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Let $\vec{B}$ be an uniform magnetic field in space. I'm having problems trying to imaginate the behavior of a wire loop inside the magnetic field. First, we got that the torque $T$ will be $T$ = $\vec{\mu} \times \vec{B}$, where $\vec{\mu} = NiA\vec{n}$, being A the area inside the wire loop, $i$ the current and $N$ the number of loops. Also, $\vec{n}$ is the normal vector here.

So, we can write the torque as $T = \mu B \sin{\theta}$ and an infinitesimal torque $dT$ will be $dT = \mu B \sin{\theta}d\theta$. So, we can write the work as $$ W = - \int \mu B \sin{\theta}d\theta$$ So $W = \mu B \cos \theta$, and since the potential energy $U$ is $-W$, $U = - \mu B \cos \theta$.

Okay. Not ill start explaining my problem: So: Let's say, when the normal vector $\vec{n}$ is at angle $\frac{\pi}{2}$ with the magnetic field. Here, the torque is maximum, and the potential energy is $0$. So here is my problem: I tried to associate this problem with the mass-spring oscillator. In the MSO,when $x = 0$, we have no force, and minimum potential energy (so lmaximum kinetic energy). However, in the wire loop, in the point where $|U| = 0$, we have the maximum torque. This is confusing me, because when the torque equals zero, it is in the point of maximum kinetic energy, isn't it? However, since the equations for torque and potential energy don't actually "converge" (by converge i mean: when one is maximum, the other also is, and so on with the minimum), it is being hard to think about the movement of the wire loop. Any help? Thanks!

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You need to do a definite integral from $0$ to $\theta$.

This is not equivalent to a MSO since the angle only has a finite range.

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  • $\begingroup$ The book i'm using doesn't do a definite integral. Why? $\endgroup$ – Vitor C Goergen Nov 2 '16 at 22:06
  • $\begingroup$ @Dovah-king It will give $W(\theta)-W(0)$ and $U(\theta)-U(0)$. You want to define $U(0)=0$. Not necessary, just an added constant, but less confusing. $\endgroup$ – Keith McClary Nov 2 '16 at 23:54
  • $\begingroup$ Yeah. About the finite angle range: Does this means that he won't complete a $2 \pi$ degrees turn? If not, what is the maximum range and how do i calculate it? $\endgroup$ – Vitor C Goergen Nov 3 '16 at 0:12
  • $\begingroup$ @Dovah-king If I'm understanding you correctly, this is equivalent to a pendulum. If it has enough KE it will swing over the top. $\endgroup$ – Keith McClary Nov 3 '16 at 0:45
  • $\begingroup$ But, in the pendulum, imagine the point where $U=0$, we have that $K$ and the force that causes the oscillatory movement is $0$, why this doesn't happen here? $\endgroup$ – Vitor C Goergen Nov 3 '16 at 1:15

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