I disagree. I specifically mentioned it near the end of the class prior to the exam when I mentioned the "interesting" motional emf application. But I also recall speaking about it at some other point in the semester; sure enough, using the PHYS208 Search Engine, I find the following mention of terminal velocity.
The main thing is that the loop will experience a net force. Previously when we considered a flat rectangular loop in a uniform magnetic field, we determined that there was no net force acting on the loop but that there was a torque present from the force couple. Problem 60 of Ch 29 in HRW is the simplest example of a non-uniform field (resembling that from a magnetic dipole) for which a net force could be readily evaluated. Note that this does not require a changing magnetic flux to generate this force; if the loop has a current, it will experience a force.
In this circuit at the instant of switch closure there is no current flowing through the inductor. As you stated, the effective circuit at t = 0 is the two resistors in series with the battery; the emf across the inductor at this instant will be the voltage across the resistor since they are in parallel.
A long time after switch closure the inductor will be fully energized. That means that the current in the inductor will be steady, and thus no emf will be across the inductor. Still being in parallel with the second resistor, it will have no voltage difference across it and accordingly there will be no current in that resistor. So after a long time the inductor effective shorts out the second resistor. The battery then "sees" an effective resistance of just one R in the circuit.
An electric monopole is just the ordinary charge of either sign that we have been discussing for several months. We find that electric field lines emanate from positive charges and terminate on negative charges.
Any magnet that we are able to create, be it a current loop or a bar magnet, always comes with both a North and a South pole. An isolated magnetic pole (the monopole) that is just a source or sink of B lines is never found. Poles always come in pairs and field lines circulate without a starting or ending point.
The possibility of magnetic monopoles has intriqued physicists for a long time. One outcome of their existence would be to make Maxwell's equations more symmetric. Unfortunately there has been no definitive experimental observation ofa magnetic monopole to date!
I should probably dodge this question, but here goes... I do not believe that the laws of physics would preclude such an event. I would propose that it is just a strength of materials issue. Since the cartilage of the nose is weaker than the bone of the skull, I could envision a situation where the head would be moving fast enough and encountering an object with just the right rigidity that the acceleration involved would not break the skull but _would_ be adequate to damage the cartilage of the nose (probably more likely if the nose happens to be more massive than average). However I would also have to assume that this acceleration could be adequate to cause a concussion or worse damage to the brain enclosed. Reminds me of the old joke "What's the last thing to go through a bug's mind when it hits the windshield of a car..."
Consider the configurations above, each current labeled 1 or 2. On the left the currents are parallel, antiparallel on the right. The directions of the magnetic field created by each current are indicated, either up or down. In addition, the forces experience by each current in the field of the other current is also indicated, either left or right. On the left the force on each current tends to pull it toward the other; on the right the force tends to push it away. Make sure that you can apply the right hand rules for getting the field and force direction in each case!
A voltage difference (potential difference) is directly related to the potential energy difference between two points; simply multiply the voltage difference by the charge to get the difference in potential energy. This difference in potential energy becomes the kinetic energy of the accelerated particle.
You may also think of the voltage difference as being associated with an electric field that accelerates the particle through the electric force. Either viewpoint works, although the first one is simple for calculating the final velocity in these situations.
I should also comment on the unit of energy often encountered when working with
fundamental particles, the electron-volt or eV.
The electron volt is the energy acquired by a particle with the fundamental charge e,
the magnitude of the charge on an electron, when moved across a potential difference of one volt.
That is,
1 eV = | (1 e)(1 V) |
= | (1.602 10^{-19} C)(1 V) |
= | (1.602 10^{-19} J) |
In general, charge will be distributed over a volume; if it is distributed uniformly, the volume charge density rho will be a constant. For example, if charge Q is distributed uniformly over the volume of a sphere of radius r, rho = Q / (4/3 pi r^{3}).
When dealing with conductors, all excess charge resides on the surface, so the distribution of charge will most generally be described by a surface charge density sigma. For an isolated conducting sphere the charge density is uniform, sigma = Q / (4 pi r^{2}). If the conductor has varying radius of curvature over its surface, the surface charge density will vary inversely porportional to the curvature.
When charge is spread over a line distribution, a linear charge density lambda is considered. You may also encounter lambda as describing the distribution along a cylindrical object, i.e. how much charge per unit length the cylinder posesses.
Direct integration to find the electric fielf from a charge distribution will always work -- sometimes the integral may be a bit messy. When the symmetry of the charge distribution permits an identification of the direction or the electric field everywhere and identification of a surface over which the electric field magnitude is constant, Gauss's law provides a quick, convenient way to calculate the functional dependence of the electric field. Of course, Gauss's las is true for any charge distribution and for any closed surface; however, it may provide little utility in evaluating the function dependence of the electric field.
Looking at previous exams, sometimes I ask that Gauss's law be used to solve the problem. Sometimes I leave it to the discretion of the student. If the charge distribution is spherical, a long cylinder, or extended sheet of charge, application of Gauss's law would be recommended.
Ammeters measure current and voltmeters measure voltage. You have been using them in the lab (although in the form of a digital multimeter which may operate in either mode, plus as an ohmmeter).
Maybe I am misunderstanding your question...
I was wondering more how to handle them in a circut or in a problem. I know what they are! [10/5 8am]
OK. To evaluate their influence in a circuit, just replace them with their effective resistance (for ammeters it's usually low and for good voltmeters the resistance is high). Then just work out like a typical multi-resistor circuit. The point of the homework exercise involving meters was that the act of measuring circuit values with these meters introduces a change, because an extra circuit element is introduced.
Hope that helps more than my last answer...
No. When we examine the concept of inductance, I will try to remember to highlight the operation of the stun gun.
I have added a discussion of the binomial theorem to the course pages. Let me know if you further discussion...
I posted 30 x 10^{9} instead of 3 x 10^{10} because I lean toward using engineering notation which uses only powers of 10 that are multiples of 3. With true scientific notation the number (Is it still called the mantissa?) must be between 1 and 10; with engineering notation you employ mantissas between 0.1 and 100 and by keeping the powers of 10 as multiples of 3 you may use reasonable metric prefixes such as micro, milli, kilo and mega (and in this case giga). I would prefer to say 30 GJ rather than 3.0 x 10^{10} J; of course the tradeoff with this kind of talk is that you may not be sure if the 0 in 30 really has significance. With scientific notation, there is no doubt as to each digit's significance.
However if HRW were to claim their answer is better than mine, I would have a basis for an argument. After all, they report the charge as 30 C so I do not actually know whether the 0 is significant there! Thus 30 GJ may actually be a better statement of the answer than theirs!
I would like you to memorize neither, if you can believe it. What I would like you to be able to do is to look at the distribution of charge, make a good choice of infinitesimal elements, and set up the integral for the E-field evaluation. The rest is calculus and the final results need not be memorized for instant recall.
a) Two resistors are in parallel if both ends of each resistor are connected directly together; each resistor has the same voltage difference across it when in parallel.
Two resistors are in series if one end of each resistor is connected together, with no other circuit element connected at that point; each resistor has the same current when in series.
b) If you have a circuit constructed from numerous resistors, some pairs in series and some in parallel, you may replace each pair with its equivalent resistance. It does not matter in which order this is carried out. But I have already shown you circuits where a progression of replacements must be made in reducing to the simple equivalent circuit. For example, replacing two resistors in series with their equivalent, then seeing that it is in parallel with a third resistor. Be careful!
c) The equivalent resistance is placed in the circuit to replace the two real resistors. If it is going in for two parallel resistors, it should experience the same voltage difference; it should go in where each of the two ends of the resistors were joined. If the equivalent resistance is going in for two resistors in series, it should experience the same current; one end of the equivalent resistance should replace the one end of R1 and the other end of the equivalent resistance should replace the end of R2 which was not connected to R1. (Hope that's clear...)
Yes, in general it is safest to specify a current for each element in the circuit. Of course for elements in series, the current in each will be identical, so the same current symbol is typically used for each.
A junction is any point in a circuit where current may split. Different branches of a circuit are said to join at a junction. A charge carrier approaching the junction has a choice through which path it will go. At any point in the circuit, if the current has no choice but to continue to flow along a specific path, it has not encountered a junction.
Overall, do you take into account the charges and do you just deal with one charge at a time? [9/13]
Yes to all questions. You consider the potential charge by charge and sum the contributions from each. See Eqn. 25-27 in HRW. Not sure if I have really anwsered your questions though...
Some confusion often does result when we have several equations involving similar quantities. The relationship between E and V involving the line integral relies on a specified path, along which we define the infinitesimal displacement ds. In the other equation for E about a point charge, the distance r is the distance between the charge and the point of evaluation (where the positive test charge would be). So both quantitites have dimensions of length but represent very different quantities.
Just as the electric field is related to the electric force per unit charge, the electric potential is related to the electric potential energy per unit charge.
Referring to the earliest reference to superposition in HRW
(page 324) states:
Given a group of particles we find the net (or resultant) gravitational force exerted on any one of them by using the principle of superposition. This is a general principle that says a net effect is the sum of the individual effects. Here, the principle means that we first compute the gravitational force that acts on our selected particle due to each of the other particles, in turn. We then find the net force by adding these forces vectorially, as usual. |
We will again be relying on the superposition principle next week when we calculate the electric field arising from a continuous distribution of charge by dividing it into infinitesimal elements and integrating over the infinitesimal contributions to the electric field that result. So it is fair to say, that without the superposition principle, calculus would be much less useful!
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