Is it possible to have torque on coil or magnetic field phenomenon by vector addition of magnetic field due to rectangular coil and external uniform magneyic field
@123 you could do the calculation that way, but it would be complicated. The energy density of a magnetic field is ½μB² so you could vector add the fields then integrate the total field to get the total energy. Then differentiating this energy wrt the coil angle would give you the torque on the coil.
It would be far simpler to use the Lorentz force on the coil, or approximate the coil by a dipole field and use the equation for the torque on a dipole in an external field.
@JohnRennie Thanks a lot for clear the confusion. These answers really helped me a lot to understand the phenomenon.
As coulombs force can be explained by two charges and distance between them. We create the idea of electric field from coulombs force.
But in lorectz force we already have magnetic field proportional to force. Why it is the case?? Why we won't define idea of magnetic force and magnetic field separately as in electrostatic??
One more question about magnetic field. How do we know the F = q(v x B), magnetic force is perpendicular to both magnetic field and direction of flow of charge.
Because by experiment we can observe only magnetic force not the magnetic field. How do we know field is perpendicular to force??? Pls pls explain.
@JohnRennie you are right but in the case of magnetic field it is perpendicular to the force. How do we know where needle of compass moving is not the direction of magnetic filed?
Because needle is moving due to magnetic force not due to magnetic field. In my opinion
Let there exist a multivariable function $f(\mathbf x)$, where it's domain is $D\subseteq \mathbb R^n$. Let there be an open set $S \subset D$. It is given that $\partial f/\partial x_i$ is defined at every point $P \in S$, for all $i$ from $1$ to $n$, where $x_i$ are the $n$ orthogonal coordinates forming up the $\mathbb R^n$ space. From the above given information, can we conclude that $f$ is continuous at every point $P\in S$? I think yes, we can.
@MoreAnonymous I also asked this in that chat room. But many times I have found more satisfying and understandable answers form this chatroom regarding such questions. In case this is inconvenient for anybody, I will consider not posting such questions here.
As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including:
ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room.
An altern...
@JohnRennie I'm kind of confused about this but in SR can a point like object spin faster than the speed of light? I mean intuitively I suspect it would lead to some kind of contradiction. I mean if $ds^2 = -c^2 dt^2 + r^2 d \theta^2 $ but when $r=0$ then $\dot \theta $ can be anything
@FakeMod No, existence of the partial derivatives is not enough: $f(x,y) = \frac{2xy}{x^2 + y^2}$ with $f = 0$ at the origin has all partial derivatives, but is not continuous at the origin
you need continuity of the partial derivatives for them to say something about $f$, and if all partial derivatives exist and are continuous, then $f$ is not only continuous but differentiable
@FakeMod This is not true. Consider $f(x, y) = (xy)/(x^2 + y^2)$ for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. Around any neighborhood of the origin, both the partial derivatives exist.
The discussion in the MathSE chat starts from here ^
@MoreAnonymous A point cannot "spin" classically. There is no physical difference between a point to which you attach a rotating frame and one to which you attach a non-rotating frame. Only extended bodies can spin, and there are no ideal rigid bodies in GR.
@MoreAnonymous What does a "spinning point" mean? Angular velocity is proportional to $r\times v$, where $r$ is the distance to the center of rotation and $v$ the velocity of the rotating object, and $r=0$ always for a point w.r.t itself as the center of rotation.
You wrote down a metric, $\theta$ is one coordinate, $t$ another, and they don't depend on each other, so differentiating $\theta$ w.r.t. $t$ doesn't make any sense
it's as nonsensical as writing $\partial y / \partial x$ for the $(x,y)$-coordinates of 2d space.
What if I had $ds^2 = - c^ dt^2 + dr^2 + r(t)^2 d \theta^2$ and $r(0) = 0$ and $r(t \neq 0) \neq 0$ and it is continuous then can I take limits at $t=0$?
@MoreAnonymous What does the metric you wrote down have to do with a point spinning? You just wrote down a generic FLRW metric in hyperspherical coordinates.
well, with $t$ dependence instead of $r$ dependence, but I don't understand why you think this makes anything spin
@MoreAnonymous It's the metric. You have to solve the geodesic equation to find the geodesics. Are you trying to run (=find edge cases to the formalism of GR) before you can walk (=properly apply the formalism of GR) again?
@Slereah sure, but the author seems to think this means only "Diff(M)-invariant quantities" are observable, although he also mentions the nature of an observable is contentious - but I think it is uncontroversial that the values of a scalar field would be observable, even though it is clearly not invariant under a diffeomorphism like translations
I still think what I wrote here is essentially correct, and the confusion arises because the actual gauge transformations and the action of a diffeomorphism just look really similar
@satan29 Please don't post your questions here directly after you asked them; interested people watch the main site anyway, and if everyone did it, the room would be flooded with new questions.
Aaahh I see. All the electric field cancel each other.
In case of electric current in wire.
@ACuriousMind what is the effect of magnetic field if wire has current and we shoot free electron parallel to the wire. What is the trajectory or motion of electron
by measuring the magnetic force on some current running through it - our best instruments for measuring magnetic field do that indirectly via the Hall effect
the fields are essentially defined by the forces they exert on test charges/current - you measure an electric field by observing the force it exerts on charges at rest inside it, and you measure a magnetic field by observing the force it exerts on current inside it
Electric force and electric field is very easy and simple to understand. Because both act in same direction. But magnetic field and magnetic force is too much confusing.
@ACuriousMind your and Qmechanic's trademark phrases are rubbing off on me: I actively resist trying to say things like "What makes you think that...?" in a colloquial setting.
Plus I also ask myself, "Who are we?" at 2 in the morning