Mon, May 7, 7:04 p.m. - Can you explain why circulation is negative in number 5 from Exam 1 2018?

Circulation of a field is the dot product of the field and the path. In #5, the path runs CW, while the B-field (by RHR) goes CCW. Thus, the dot product for #5, Path C, is negative.

Fri, May 4, 7:18 p.m. - For Gaussian surface problems and when dealing with q(enclosed), when do you compare total volumes versus total areas? For example, concept test 3.5 asks to define q(encl), but to do so you determine the ratio of volumes of what is enclosed by the Gaussian surface to the whole circle. When would you use areas? Is there a rule for knowing to use volumes or surface areas for Gaussian problems?

When calculating $q_{enc}$, you need to consider the charge distribution.
If the charge is spread out through a volume, the charge density would in charge per unit volume.
If the charge is spread out in area, the charge density would be charge per unit area.
If the charge is spread out in length, the charge density would be charge per unit length.
To get $q_{enc}$ you need to multiply the charge density by either a volume, area, or length depending so that you wind up with just "charge" enclosed.

Mon, Mar 5, 7:20 p.m. - For diffraction, why does the intensity of light change based on the number maxima you are at?

Are you asking about a single slit? If so, intensity goes down because the phasors continue to wrap in on themselves, making each subsequent "max" more wrapped and thus lower total amplitude. Try playing around with the simulation (scroll down for a single slit) that is posted on the lecture 12 calendar page to see this effect. If you're talking about a diffraction grating, the answer is MUCH more complicated, and beyond the scope of what we've learned in PHYS212-it's something students learn about in our upper division Optics course!

Sun, Feb 18, 3:12 a.m. - How does the Earth have magnetic poles? Why are the magnetic poles on a magnet opposite of the magnetic poles on Earth? Why is the South Pole at the North Pole? Why didn't they just make it so the names were the same?

Geographic "North" was established before we knew anything about magnets. Magnetic "north" was so named because the north pole of a compass magnet is the side that points toward geographic north. The Earth generates a magnetic field in its spinning metallic liquid outer core. As we learned in Unit 1, the motion of charged particles (aka current) creates magnetic fields. In the Earth's outer core, heat escapes from the core via convection. These convective currents contain charged particles and thus create the Earth's magnetic field.

Tue, Feb 6, 11:11 p.m. - I'm really confused on chapter 20 #36. I started the problem by first calculating the force using F=qE Then I used the F value to calculate the acceleration using F=ma I know the initial and final velocities but I don't know how the distance fits in here

You actually don't want to calculate the acceleration. Since the electric field is uniform, the electric force is constant for the proton, so the work is $\vec{F} \cdot \Delta\vec{r}$. You can then use the work-kinetic energy theorem to find $\Delta\vec{r}$.

Tue, Feb 6, 8:02 p.m. - For the previous year exam, question #7, how do you determine that F(mag) points up?

The protons are traveling at a constant speed in a straight line. Thus, $\vec{a} = 0$ The electric field in this region points downward, which means that there is a downward electric force (since $\vec{F} = q\vec{E}$). Since there is no net force on the proton, the magnetic force must be balancing the electric force (to make $\sum F_y=0$). Thus, the magnetic force must point upwards.

Fri, Feb 2, 1:10 p.m. - How do you know when to use Biot-Savart or Amperes law?

Start by drawing the B-field. To use Ampere's Law you need to have a B-field with symmetry that allows you to simplify the circulation $\oint \vec{B} \cdot d\vec{l}$. This means having an amperian loop where each side of the loop either has a line integral of 0 ($\vec{B} \perp d\vec{l}$) or $\vec{B}=0$) or where $\vec{B}$ can be pulled out of the line integral (because it is constant on that particular segment of the loop). Biot-Savart should be used for everything else. And the special cases of a long straight wire or a single loop are fine for you to use if justified.

Sun, Feb 4, 6:00 p.m. - For a solenoid made out of a fixed amount of wire and a fixed amount of current is the magnetic field the same in the center regardless of the amount of turns?

For an ideal solenoid, the # of turns per meter and the current are what determine the magnetic field in the center of the solenoid. So, depending on how you wind the wire, you could get different B-fields.

Fri, Feb 2, 7:26 p.m. - For CH 26 question 32, can I ask a verification question. When it says that the B-field 1.2 cm from the wire's axis is 67 microT, does it mean that the entire wire has a radius of 1.2 cm? Or after I find the enclosed current in 1.2 cm, would I need to use the ratio to find an equation to express the current of the wire?

Assume that the wire has a radius $<<$ 1.2 cm.

Fri, Feb 2, 12:07 a.m. - I am really confused about CH 26 #31 (me again). The length is 2.2m. Dividing 2.2m by 2*pi*(.025m), where .025 m is the radius (.05 m diameter; 5 cm diameter given in the book). The answer is 14 (n = 14 coils). Please explain!

N = 14 coils but we don't have a length for the solenoid itself. $n$ should be the total # of coils divided by the length of the solenoid. But for problem #31, we don't have the length of the solenoid (the length of the wire is not the same thing)!

Mon, Jan 29, 12:53 p.m. - p. 481 Figure 26.23a (Wolfson). How did we get the direction for B? If I try placing my thumb along the bottom along the direction of I, doesn't my hand curl towards the opposite of the currently denoted direction of B?

For this figure, the magnetic field isn't created by the loop. So, the direction of $\vec{B}$ is given.

Thu, Feb 1, 6:25 p.m. - Isn't CH 26 #31 a solenoid? Why can I not use B = (mu not)*n*I?

It's not really a solenoid, because it doesn't have a length, so it is more just a case of multiple loops. In the equation for a solenoid, n, is the # of turns per unit length. For # 31, we don't know n!

Mon, Jan 29, 12:57 p.m. - Why does a dipole experience a net force in a nonuniform field? Is this because a uniform field (I am assuming magnetic) implies that the forces cancel one another out?

Yes! As we discussed in CT#1 in lecture05, the force on the right side of the loop is equal in magnitude but opposite in direction to the force on the left side of the loop. If the field weren't uniform, it would be possible for the magnitudes of the forces to be different because $I\vec{L}\times\vec{B}$ could have a different value of B for each side.

Tue, Jan 30, 6:24 p.m. - When talking about two parallel wires and their magnetic fields and forces, the reading quiz for Tuesday states that "the current in one wire produces a magnetic field at the other wire, and that causes a magnetic force on the current in the second wire)." How does the magnetic field from one wire cause a force on that wire towards the other wire? Is this found by using one of the right hand rules?

It helps to consider the two wires separately. Take a look at CT#5 from the lect05 notes. Let's just consider the top wire for a second. All by itself, this current carrying wire produces a magnetic field, and we could figure out the strength of this magnetic field for a distance, $y$, away from the wire. Now, let's add back in the bottom wire. This bottom wire is sitting in the magnetic field created by the top wire. And we can calculate the force on the bottom wire as $I_{bottom}\vec{L}\times\vec{B}$ where $\vec{B}$ is the magnetic field created by the top wire! This is sort of like charges, where we talked about source charges and test charges, but in this case we have source currents (the top wire) and test currents (the bottom wire).

Tue, Jan 30, 6:20 p.m. - How do you know which formula to use when trying to find the magnetic field? For example, when trying to calculate the magnetic field for ch.26 number 31, how would you know which equation to use? B=UoI/2a or dB=UoIsin(theta)/4pir^2? Or are you not supposed to use either of these? I'm a little (lot) lost..

For 26.31, the easiest way is to use $B_{loop} = \frac{mu_0 I}{2 a}$. This is the expression for the B-field due to a single loop of current. So you need to multiply it by the number of loops that you get if you wrap the given length of wire into loops of the specified radius (hint: one loop requires a length of wire equal to the circumference of the loop)

Mon, Jan 29, 12:41 p.m. - Wait, how does a permanent magnet have moving charge? It doesn't shock you...

We'll talk about this in the quantum mechanics unit, but essentially, electrons and protons have magnetic moments because they are rotating charges!

Sat, Jan 27, 6:11 p.m. - Let's say we're trying to calculate the potential difference across a wall. Should we be calculating for $|\Delta V|$ or $\Delta V$ (with or without absolute value)?

This depends on what the question is asking. In general we use $|\Delta V|$ when dealing with circuits, ohms law, electrical power etc. basically things where the direction of current or sign of $\Delta V$ doesn't make a difference. If, however, a question is asking about the direction of the electric field or how the energy of a charged particle will change as it moves across $\Delta V$, then the sign matters!

Sat, Jan 27, 9:05 a.m. - What is the relationship between energy and electric field and voltage and electric field?

For a case where the electric force is the only force we're dealing with: $q\Delta V_{AB} = \Delta U_{AB} = -W_E = -\int \vec{F} \bullet \vec{dr} = -\int q \vec{E} \bullet \vec{dr}$. If we're in a spatially uniform electric field: $q\Delta V_{AB} = \Delta U_{AB} = -q \vec{E} \bullet \Delta\vec{r}$

Sat, Jan 27, 9:14 a.m. - Is the whole point of A12a to see that no light is produced unless you complete the circuit with the silver wire? Isn't the nichrome wire just a short? Should it be giving extra resistance? I don't understand A12b.

for 12a) You are comparing two completed circuits. The first circuit has the two bulbs connected in series to the batteries. The second circuit has two bulbs and a length of nichrome wire connected in series to the batteries. The bulbs should glow in both cases, allowing you to compare the brightness. for 12b) the nichrome wire doesn't short out the circuit for 2 reasons: 1) nichrome has fairly high resistivity (compared to a normal wire) and 2) it is connected in series, so isn't providing a direct current path between the two terminals of the battery.

Sat, Jan 27, 11:36 p.m. - I am still VERY confused about work on versus work by. An electron is being slowed down. Is work on q*delta-V? Or work by? How about kinetic energy? Is W=delta-K (work-energy) where W is work on or work by? I am thoroughly confused. I asked about this on problem session on Friday, and I still can't seem to wrap my head around this. I really would appreciate some clarification (thank you in advance).

The work done on a charge by the electric force is equal to $\Delta K$, so, if the particle sped up, positive work was done *on* it by the electric force. When we write: $\Delta U_{AB} = -W_E = -\int \vec{F} \bullet \vec{dr} = -\int q \vec{E} \bullet \vec{dr}$, $W_E$ is the work done on the charge by the electric field.
So, let's take an example case. Let's consider a uniform electric field, and we're going to take an electron and a proton and move them along (in the same direction as) the electric field.

• If a proton moves along $\vec{E}$, then $\int q \vec{E} \bullet \vec{dr}$ is positive, $W_E$ is positive, and the proton will have sped up (makes sense) because the electric force did positive work on the proton.
• If an electron moves along $\vec{E}$, then $\int q \vec{E} \bullet \vec{dr}$ is negative, (because $q$ is negative), $W_E$ is negative, and the electron will have slowed down because the electric force did negative work on the proton.

Thu, Jan 25, 1:55 p.m. - When using the right hand rule for number A18, I get that the force due to the magnetic field is downwards, and I believe that the bar is moving to the left. Because F and R are perpendicular, why isn't the work done by the magnetic force equal to 0? I'm pretty sure it cannot be because otherwise the final speed of the bar would be equal to 0 by work-kinetic energy.

To get the direction of the force, you'll need to use $\vec{F} = I\vec{l} \times \vec{B}$. $\vec{l}$ for the bar points $\uparrow$ (the direction the current is flowing through the bar) and $\vec{B}$ points into the page, so the force is not downwards!

Sun, Jan 21, 2:30 p.m. - I am having problems with CH 20: 57. Could you point me in the right direction regarding how to replace r with a constant? Would it need to be replaced by a constant?

First off, there is improved wording for #57 posted on the homework sheet and the calendar page. That might clear things up. Remember that r needs to be expressed in only constants (like d, or L) and/or the integration variable. r should be the distance between point P (which for the improved wording is at x=d) and your tiny piece of charge $dq$ (which is at x=x!).

Sun, Jan 21, 12:27 p.m. - I don't understand how the electric force (exerted by the charge) causes the balloon to expand/contract? I KNOW that like charges repel. But WHY does this mean that the balloon is changing in size? This is what I am not able to understand force-wise.

Draw a balloon. Imagine that charge is uniformly distributed over the surface of the balloon. Pick any charge at random, let's call this charge Fred. If you think about the positions of all of the other like charges pushing on Fred, they will contribute a net force outward on Fred. Each charge on the balloon experiences such an outward force, thus the balloon expands a little bit.

Sat, Jan 20, 4:27 p.m. - Regarding CH 20 #55 (Wednesday's Assigned Problems), I'm a bit confused on how the signs work out for Coloumb's law. I end up getting q1 = q2 after solving the quadratic. But, my expression for the the force on q1 before has a negative sign because q1 and q2 have opposite signs (attracted to each other). But, the value of this expression is 2.5N [F before on q1] = -(k|q1||q2|)/r^2 = -(k|q1||q2|)/1^2 = -k|q1||q2|=2.5N -k|q1||q2|=2.5N k can't be negative (neither can the pairs of absolute value symbols); therefore, the left side has to be negative. However, 2.5N is positive. The two sides are not equal. I'm really having trouble with the signage in Coloumb's law. Do you mind guiding me through this?

For #55, the magnitude of the force is the same before and after. So, we can say $\frac{k|q_1||q_2|}{r^2} = \frac{k|q_{final}||q_{final}|}{r^2}$. We know that in the "before" picture the spheres are attracted to another, so $(q_1 q_2)$ must be negative. Thus, $\frac{-k q_1 q_2}{r^2} = \frac{k q_{final}^2}{r^2}$ (note we are not saying the directions are the same here, just the magnitudes). Now, write an equation for conservation of charge and put it all together.

Sat, Jan 20, 3:41 p.m. - Question on handin problem: A109: I assume Q is positive (+Q)? I am making this assumption because the particles denoted q in the diagram appear to be +q, whereas those denoted -q in the diagram appear to be -q. I just wasn't sure about Q.

Yes, you can assume that Q is positive.

Thu, Jan 18, 3:37 p.m. - Regarding field lines, how did we know that q would have 4 and thus 2q would have 8? Was the 4 just an arbitrary number we picked? Would we need a reference giving us the q and number of field lines to be able to draw a 2q or -q or q/2, etc. situation?

The only rule here is that the 2q charge needs to have twice the number of lines as the 1q charge. So, we could have had the 1q charge have 100 lines and the 2q charge have 200. Think of it kind of like a map...there isn't a hard and fast rule on what the scale of a map needs to be, only that relative distances need to be correct. Same thing with field lines. The relative number and spacing needs to be correct, but we choose the absolute number of lines so that it best suits our view of the problem.

Thu, Jan 18, 6:52 a.m. - Regarding CH 20 #19, I am somewhat confused. Is the proton above the electron, which is supposedly as "low" as the particle can go? Isn't the gravitational force produced inside Earth rather than by the proton? Can you please clarify or provide a bit of guidance?

This problem is asking for you to compare is the gravitational force between the Earth and an electron with the electric force between a proton and an electron.

Thu, Jan 18, 6:49 a.m. - I'm not getting the answer for CH 20 #13. Suppose the mass is 70 kg. 70 kg * 1 nucleon per 1.67 * 10^-26 nucleon is 4.192*10^27 nucleon. There are 6.287 * 10^27 parts (multiply by 1.5 to include electrons on top of protons and neutrons which are the nucleons). 6.287 * 10^27 parts * (1.602*10^-19 C / 1*10^9 parts) Why does this get me 1 C? Definitely not right.

The thing is that those nucleons will be protons & neutrons. We're only worried about the protons for this problem. So, we're talking about half your mass being composed of protons. Figure out the number of protons in your body ($m_{proton} = 1.67 \times 10^{-27}~kg$). You will also have an equal number of electrons in your body (we neglect their mass, since it is negligible compared to the protons). The net charge on your body would then be $n_{elec}q_{elec} + n_{pro}q_{pro}$. And we're told in the problem that $(q_{elec} + q_{pro})/q_{pro} = 1/10^9$.

Wed, Jan 17, 10:19 p.m. - For A2b, I assume the balloon being pushed has an additional Fpush, which the other balloon (just tension, electric force, and gravitational force) does not. Should the tension for both of these balloons be directly upwards or slightly northeast? I'm trying to draw the FBD to better understand the problem.

For A2b, draw the FBD for balloon that you are *not* holding onto. The non-held balloon will have 3 forces on your diagram, gravity, tension & the electric force (for both parts (a) and (b)).