# Equivalence of two equations of force on a magnet in a magnetic field

According to Biot and Savart's experiment, force on a magnetic pole m due to a circuit s' is: $$\vec{F}=m\oint\dfrac{Ids'\times\hat{r}}{r^{2}}=m\vec{B}$$

Now if we consider the magnetic pole m as a circular circuit s, then according to Ampere Force Law if we integrate force on ds$=I\vec{ds}\times\vec{B}$ around the circular path of electron's orbit inside the magnet, then how do we get the same force as $m\vec{B}$

i.e.please show by calculation how $\oint I\vec{ds}\times\vec{B}=m\vec{B}$

where integration is taken around the circular path of electron's orbit inside the magnet

Note:- Both forces should be equal because both tells the same thing, i.e.force on a magnetic pole due to circuit s'. We are only looking at the same force in two different ways.

My calculation:-

Since we consider the $\vec{B}$ around the magnetic pole as constant,we also consider the $\vec{B}$ around circular circuit s as constant. Thus $\vec{F}$ reduces to:

$$\vec{F}=IB\oint\vec{ds}\times\hat{B}$$

Since magnet points in the direction of magnetic field $\vec{B}$ and the orientation of magnet is always perpendicular to the plane of circular circuit s inside it. Therefore $\vec{B}$ points in the direction perpendicular to $\vec{ds}$ .Thus $\vec{F}$ reduces to: $$\vec{F}=IB\oint \text{ds sin90} \hat{a}$$ (where $\hat{a}$ is a vector perpendicular to $\vec{ds}$ and $\vec{B}$. Therefore $\hat{a}$ points in the direction of radius of circular circuit) $$\tag{1} =IB \oint \text{ds} \hat{a}$$

Now for every $\hat{a}$ from 0 to $\pi$ of circular circuit s, there is a negative $\hat{a}$ in the exact opposite direction. Therefore:

$$\tag{2} \oint \text{ds} \hat{a}=0$$

By (1) and (2):

$$\vec{F}=0$$

Since the electrons are moving in a circular path inside the magnet, pole strength of the magnet is non-zero. Magnetic field due to the wire is also non- zero. But the force on magnet is coming out to be zero.

But according to Biot-Savart equation, when pole strength (m) and magnetic field ($\vec{B}$) are non-zero, force on magnetic pole would also be non-zero.

Why is this contradiction happening? Am I missing something or somewhere wrong in my calculation??

• Can somebody answer??? – user130324 Oct 6 '16 at 14:05
• " consider the magnetic pole m as a circular circuit s" I think here you got confused. Circular circuit with current is equivalent to magnetic moment (also called magnetic dipole), not magnetic monopole. – Ján Lalinský Oct 6 '16 at 20:47
• Is it like this "force on a magnetic pole m is $m\vec{B}$ and the force on the whole magnet is $m\vec{B}+(-m)\vec{B}=0$" If this is the case, then how does a magnet move in a megnetic field – user130324 Oct 7 '16 at 5:05
• Magnetic pole theory works well only for long thin rod magnets; in general, force on a magnet cannot be expressed as sum of forces on two magnetic poles. In special case where a thin rod is magnetized so the poles are its ends, and the magnetic field is uniform, the net magnetic force on the rod would indeed be zero. In general, field is not uniform and the magnet is attracted to places with higher magnetic field. – Ján Lalinský Oct 7 '16 at 21:38
• I have modified my question and posted it as a new question here. Please try to answer it. – user130324 Oct 8 '16 at 14:40