Blue vector is the magnetic field vector. Red vector is the velocity vector of the point charge (light grey) q. Green vector is the force vector (force on moving charge q due to the magnetic field). I have heard that workdone by a magnetic field on a moving charge is zero. Let's say that a charge q which was already moving with a velocity 'V' came under the influence of a magnetic field 'B'. It started experiencing a force 'F' due to the magnetic field given by F = q(V vector X B vector). Now due to this it will also start moving along the 'y' axis with some acceleration. So wouldn't we say that the work-done = F x displacement in y ?  Edit: What I am trying to say is that the displacement of the charge will have a 'y' component and the force here is also acting in the 'y' direction', so shouldn't the expression I wrote above be valid?
According to my text book:  The Lorentz Force F on a charged particle moving in an electromagnetic field is given by:  $$F= qE + q(v\times B)$$  then it states that "it is to be pointed out that only the electric force does work, while the no work is done by the magnetic force which is simply a deflecting force."  How can this by true won't the the magnetic force have a considerable magnitude, will it now impart kinetic energy to the charge? Any explanations will help a lot. Please keep the terminology simple, thanks.
Use Stokes's Theorem to show $\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2$ I am a little stuck on the following problem:  Use Stokes's Theorem to show that  $$\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2,$$  where $C$ is the suitably oriented intersection of the surfaces $x^2 + y^2 + z^2 = a^2$ and $x + y + z = 0$.  OK, so Stokes's Theorem tells me that:  $$\oint_{C}\vec{F} \cdot d\vec{r} = \iint_{S}\operatorname{curl} \vec{F} \cdot \vec{N} ~dS$$  I have calculated:  $$\operatorname{curl} \vec{F} = -\vec{i} - \vec{j} - \vec{k}.$$  I then figured that on the surface $S$ we must have:  $$\vec{N}dS = \vec{i} + \vec{j} + \vec{k}dxdy$$  since this follows from the equation of the given plane.  However this will then give me:  $$\operatorname{curl} \vec{F} \cdot \vec{N} = -1 -1 -1 = -3$$  And thus I would get, if project this onto the $xy$-plane:  $$\iint_{S} \operatorname{curl} \vec{F} \cdot \vec{N} ~dS = -3 \iint_{A} dA = -3 \pi a^2$$  which is obviously not correct.  I would greatly appreciate it if someone could help me with this.  I actually had multivariable calculus a few years ago, and I know that I knew this stuff then.  However, now that I need it again I notice that I've become quite rusty.  Thanks in advance :) I am having trouble understanding the estimation of an AR process. In some textbooks, the AR(1) process is defined as follows: $y_{t}=\theta y_{t-1}+ϵ_t$ (which does not contain a constant). So the OLS estimator is biased. I am confused about the cause of the bias. It is explained that $y_{t-1}$ is dependent on $ϵ_{t-1}$ although it is independent of $ϵ_t$. However in linear regression, if the equation does not contain a constant, we cannot make sure the expectation of disturbance is zero. So the OLS estimator is bias without a constant. Does it mean that the OLS estimator is unbiased if I add a constant in AR process? Remove sudo privileges from a user (without deleting the user) Is there anyway to get tuple operations in Python to work like this:  &gt;&gt;&gt; a = (1,2,3) &gt;&gt;&gt; b = (3,2,1) &gt;&gt;&gt; a + b (4,4,4)   instead of:  &gt;&gt;&gt; a = (1,2,3) &gt;&gt;&gt; b = (3,2,1) &gt;&gt;&gt; a + b (1,2,3,3,2,1)   I know it works like that because the __add__ and __mul__ methods are defined to work like that. So the only way would be to redefine them? I have a larger, external monitor connected to my MacBook Pro and have arranged it such that the dock is in the external monitor (via System Preferences > Displays > Arrangement, by dragging the white bar to the larger screen). This works fine, except for the fact that every now and then, the dock will suddenly shift back to my MacBook screen, but when I open up Displays to change this back, the white bar is still on the external monitor. So I have to drag it to the smaller screen, and then drag it back to the external monitor.   It's a minor issue, obviously, but I've noticed that it has been happening more and more frequently and it's getting quite annoying. Perhaps there's something I've been doing unintentionally for this to happen?   I'm using Mavericks, 10.9.5. I remember after Stargate: Universe was cancelled, that I read, somewhere on the web, either part of an interview or an article by one of the producers about what they were planning if there had been a third season.  (He said they weren't sure whether they'd set it soon after the ending of the second season, or later, or were going to do something else.)  He also wrote that, while he felt okay discussing that, he did not feel like he could reveal how the story about the superstructure the Destiny had found would turn out.  He said that was something the other two producers had worked out and he didn't want to reveal it and would leave it up to them to reveal it if they wanted to.  Was this ever revealed?  Did the other producers ever discuss or reveal anything about what was going to happen with that storyline?    Addition: I'm referring to the superstructure that is the message found in the background radiation throughout the universe, not anything from the planet builders or other easily seen physical structures. How to disable USB storage device support I want to disable a particular USB storage device (a sandisk flash drive). I read somewhere that you can use the .config file. How do i use it to disable that? In Illustrator is there a way to precisely scale to a pivot point or "reference point" that is off the screen? I mean when you want to zoom in somewhere far away from where you set that point and then scale precisely to align something.   Here's :     To scale relative to a different reference point , click where you   want the reference point to be in the document window, move the   pointer away from the reference point, and then drag until the object   is the desired size.   Well, I tried using this click-drag scaling method as well, but the level of precision is too low to be usable. It overshoots or undershoots the guide I'm aiming for.  The Scale tool's pop-up dialog isn't useful either. It only offers percentages. And moving/zooming are disabled while the dialog window is up, meaning I cannot navigate to the guide I need to see.  For these reasons, I need to be zoomed in on the guide, tweaking the measurements from the top Control Panel.  Some images to help illustrate what I mean:    ▲ The circle marks where I would like the center of my scale transformation to be.    ▲ Now zoomed in somewhere else, I am trying to get the black border of this bitmap image to align perfectly with the vector guide by scaling it.    ▲ I nudge the numbers in the Control Panel until the image is perfectly aligned.  One would think all would be perfect, but then I look and my reference point has vanished!   It seems the scaling has been executed relative to one of these limited nine reference point choices  instead of the reference point I specified.  There must be a way to scale precisely to a specific reference point in illustrator. Can somebody please enlighten me? Are there cases where a period is placed inside parentheses? In which cases the period (or a comma) should be placed inside the parentheses? What is the association bonus, and how does it work? Where is "the" used? Is there a rule? Why no definite article in "Shops are open late in summer"? Notice these sentences:   Shops are open late in summer.   Summer is a traditionally viewed as a slow season for games.   At the summer solstice, the days are longest and the nights are shortest.   it's time to start thinking about what 2016's Song of the Summer will be.    In English class, they told us that "the" is used for the things that are known.  One example for this was a phrase like "The man who killed his wife".  So when we are speaking of that man, we know what he did. Somehow we know something more than nothing about him. He's not just every man in the world.     So Now I'm confused about why "the" is not used in the first and second sentences. Can anyone please explain it for me? Basically, I acidentally spawned too many ArmorStands due to a wrong code writing , resulting in the game crashing, and wanted to know how to delete those armorstands from the World save folder. They hava names and I belive they also have a scoreboard set, if that helps.   I have already tried to open the world but it crashes, I tried to open a server (Since it worked the last time I tried it) and killing it all from the console but, different from the last time, it crashed before killing them and now I just tried MCedit but it crashes as well. Convert NSDate to NSString How do I convert, NSDate to NSString so that only the year in @"yyyy" format is output to the string? No sound ALC294 Asus ROG Strix 512 Ubuntu 20.04.01 How do you get a submit button in the Windows Phone keyboard? How can I get maximum DoF with no distance markers? I am aware of what hyperfocal distance is and the general principles behind it. However, most of the tutorials I have seen and read regarding it make use of lenses with focus distance indicators. Unfortunately, neither of my lenses (Nikkor 18-55mm &amp; 35mm f1.8) have indicators.  I have read that you can 'cheat' and simply focus a third of the way into the shot (which I have done previously simply by spot-focusing on the lowest focus point in my VF). I also assume you could pace out the distance and note an object to focus on when you get back to the camera (though that might be easier said than done).  Does anyone have any suggestions on how to achieve maximum sharpness from front to back without focus distance markers. A new lens is not an option; global economic crisis and all that! Best Algebraic Geometry text book? (other than Hartshorne) Proof of orthogonal matrix In the following question we regard vectors in $\Bbb R^n$ as columns and define the dot product in the usual way which means that $x · y$ = $x^{t}y$  a) If A is an n × n matrix show that $A_{ij}$ = $e_i \cdot A e_j$ where $e_i$, $i =1, \dots , n$ are the standard basis vectors in $\Bbb R^n$  b) Show that a matrix $P$ is orthogonal if and only if ($Px$) · ($Py$) =$x · y$ for all $x, y$ ∈ $\Bbb R^n$  Working -   I got part (a) alright but have no idea of where to start with part (b) except to use the information that an n x n matrix $P$ is orthogonal if $P^{-1}$= $P^{t}$ and that $P^{t}$$P$=$I$, $P$$P^{t}$=$I$, where $I$ is the identity matrix. Any suggests are gracefully appreciated. Feel free to edit my question for clarity. Thank you Random forest regression not predicting higher than training data I've noticed that when building random forest regression models, at least in R, the predicted value never exceeds the maximum value of the target variable seen in the training data. As an example, see the code below. I'm building a regression model to predict mpg based on the mtcars data. I build OLS and random forest models, and use them to predict mpg for a hypothetical car that should have very good fuel economy. The OLS predicts a high mpg, as expected, but random forest does not. I've noticed this in more complex models too. Why is this?   &gt; library(datasets) &gt; library(randomForest) &gt;  &gt; data(mtcars) &gt; max(mtcars$mpg) [1] 33.9 &gt;  &gt; set.seed(2) &gt; fit1 &lt;- lm(mpg~., data=mtcars) #OLS fit &gt; fit2 &lt;- randomForest(mpg~., data=mtcars) #random forest fit &gt;  &gt; #Hypothetical car that should have very high mpg &gt; hypCar &lt;- data.frame(cyl=4, disp=50, hp=40, drat=5.5, wt=1, qsec=24, vs=1, am=1, gear=4, carb=1) &gt;  &gt; predict(fit1, hypCar) #OLS predicts higher mpg than max(mtcars$mpg)       1  37.2441  &gt; predict(fit2, hypCar) #RF does not predict higher mpg than max(mtcars$mpg)        1  30.78899 How do I extract all the external links of a web page and save them to a file?