Timeline for Work done by magnetic field on magnetic dipole
Current License: CC BY-SA 3.0
18 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Aug 10, 2014 at 0:27 | comment | added | User | $\Delta U=+W$ is used when the work is done ON "the system" which provides the external forces on the actual object (in this case, the actual object is the magnetic dipole) and thus, "the system" is "the magnetic field" or "the system who creates the magnetic field", NOT "the magnetic dipole". Therefore, your Third, Fourth, and Fifth comments are totally wrong. You should carefully examine "the system" when you calculate a work done by whatever on whatever. | |
Aug 10, 2014 at 0:13 | comment | added | User | Also, about the "ON" and "BY" things, i found that you are wrong. You said that "Again, using $\Delta U=−W$ is assuming that $W$ is the work done BY the dipole." But actually, $\Delta U=-W$ is assuming that $W$ is defined to be the work done BY a conservative force, and hence in this case, BY the Magnetic Force, (since the Magnetic Force is a conservative force) and hence, By the Magnetic Field, On the Magnetic Dipole. | |
Aug 9, 2014 at 23:59 | comment | added | User | $\int_{\theta}^{\pi/2}\mu B\sin\theta d\theta=-\left.\mu B\cos\theta\right|_{\theta}^{\pi/2}=\mu B\cos\theta$ | |
Aug 9, 2014 at 23:55 | comment | added | User | I think if we define the direction of $\vec{\theta}$ between $\vec{\mu}$ and $\vec{B}$ to be positive when the vector is coming out of the screen with the right hand rule in the counter-clockwise direction, then since the torque $\vec{\tau}=\vec{\mu}\times\vec{B}$ on the dipole tends to make $\vec{\mu}$ to line up with $\vec{B}$, (in other words, the torque makes the angle between $\vec{\mu}$ and $\vec{B}$ decrease) we will have $W=\int_{\pi/2}^{\theta}\vec{\tau}\cdot d\vec{\theta}=\int_{\pi/2}^{\theta}(-\tau d\theta)=\int_{\theta}^{\pi/2}\tau d\theta=$ (...continue in the next comment...) | |
Aug 9, 2014 at 23:21 | vote | accept | User | ||
Aug 10, 2014 at 0:53 | |||||
Jun 25, 2014 at 0:19 | comment | added | User | ;;;;; Slide is not wrong...;;;;; It's from smart physics which is a kind of a textbook for a elementary physics... I think you are the one who do not "pick one and stick to it"... | |
Jun 25, 2014 at 0:04 | comment | added | ArbiterKC | Exactly; the slide is wrong. You can't use both $\Delta U=-W_{by}$ and $\tau_{on}=\mu B\sin\theta$. You need to pick one and stick to it. This is my last comment on the topic. | |
Jun 24, 2014 at 19:50 | comment | added | ArbiterKC | Work done ON dipole: $$\Delta U=W_{\text{ON d}},\;\tau_{\text{ON d}}=\mu B\sin\theta,\,W_{\text{ON d}}=\int_{\theta_i}^{\theta_f}\mu B\sin\theta\,d\theta=\mu B(\cos\theta_i-\cos\theta_f)$$ If $U_i=0$ @ $\theta_i=\pi/2$ and $U_f=U$ @ $\theta_f=\theta$, then $U=-\mu B\cos\theta=-\vec{\mu}\cdot\vec{B}$. Now, for work done BY dipole: $$\Delta U=-W_{\text{BY d}},\;\tau_{\text{BY d}}=-\mu B\sin\theta,\,W_{\text{BY d}}=-\int_{\theta_i}^{\theta_f}\mu B\sin\theta\,d\theta=\mu B(\cos\theta_f-\cos\theta_i)$$ Same assumptions on $U_{i,f}$ and $\theta_{i,f}$ also give $U=-\vec{\mu}\cdot\vec{B}$. | |
Jun 24, 2014 at 14:00 | comment | added | User | I don't know why you think that way, yet. But, what I was saying is that If you let $\Delta U=+W$, then when you derive the Potential energy from that work, you will get, $U(\theta)=\Delta U_{\frac{\pi}{2} \to \theta}= -\int_{\frac{\pi}{2}}^{\theta} \tau d\theta=\int_{\theta}^{\frac{\pi}{2}} \tau d\theta= W_{in your answer} = \mu B cos(\theta)= \vec{\mu}\cdot\vec{B} \neq -\vec{\mu}\cdot\vec{B}$ Which is not consistent with our expectation! How do you think about that? | |
Jun 24, 2014 at 3:36 | comment | added | ArbiterKC | Again, using $\Delta U=-W$ is assuming that $W$ is the work done BY the dipole. As the magnetic field exerts a torque on the dipole, the dipole exerts a torque back on the magnetic field (Newton's 3rd law), and it is then that torque that we should use to calculate $W$ (which is opposite in sign of the one we usually use for the torque that the field exerts on the dipole). | |
Jun 24, 2014 at 3:13 | comment | added | User | I think $\Delta U=+W$ is not true since if it is true, then, it would not consistent with the result above, $U=-W=-\mu\cdot B$... | |
Jun 24, 2014 at 1:31 | comment | added | ArbiterKC | Perhaps the easiest way to look at it is: it's the same problem with the physics vs. chemistry definitions of $W$ in the 1st Law of Thermodynamics; $\Delta U=Q\pm W$. The plus sign is used by chemists, who talk about $W$ being the work done ON the system in question, while physicists use $W$ as the work done BY the system. In this case, the system is the dipole, and the magnetic field exerts a torque ON and does work ON the dipole, so we should really use $\Delta U=+W$, in which case your calculus intuition gives you exactly what you'd expect. $U=0$ is then convention. | |
Jun 23, 2014 at 19:07 | comment | added | User | Ok. So, we just define $U=0$ at $\theta=\pi/2$ without any reason, right? But then, why do you calculate the work done by the magnetic torque from $\theta$ to $\pi/2$ not from $\pi/2$ to $\theta$? It seems that you are calculating work done by an external force to oppose the magnetic torque, not by the magnetic torque which is opposite to the work done by an external force to oppose the magnetic torque. | |
Jun 23, 2014 at 4:02 | comment | added | ArbiterKC | In general, the work needs to be defined from some starting position/angle to some final position/angle: $$W=\mu B(\cos\theta_i-\cos\theta_f)$$ Then we have $\Delta U=U_f-U_i=-W$, but we are free to choose the zero at any point we want since it's a potential energy (let's say $U=0$ when $\theta=\pi/2$). This gives the desired result, but of course you could choose any other zero point you wish. | |
Jun 23, 2014 at 3:27 | comment | added | User | I'm not so sure why we define that the potential energy is zero when $\theta = \pi/2$. I thought we are deriving the potential energy from the work, not the work from the potential energy. Then, how do you know the potential energy is zero at $\theta=\pi/2$ even though we haven't derived the work done by the magenetic field on the magnetic dipole? | |
Jun 23, 2014 at 3:05 | history | edited | ArbiterKC | CC BY-SA 3.0 |
deleted 2 characters in body
|
Jun 23, 2014 at 2:58 | history | edited | ArbiterKC | CC BY-SA 3.0 |
deleted 10 characters in body
|
Jun 23, 2014 at 2:26 | history | answered | ArbiterKC | CC BY-SA 3.0 |