you can treat the two systems as being equivalent (large here being the point charge definition from lecture 1).  At a position that is inside the sphere, the electric field is of course zero.

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Following comments from some of you, I've cleaned up the sound on lecture 16. Everything should be louder and clearer now. Inevitably that'll mean it's *too* loud for some of you, but you can at least reduce the volume!

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Answer - Excess Charge Expression on the Surface of a Conductor.

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Answer - Excess Charge Expression on the Surface of a Conductor.

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I’ll try and answer your question as best I can because I’m not sure exactly what you mean by “maximal circle” packing! Drop me another message if the below is not clear!

The easiest way (at least for me) to think about is to consider the potential energy of the whole system. If you have excess charge in the centre you have excess potential energy because the sum of your distances is not minimised.  This is energy argument I referred to in the lecture.

You can also think about the problem in terms of electric fields.  Any excess charge will lead to an electric field.  If you’ve watched to the end of the lecture you’ll see that when there is a net field on a conductor, it creates a counteracting field through rearrangement of charge.  An excess field will always induce forces and separation of charge.  This cause the separating charges to create a counteracting field.  This will continue until there’s no net field inside the conductor.  In other words there is no excess charge inside the conductor either —> all the charge is on the surface.

Finally you can even think about it in terms of potential.  If you have an excess charge and you want to introduce it to a conductor, the potential is lowest at a position furthest away from the centre.  This is the simplest explanation conceptually but it doesn’t really explain the movement of charge, which is why I don’t use it.

Your example of a charged ring is an interesting one.  Here the inside surface of the ring is closest to the centre of the ring itself so you indeed get a surface charge on both the inside surface and on the outside surface.

When you say:

“excess charge being expressed on the surface would be due to local movements motivated by the different charges in the medium moving inward to cancel the excess in the inside”

This is certainly a correct way of conceptualising it but is not my preferred way.  You start with excess charge in the centre (say positive).  The electric field causes negative charge to move to the centre so that the net internal charge is 0 and as a result the charge at the surface is not neutral any longer.  While it makes sense, it focusses on the movement of (in this case) the negative charge.  I find it easier to conceptualise it in terms of the movement of net charge.  You should go with what is most intuitive to you but make sure you don’t confuse yourself or an examiner when describing it!

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The limits when integrating over a loop are 0 and 2πr.  Integrating dL gives you L and you're integrating over the whole length of the loop.  The zero point is arbitrarily defined, it can be any point on the loop.  The limit of 2πr just reflects that you are integrating over the entire loop, so beginning (and ending) at your chosen zero point with always give you the circumference.

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Answer - Exceeding limit of proportionality

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Answer - Exceeding limit of proportionality

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Nice question!  There's absolutely a limit to the problem I set because it only works for *small*  displacements around a point.  If you go to larger displacements then the relationship between restoring force and displacement will certainly deviate from a linear one because if you superpose the two fields, you'll get an answer that more closely approximates the inverse square law.  Consider the limit when we the distance to one charge is much less than distance to the other.  The force here is dominated by a single charge and the 1/r^2 relation holds.  So both have their limits - the hookean spring because it begins yield under force - the charge because it deviates too far from the centre.

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Q - How can I view the worked examples more easily - the thumbnail screen is too small!

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Q - How can I view the worked examples more easily - the thumbnail screen is too small!

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A - I've added an explainer to the course documents. Check out:
Course Documents >>> Electromagnetism >>> Viewing Lectures in Panopto

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Q - Where is the solution to the problem in part 4 - has it not been uploaded to DUO yet? 

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Q - Where is the solution to the problem in part 4 - has it not been uploaded to DUO yet? 

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A - I have done a recorded solution in lecture 5 (online at 11am, 4th December). A scan of the solution is also included in the lecture 5 folder.

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On slide 46, why do the first and last diagrams look the same but one has negative potential energy?

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On slide 46, why do the first and last diagrams look the same but one has negative potential energy?

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They shouldn't! This is a typo and has now been corrected on the slides and in the video. This is what it should look like.  So when the dipole is parallel with the field, the potential energy is positive.  When its antiparallel, there is a negative potential energy.

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Q - In the vector torque derivation, why isn't the dipole vector torque equal to -qdEsin(phi)? 

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Q - In the vector torque derivation, why isn't the dipole vector torque equal to -qdEsin(phi)? 

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A - It should be negative.  I have updated the slides and the derivation scan.  Here's the correct form:

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The solution that appears once you've answered questions 2 seems to contradict the answer it says is correct? Should U=0 be the correct answer or am i misunderstanding the question? Thanks.

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Thanks, I've fixed this now. U=0 is the correct answer, as indicated by the explanation.

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Good afternoon,
I'm having some problems with the signs, as shown below. They seem to keep switching? Am I missing something?

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Answer - sign problems with rubber band conundrum 

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Answer - sign problems with rubber band conundrum 

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I think I answered you by email but I'll post the answer here for everyone else's benefit.

The sign change comes about because the point at which we’re evaluating the field  is centred between two charges. The distance to each charge has a direction and a magnitude.  We define the direction by the unit vector i, whose direction points from left to right.  The distances are then (d+x) and (d-x) but they’re pointing in opposite directions so we denote them +(d+x)  and –(d-x).  If you substitute those two distances into the top equation you get the bottom equation.

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Relationship based on Newton's Third Law

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Relationship based on Newton's Third Law

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Since F 1,2 = -F 2,1 does that mean that the order of the numbers 1,2 with the unit vector should also swap when using this relationship?

Thank you

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'Flux of a field parallel to a surface is positive'
I am a bit confused as to which surface is being considered when the dipole is in the enclosed box on the slide. Is it the two vertical sides of the box or the horizontal ones?

Thank you in advance.

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Diagrams showing the potential energy of a dipole conventions

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Diagrams showing the potential energy of a dipole conventions

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In the derivation of the formula for U depending on phi, the positive charge is on the right and the negative charge on the left in the diagram. This derivation results in U = - p . E. The following diagrams displaying the sign conventions confuse me a little as I would expect that the leftmost diagram would show U = + p . E and the rightmost one would show U = - p . E. Why is this not the case?

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When I say that flux of a field parallel to a surface is positive, I'm referring to field lines.  So if a field line leaves the box, its parallel because the surface normal points away from the surface (so field line and surface normal point in the same direction). Such a field line would therefore contribute a positive flux.

A field line that is anti-parallel to the surface  enters the Gaussian surface, and therefore contributes a negative flux.

The point that I'm making with the dipole is that any Gaussian surface that encloses a dipole will have an identical number of field lines leaving and entering and so the total flux is zero.  This is consistent with Gauss' law because the total enclosed charge is also zero.

So to answer your question, it applies not to a single face of the cube but to the entire enclosing surface. That said, its easy to see if you look at one of the fields coming out of the top surface, that the same field line also re-enters the same surface.  The same will be true of all field lines.

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I don't understand why the two circles at the end of the cyclinder aren't considered when we integrate to find the surface area of the cyclinder. You mentioned that it was because they were perpendicular to the line of charge but wouldn't this mean that the flux = EA and not 0 (like it would have been if they were parallel)?

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The key point here is that the line of charge is infinite.  So our ends of the cylinder that we chose do not denote the end of the line of charge - it continues to infinity either side of the cylinder.  You can think of the field lines extending radially outwards towards as infinity as well.

Another way of thinking about it is to consider just one of the two circular surfaces as shown in the sketch below (you can click to enlarge). At the circular given surface there is an equal (and infinite!) charge on either side so its perfectly symmetric.  So the field at the surface has to be zero - remember what field lines of two positive charges look like in lecture 2.

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Answer - Diagrams showing the potential energy of a dipole conventions

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Answer - Diagrams showing the potential energy of a dipole conventions

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It seems that by correcting one typo above.  I've introduced an extra one.  Sorry about that!

To be absolutely clear, you are correct.  When the dipole is aligned positive sign to the right then the energy is - p.E as in the example.

The point of course is that these are arbitrary conventions but as I said in the lectures, we should always be consistent  - apologies for not being so!

Pdfs and slides in the presentation have been updated.

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What happens to an electric field when an insulator is placed in its path?

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What happens to an electric field when an insulator is placed in its path?

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While I understand that an electric field is not disturbed when a neutrally charged insulator is placed in its path, wouldn't the electric field be affected if the insulator were charged?

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Hi, someone else already asked about the signs in the rubber band conundrum, but going back over it now I still don't understand why, when the Taylor expansion for 1/(d-x)^2 is subbed into the equation for E(p+x), the sign changes to a negative. This might be stupid but if that truly is the case doesn't everything cancel? Surely it must be ()-(1/d^2 + 2x/d^3) in order to get the -4x/d^3 in the final answer? Thanks

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Does the potential difference across a battery decrease as the potential difference across the capacitor increases, until the potential difference across both is equal?

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Answer - While I understand that an electric field is not disturbed when a neutrally charged insulator is placed in its path, wouldn't the electric field be affected if the insulator were charged?

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Answer - While I understand that an electric field is not disturbed when a neutrally charged insulator is placed in its path, wouldn't the electric field be affected if the insulator were charged?

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If the insulator is charged, then yes you would compute the net electric field through superposition as usual. The point is that the  internal electric field of the insulator does not change.

Consider any position inside the insulator.  The electric field at that point in the absence of an external field can be termed E1.  If we apply an external field of magnitude E2 the total field E_total = E1+E2.  This is true irrespective of any excess charge.

Consider the same point inside a conductor. In the absence of an external field, E1 = 0.  If we introduce an external field of E2, then the internal field will change to counteract it so that E1 = -E2.  So the total field  E_total =  E1+E2 = 0

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There was previously a typo in the slides that caused some confusion. I've since the corrected it.  You have now successfully found a second typo, which I did not spot before!

To avoid any further confusion, what I’ve done is uploaded a full derivation just now which goes through each step.  Its in the lecture 3 folder and the following link:

https://duo.dur.ac.uk/bbcswebdav/pid-6182201-dt-content-rid-21787448_2/xid-21787448_2

In the (hopefully unlikely) event that someone spots a typo in this - please let me know!

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Shouldn't the area enclosed be given as an expression in terms of R rather r, because of the fact that R is the radius for the smaller circle?

Thank you

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Answer - Relationship based on Newton's Third Law

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Answer - Relationship based on Newton's Third Law

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The answer is, you can write it in different ways.  If you want to write everything in relation to r_{1,2} then the signs of the two forces have to be opposite, because they are equal and opposite.  You can also write the same expression without the minus sign but change one of the unit vectors to r_{2,1} and it has the same meaning (but its a bit more subtle). Think of the unit vector as being a reference direction: If there's a minus sign in front of the unit vector it means the force is applied in the opposite direction.  See the sketch below

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In part 2: introducing dieletrics at around 11:15 you say that the net flux through A_2 is zero becuase the surface is perpendicular to the field lines. Wouldn't that mean that the flux is at its maximum value not its minnimum? Should it be that the flux zero because the surface is parralell to the field lines?

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When I said that the surface is parallel to the field lines, what I was referring to (and what I generally refer to) to was the area vector. I’ll try to be more explicit about this in future lectures!

So to be clear - in that diagram, the field lines are horizontal (left to right), while the area vector of the long cylindrical surface A2 is always orientated at right angles to the field lines (in relation to the page, that means either up, down, out of the page, into the page or somewhere in between).  The point is that no field lines enter or leave that particular surface and therefore that the flux density is zero.

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Yes, as the electric field of the two equalises you will see a corresponding change in potential difference.  You can think about it in terms of the redistribution of charge from the electrodes to the capacitor plates.  Each plate and its corresponding battery electrode become an equipotential, so as a result the potential difference been them has to be the same —> potential difference between plates = potential difference between electrodes.  The consequence of this is of course that the potential difference across a battery will decrease as it discharges.  So a battery that is almost empty has a lower voltage than a fully charged one.

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Your welcome! This is a really good question.  The answer is actually “no” but the reasons probably probably aren’t immediately obvious! The point is that we are relating three quantities: 1) Electric field2) Enclosed charge3) Area
The issue is that for the regime you’re asking about, r>R, the enclosed charge is always the same.  However, as we increase the enclosed area, the electric field is decreasing.  If that statement sounds confusing, do a sketch and look at the field lines: bigger radii lead to bigger circles and field lines that are more spaced apart signify a weaker electric field.  When r=R the electric field is highest and as r increases, the electric field decreases radially.  If we were to subsitute R instead, we’d see a constant electric field with increasing radius because R and Q are constants.

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The answer and explanation to the first quiz question was not correct and has been updated.  Explanation should now read:

Potential difference is expressed as V=Ed.  Since the capacitor is isolated, the amount of charge does not change - E is therefore constant and the change in potential difference is only determined by the change in plate separation.  It therefore doubles.  (It does *not* stay the same - apologies for the mistake in this question previously).

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Why is there a field through A3 if the dipole is negative end of the dipole cancels the field out?

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Just to clarify, on the slide talking about charge density, all the units seem to be cm^-n, depending on the dimension. You mention this is centimetres - but would this not be coulombs/m? So C m^-n? I can't really see a reason why it would just be a unit of length!
Thanks

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The enclosed charge A3 will always be net negative because there is slight net-negative charge on the surface closest to the conductor.  If you want to consider the problem in terms of field lines, try and draw a sketch of A3 and consider what happens in you enclose just half of the dipole on the molecular scale (so that the positive half of the dipole is outside the gaussian surface).  On a molecular scale, this will have a net negative flux because the net charge is negative.

Technically, if we drew a really tiny surface that just perfectly enclosed a single dipole and didn’t overlap into any neighbouring dipoles then the net flux would be zero as you suggest.

However, we’re considering this on the macroscopic scale, the net enclose charge will be the sum of all of the enclosed charge. While we could be enclosing some complete dipoles that are really close to the surface (so for these the net contribution to flux is zero), if we add up all of the charges in our surface A3 it will always be negative.

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When I solved the weekly problem, I considered the x and y directions separately. This ended up with me having two different quadratics to solve, which left me with four possible solutions for the equilibrium position. While I understand why one of the answers can't be a solution (the one lying between the two charges), why does the equilibrium position necessarily need to lie along the line joining the two charges? Also, why did the method I used end up giving answers which don't make sense? Do they have any physical significance/ are there any situations in which they would make sense (for example, would some of the solutions make sense if we considered a positron instead of an electron)?

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You have two questions, so I’ll answer them both.

1) Why does the position need to be on the line between the charges?  As with so many things, this question is most easily answered by drawing sketches!  Draw the two charges and, for clarity draw them both on the x-axis (this just makes it easier to visualise). Now pick any position for your charge that isn’t on the line.  What are the forces on that charge, by superposition?  You’ll see that for any positions you’ll get two forces (one from each charge) and that if you superpose them you will always get a net force in the y-direction (if its easier, try superimposing the x and y components of your two vector forces).  So the charge is not at equilibrium.  The only exception is if the charge is on the x-axis and in this case the forces always have net y-direction force of zero (and a net force in the x-direction that depends on the position in x).

2) Why can’t the 4 answers in the quadratic all form a solution?  The short answer is that without seeing your working, it’s difficult to know.  But you can work it out by considering point (1) above.  You probably have positive and negative roots in x and y and you’ll need to deduce why some of those positions aren’t physical because x and y both need to be negative. If you (or anyone else reading this!) want to convince yourself of this, just put your results into coulomb’s law and work out the net force (be careful about using the correct vector forms).  You’ll find only one solution gives you a zero force!

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Charge density calculation for dielectric surface

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Charge density calculation for dielectric surface

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I cannot understand Gauss’s law in dielectrics: I do not understand why is the flux 0 in the conductor, while it is equal to the part of Gaussian surface in the dielctric only contains the free charge and not the bound charge.

My understanding is that a conductor cannot contain charges inside but I do not get why the charge in the part of the Gaussian surface in the dielectric is not the difference between the free charge of the conductor and the bound(induced) charge, but rather only the first one.

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Answer - Charge density calculation for dielectric surface

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Answer - Charge density calculation for dielectric surface

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Your question is about why we use the difference between charge densities (sigma-sigma_d) rather than just charge that’s in the dielectric. The answer is that we’re evaluating Gauss’ law across the entire surface: it’s not just the charge in the dielectric, it’s the total charge.  When we consider the flux however, we see that there’s only flux through one surface A3.  The charge enclosed by the *entire* surface is the net charge, defined by the two charge densities.  We multiply that net charge by the area through which there is a flux A3.  We ignore the areas A2 and A1 because for Gauss’ law the definition of the area is the area through which the flux is passing.  At this point you might be asking, why can’t we just use the whole Gaussian surface?  Why do we need to express it through just one of the three surface where there’s flux? If that sounds a bit unconvincing consider the following:

1) What if we increase the total area (A1+A2+A3) by doubling the length of the cylinder.  Therefore the total area doubles but the area A3 remains the same.

2) Therefore twice as much net charge is enclosed by the total surface and there will be twice as many field lines exiting the surface.

3) The net flux across the surface will therefore also double.

4) But the net flux is only non-zero at A3 so we would have twice as much electric flux at A3 as well.

5) The electric field is double and so is the electric flux, therefore from phi = EA we can say that the area must be constant.

6) The area to consider is for Gauss’ law is therefore A3 (which remains constant when we double the cylinder size) and not the total area.

This is a subtle distinction and one that only makes sense if you understand the underlying physics – simply remembering the equations doesn’t help here!  Sketching the above also helps!

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I'm still a bit confused as to how the enclosed charge is calculated in video 2. In the video it was the area multiplied by the difference in charge density, but should it not be the volume?

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Answer: Net Charge for the Dielectric Surface

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Answer: Net Charge for the Dielectric Surface

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Within the volume that we're evaluating all of the charge can be said to be on two surfaces, the surface of the capacitor plate and the surface of the dielectric. The charge densities (sigma and sigma_d) refer to charge per unit area on those two surfaces.  So we multiply by the total area of the surface to get the total charge enclosed by the cylinder.

I hope that answers your question. If its still confusing, drop me an email or we can discuss at next week's office hour.

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Hello, in video 2 I'm not sure how we got to the first line of the derivation: specifically the part with (1-1/k). I follow the rest of the working, but have tried expanding the first line and it doesn't make sense to me where that kappa comes from. Thank you!

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Answer: Deriving the dielectric constant

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Answer: Deriving the dielectric constant

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We're not deriving the dielectric constant from first principles here, so you don't need to understand where the expression comes from (I just included it to give you a chance to do some rearrangement of equations). The important bit is in the next slide where we define kappa as the ratio between the external and the net fields. This, and the definition of the capacitance of a dielectric are the two important aspects to understand here.

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I can understand this, when one magnet attracts another toward it is it not doing work on that magnet?

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That's a nice question!  I'm afraid that you'll have to wait for the answer until lecture 13!

For the time being, consider in the lecture that we're talking about individual charges and their fields. In a magnet, we're talking about arrangements of charge. So the "lecture 9" argument is that the magnetic field from a moving charge does no work. For the more nuanced argument, wait and see!

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Please could you explain how in question 27.17 (Question section 27.4) the force's direction is south... I crossed the vectors and calculated it to move in the k direction (into the page) but the answers say it just moves south... How can it move south when the charge itself is moving downwards and the field is moving east?

Calculating the magnitude of the force itself was fine.

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Yes, you're absolutely right. To be honest, I'm not sure why I said centimetres. It's Coulombs per metre etc. I've updated the slides and the recording to make that clear.

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Doesn't the direction of the magnetic field produced by a moving charge depend on the sign of the charge? Also, is the direction the magnetic field is pointing in on the second slide of the second video incorrect?

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You're correct about the right hand rule being valid only for the positive charge (I added an explainer to lecture 11 to clarify).

You are also correct about the field lines on the sketches and they've been updated on the video and the pdfs. You have a good eye for detail, I didn't spot this!

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In the first slide of part 4 you specify that for 'uniform' electric fields, F and E are parallel and related by the equation F=Eq and similarly that for 'uniform' magnetic fields, F and B are perpendicular to each other. Similarly, in the next slide is written 'a uniform magnetic field does no work on a moving charge'. Must the fields be explicitly 'uniform' for the statements to be valid? What would happen if the fields weren't uniform?

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My interpretation of this is that you are (probably) correct and the question is not particularly well phrased!

I think the interpretation should be that the ball is falling in the (-j) direction and "east to west" corresponds to "from (+i) to (-i)". Assuming this is the meaning, the field direction would be (-k), in other words into the page, as you say. Either way you can define the direction as south, but it's a confusing way to write it.

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Part 1 - Magnetic Field along the Axis of a Current Loop

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Part 1 - Magnetic Field along the Axis of a Current Loop

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When doing the integral for dBz, the (4*pi) constant disappeared when we wrote the equation for dBz from the Biot Savart Law, but in the final equation for the magnetic field along Z there was a 2 in the denominator. I'm not entirely sure how this happened.

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Several times you defined the force, dF, on a segment, dL, as being equal to dB = mewnought etc etc. Should this say dF is equal to the expression or is it dB and I am missing something. Thanks

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It should say equal to dF - it's a force after all! I believe i have changed it now.

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Charge/ wire moving in and out of page convention

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Charge/ wire moving in and out of page convention

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Hi, please could you just clarify what the convention is (dot or cross) for charge in or out of page, is the wire then the opposite of this?

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Answer: Charge/ wire moving in and out of page convention

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Answer: Charge/ wire moving in and out of page convention

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ⓧ - The direction is "into the page"
⊙ - the direction is "out of the page"

The anti-parallel wires are therefore depicted as a ⊙ for one current and a ⓧ for the other. Parallel wires have the same symbol for both.

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When you use the cross product idenitty for the vecotrs which are parallel, ie you interchange them. Is the cross product of 2 parralel vectors not just equal to zero regardless of the order?

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I may be wrong but should the magnetic force not be perpendicular to the plane's normal/ parallel to the plane of the loop in order to generate a torque?

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I think maybe there's a confusion around the orientation here (the panopto video platforms don't allow me to add sketches which would make things much easier!).  The point is that the loop lies flatly in the x-y plane.  The field is aligned along the y axis (going in the positive direction).  So:
1) The field is parallel to the plane 2) The field is perpendicular to the surface normal of the square loop.
3) Two of the four wire sections that make up the square loop are perpendicular to the field (the two aligned along the x-axis), resulting in a torque.

If we rotated the coil into the x-z plane:
1) The surface normal would be parallel to the field
2) The square loop would be perpendicular to the field.
3) All four sections that make up the loop would be perpendicular to the current flow (they're aligned in x and z and the field is in y), resulting in either expansive or contractive forces.

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torque= force x distance, so total torque for the square loop in a magnetic field is t= IWBxH or IWBHsin(fi). But why if we say that A= WxH then

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I think your question got cut off but if you're asking about the statement A = WxH, this is just a scalar calculation (the x is a multiplication sign not a cross product, since W and H are not vectors). So this does not affect the cross product in the vector form.  The vector form is derived from the cross product between force and distance (the definition of torque) and the cross product between dL and B (the direction of force).  If you like you can do a full vector calculation of these terms to get to the the IAxB definition but I think its easier to think about it phenomenologically:

1. If the area vector of the square coil is parallel to the field, there’s no torque

2. If the area vector of the square coil is perpendicular to the field, there is maximum torque

Therefore it’s intuitive that the cross product should be between the area vector and the field, which is indeed what we see.
(drop me an email if I've misunderstood what you meant to ask)

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A magnetic field never does work on a moving charge, uniform or otherwise. If the field wasn't uniform, force and velocity would still be perpendicular but their directions would change with the field. This makes the maths very tricky but the conclusion is the same!

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It's not clear what you're referring to in your question but yes the cross product of two parallel vectors is zero,

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When doing the integral for dBz, what happens to the unit vector from the dB biot savart formula (as in the dL x r^)? Why does that disappear? Thank you

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Answer - Magnetic Field in a Current Loop

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Answer - Magnetic Field in a Current Loop

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The cross product helps us to understand the direction of the magnetic field - it tells us that it’s at right angles to the the unit vector (r-hat) and at right angles to the segment of the loop (dL).  From examination of the symmetry in the problem, however, we can see that the x and y components of B will always cancel out, which tells us that the net magnetic field must point in the z direction. From this we just need to work out an expression for the magnitude of dB (the total field from a segment dL) and then calculate its z-component dB_z.  Note also the corrected solution on DUO (I updated the original to clarify this).  I’m pasting it below as well for convenience.

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When working out the magnetic field at point P, why did you end up using the equation where a tends to infinity when the diagram showed a finite length from a to -a?
Thank you

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Why is the Z component taken as dB sin(angle)? Dosnt this define a distance on the y axis?

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Im a bit confused about how we know that B and dL are in the same direction here - what exactly is dL in this example is it the radius?
Thanks

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Answer - direction of B in a wire example

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Answer - direction of B in a wire example

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The direction of dL is the direction of the (blue) line integral, assuming that we're referring to the example in part 2 with the application of Ampere's law to the wire.  Consider the line as being made up of a of a set of infinitesimal segments dL. Each segment is tangential to the blue circle and therefor is parallel to the circulating magnetic field around the wire at every point.

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I initially derived the general form of the expression which includes both x and a, since its more comprehensive.  However from a practical perspective, the length of a nearby power line a, will always be much larger than the distance away from the power line x (unless we’re talking about a power line, say in a different country, where the field is very going to be very small indeed!).  So we can say that a >> x.  I expressed this as a —> infinity, which might have caused some confusion, but the intention is that we consider the power line to be very long. A typical power line might have a length of 100km so if you use the formula that includes both x and a, the final answer will be the same to any reasonable precision.

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B Field of a long, cylindrical conductor

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B Field of a long, cylindrical conductor

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When checking if the two formulae are equal by substituting R = r, the two you gave on that slide (8.45) do not equate...

I think for r > R, it should be 1/r instead.

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This comes resolving the z component of the magnetic field.  I've added some clarification to the pdf and the presentation slides.

In short, dB is at right angles to  dL and r.  This means that there is an angle between dBz and dB.

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Answer - B Field of a long, cylindrical conductor

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Answer - B Field of a long, cylindrical conductor

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Yes, thanks for spotting that. It is indeed 1/r for r>R as implied by the decrease in the plot. Video and pdfs have been updated now.

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If we are considering a square loop (shouldn't H=W). When we use the equation F = I dL x B, we substitute W for the magnitude of dL, however this wouldn't be the case if you instead consider dI3 or dI1 because their magnitudes are W instead.

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The explanation to the answer was cut off therefore we are unable to see the equation you used to figure out the minimum value for potential energy

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In the question to work out the magnetic field in a solenoid shouldn't BC and AD be zero instead of AB and CD since they are parallel to the magnetic field? Thank you.

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This comes down to the definition of dot products. Within the solenoid we have the segment BC.  We have a vector field:
\underline{B}
And a vector that denotes the line BC
d \underline{L}
We can separate their magnitude and direction to give:
\underline{B} = B.\underline{\hat{B}}
and
d\underline{L} = l.\underline{\hat{dL}}

They’re also parallel to each other, which means:
\underline{\hat{B}}.\underline{\hat{dL}} = 1
And therefore, as stated in the solution:

\oint\underline{B}.\underline{dL} = B.l \oint\underline{\hat{B}}.\underline{\hat{dL}} = Bl

For segment AD, it is outside the solenoid.  Since we're assuming an ideal solenoid, we say that the field here is zero.

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Hello,
In part 1 of the lecture you derived the expression B = mu * I * n, but in the question it asks to give your answer in term of the radius of the solenoid. Does that mean the B field inside the solenoid is completely independent of its radius?

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Answer  - B field inside an ideal solenoid

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Answer  - B field inside an ideal solenoid

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The answer to this basically comes down to Ampére's law. In the example of the lecture, we drew a rectangle around half of the coil to deduce the field. Ampére's law tells us that no matter how far the radius of that rectangle extends into the middle of the solenoid the total enclosed current is the same. The rectangle I drew enclosed 4 turns and I could draw any surface, with any values of r,  that could enclose that number of turns and by Ampére's law, the field would need to be the same.

If that seems like a bit of a cop-out, try sketching the field lines of the 4 turns as a function of radius. Close to the turns the individual field lines will be stronger but their direction will less aligned. Further away from the turns, the net fields will be weaker but the field lines will be much more aligned. The superposition of fields, however, regardless of the radial position

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Answer  - Question 2 on quiz in lecture 13

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Answer  - Question 2 on quiz in lecture 13

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Apologies for this, there seem to be some persistent problems with formatting in the panopto video software.  It should be fixed now but in case of any further errors here it is:

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For the quiz Q3, the correct answer is supposedly a, but then the explanation for which answer is correct seems as if it is describing scenario b... Which is correct?

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How did you reach Fsin(phi) and not Fcos(phi) using the little diagram just below the side view

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Scenario b is correct, I've updated the quiz question.

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In the final two steps of the derivation it states B*L = mu*I*n but the final equation for B is B = mu*I*n. Where does the L go in this situation?

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Answer - Magnetic Field in Ideal Solenoid

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Answer - Magnetic Field in Ideal Solenoid

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I’ve corrected the solution on DUO.

The full equation is:

\mu In_l l = 0 + 0 + 0 + Bl

(this is just the integrals of the 4 segments and Ampere’s law)

Rearranging this gives you the desired value
B = \mu In_l

(I missed off an l in the video, which is what has caused the confusion)

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Doesn't changing electric flux induce a magnetic (not an electric) field?

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For Gauss' Law for magnetism, slide 29 to 30, why does the dA suddenly change to dS. Is this a mistake or deliberate?

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The sin term comes from resolving the z component of the magnetic field B.   This sketched below.  dB is perpendicular to both r and dL.  The x and y components of B are zero because of symmetry so we resolve only the z component.  The geometry below shows that we must use sinθ.  I have also updated the worked solution on DUO to make this clearer - take a look!

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Yes, thanks for spotting that. It's been corrected on the video and on DUO

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Just want to make sure that I understand correctly. From the 1st quiz question, after the coils align with the magnetic field, they continue on spinning, right? Or do they just align with the magnetic field and stay there?

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I'm a bit confused about the part around 8 minutes in to the 2nd video. Here you say that in general the electric field lines 'are not close' to the lines of force, but if F=Eq, shouldn't they overlap? I understand that the direction of motion of the charges wouldn't necessarily follow the E field lines, but shouldn't the E field lines and lines of force always point in the same direction?

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It should be dS throughout and has been corrected.  FYI, you may see dA in some textbooks.  They refer to the same thing:

dA refers to "area"
dS refers to "surface"

Whichever you use, for Gauss' law for magnetism, the integral is always zero!

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I thought the Amp-Max law also depended on the rate of change of electric flux with respect to time, not magnetic... The Amp-Max equation doesn't feature the flux of B, only the flux of E?