# New Technique!!!

That moment strikes again, I have been like this: http://abstrusegoose.com/426 since yesterday.

One day, a little boy asked you what is the value of magnetic field due to a circular current loop at a point along the symmetrical axis of the loop?

You got tired as soon as you hear that question “Urghhh not this again”

Well a simple Biot-Savart’s would easily do the job. Or perhaps if you want to show him a fancier method you can calculate $\vec{A}$ and then just curl the sh*t. $\vec{B}=\nabla\times( sh*t)$

What a boring problem…

BUT

I can turn this problem into an interesting one! Because I found another method to calculate the $B$ using the concept of….. dum.. dum.. dum….duuum…. MUTUAL INDUCTANCE!

See? Now the question is changed into: “How could mutual inductance has anything to do with this problem?”

Any idea?

Okay imagine a circular current loop.. Hmm well let’s make it even more interesting, imagine an ARBITRARY closed current loop. Remember that the goal is to calculate the magnetic field $B_P$ due to this loop at a point, say point P. Now let’s turn this problem into a mutual inductance problem.

Here is the key, put a really really small closed current loop with magnetic dipole moment $\vec{m_S}=I_s\vec{A_S}$ at point P. Our plan is to calculate the magnetic flux $\phi_{SA}$ through this small loop due to the arbitrary loop and then divide this flux with $A_S$. It works because the loop small loop is so tiny that we can simply write $\phi_{SA}=\vec{B_P} .\vec{A_S}$. Here is how it looks like:

The coolest thing about mutual inductance is that it is always symmetrical, in this case $M_{AS}=M_{SA}$. If you want a proof go take a look at Feynman Lectures on Physics Vol.2 pg.17-10. So we have

$\frac{\phi_{SA}}{I_A}=\frac{\phi_{AS}}{I_S}$

Which means that we can calculate $\phi_{SA}$ by calculating $\phi_{AS}$. We know that the magnetic field due to a very tiny loop is similar to that of a magnetic dipole $\vec{m}$ . Or we can picture it as two magnetic charges $\pm q_m$ separated by an infinitely small distance $d$ so that  $\vec{m}=q_m\vec{d}$.

Let’s calculate first the magnetic flux through the arbitrary loop due to  $+q_m$ alone. Thanks to magnetic Gauss law we can choose any surface as long as it is bounded by the arbitrary loop. Since the field is radial, it is best to choose a part of spherical shell of radius $r$ as the flux surface. But in general we can’t find part of spherical shell that is completely bounded by the arbitrary loop, thus we need  additional “radial surface” to make it fit to the arbitrary loop. However it’s not a problem at all since the flux through this “radial surface” is of course zero, because the field is tangential to it. So the flux is simply

$\phi_{+q_m}=\frac{\mu_0}{4\pi}\frac{q_m}{r^2}A_+$

Where $A_+$ is the area of spherical shell portion. Then by the definition of solid angle $\Omega$, we have $\frac{A_+}{r^2}=\Omega_+$. Therefore

$\phi_{+q_m}=\frac{\mu_0}{4\pi}{q_m}\Omega_+$

Don’t forget about $-q_m$ buddy, we can calculate its contribution the flux using the same method. Let’s put them together

$\phi_{AS}=\frac{\mu_0}{4\pi}{q_m}[\Omega_+-\Omega_-]$

$\phi_{AS}=\frac{\mu_0}{4\pi}{q_m}[\Omega(\vec{r})-\Omega(\vec{r}+\vec{d})]$

Recall the definition of derivative, the $(\Omega(\vec{r})-\Omega(\vec{r}+\vec{d}))$ on the left hand side can be replaced with $-\frac{d\Omega}{dr}d$. Note that $dr$ is a small displacement element in the direction of $\vec{d}$. Thus we get

$\phi_{AS}=-\frac{\mu_0}{4\pi}{q_m}d\frac{d\Omega}{dr}=-\frac{\mu_0}{4\pi}m\frac{d\Omega}{dr}=-\frac{\mu_0}{4\pi}I_SA_S\frac{d\Omega}{dr}$

Then using the mutual inductance symmetry relation we have

$\phi_{SA}=\frac {I_{A}}{I_{S}}\phi_{SA}=-\frac{\mu_0}{4\pi}I_AA_S\frac{d\Omega}{dr}$

Looks like we get the magnitude of magnetic field at point P

$B_P=-\frac{\phi_{SA}}{A_S}=-\frac{\mu_0I_A}{4\pi}\frac{d\Omega}{dr}$

The direction of $\vec{B_P}$ that we have just calculated is the same as that of $\vec{d}$ which we have arbitrarily chosen since I had secretly used $\vec{B_P}.\vec{A_S}=B_PA_S$ without telling you. So

$\vec{B_P}=-\frac{\mu_0I_A}{4\pi}\frac{d\Omega}{dr}\hat{d}$

“Wait! How could..” Yes you are right this is in fact not the total magnitude of $\vec{B_P}$. What we have just calculated is merely the projection of $\vec{B_P}$ into the direction of $\vec{d}$ that we chose. So how do we find the real value of $\vec{B_P}$? It’s not quite a big deal, we can simply change the orientation of $\vec{d}$ until we get the maximum value of $\vec{B_P}$, and that is the actual value of $\vec{B_P}$. Notice that only $\frac{d\Omega}{dr}$ that depends on the direction of $\vec{d}$ that we chose. So we have to find the maximum value of this little $\frac{d\Omega}{dr}$. If we understand carefully the meaning of gradient of a function $\nabla f$, it actually represents the maximum vectorial derivative of a function. Therefore finally we arrived at the final result

$\vec{B_P}=-\frac{\mu_0I_A}{4\pi}\nabla\Omega$

Isn’t it cute? Excellenteee right!

I hope that this new handy skill can be used creatively to tackle some of the most awesome physics problems in the world!! Can’t wait to apply this mutual inductance method on anything!

Now I think it is the time for this song