Deflection of magnetic compass sorrounded by alternating poles What happens to the deflection of the magnetic compass if it is surrounded by south poles and north poles in alternating direction of a magnet around it in a circular pattern. Will it be deflected in any particular direction ?

 A: To add to Rijul Gupta's answer: depending on the exact symmetry of the arrangement (see below), theoretically, the needle can move freely to point in any direction. Practically, there will be some direction which is one of minimum energy, so the needle will have a weak tendency to point in this minimum energy direction. Try analysing your situation with some simple equations: static magnetic fields are made up of divergenceless dipoles. In your case you have $N$ dipoles, so the nett magnetic dipole that the needle feels is:
$$\vec{\mu}=\sum\limits_{k=1}^N (\mu+\delta_j) \left(\cos\left(\frac{2\,k\,\pi}{N} + \alpha_j\right)\,\hat{X} + \sin\left(\frac{2\,k\,\pi}{N} + \alpha_j\right)\,\hat{Y}\right)\approx \sum\limits_{k=1}^N \delta_j \left(\cos\left(\frac{2\,k\,\pi}{N}\right)\,\hat{X} + \sin\left(\frac{2\,k\,\pi}{N}\right)\,\hat{Y}\right) + \\\mu\sum\limits_{k=1}^N \alpha_j \,\left(-\sin\left(\frac{2\,k\,\pi}{N} \right)\,\hat{X} + \cos\left(\frac{2\,k\,\pi}{N} \right)\,\hat{Y}\right)$$
where the $\delta_j$ and $\alpha_j$ measure minute deviations from the symmetry that you show. These imperfections will always be there. They are very noticeable because the compass bearing is designed to have very low friction, and the tiniest torque will shift the compass needle. The needle has a dipole moment $\vec{\mu}_m=\cos\theta\,\hat{X} + \sin\theta\,\hat{Y}$, where $\theta$ defines its pointing direction. The total system potential energy is $-\vec{\mu}_m\cdot\vec{\mu}$ or:
$$U \approx -\sum\limits_{k=1}^N \left(\left(\delta_j \cos\left(\frac{2\,k\,\pi}{N}\right)-\mu\, \alpha_j\sin\left(\frac{2\,k\,\pi}{N}\right)\right)\,\cos\theta + \\\quad\quad\quad\quad\quad\left(\delta_j \sin\left(\frac{2\,k\,\pi}{N}\right)+\mu\, \alpha_j\,\cos\left(\frac{2\,k\,\pi}{N}\right)\right)\,\sin\theta\right)$$
an equation of the form $U=A\cos(\theta + B)$ which clearly has a direction of minimum potential energy.
If the $\alpha_j$ and $\beta_j$ are very small, then the torque from the nett dipole will be weaker than that of the torque owing to the Earth's magnetic field, and the direction of the needle will be towards that of the Earth's magnetic field (i.e. the system behaves somewhat like a compass), but there will be a nonzero deviation between the Earth's magnetic field and the compass. The potential energy owing to the Earth's magnetic field will be of the form $U_e\cos(\theta-\phi)$, where $\phi$ defines the Earth's magnetic field's direction: the total potential energy will then be of the form $A\cos(\theta + B) + U_e\cos(\theta-\phi)$, which will be minimum for some angle different from $\theta=\phi$.
A: It will be free to move and point towards any direction. 
Similar effect are observed when it will be completely surrounded by south or north poles. This is actually what happens at the poles of the earth!
