Force between parallel current carrying loops How to derive the force between two parallel current carrying loops?(current flows in the same sense,say clockwise,in both the loops)
Radius of both the loops is R and each has a current I flowing through it.Both the loops are at a distance of 'd' from each other.
 A: (a) Not a very profound contribution, but we'd expect (wouldn't we?) that when the separation, $d$, between the loops is much less than the radius, $r$, the force is approximately that between long parallel straight wires a distance $d$ apart. So...
$$F=\mu_0 I^2 \frac rd\ \ \ \ \ \ \text {if}\ \ \ \ \ \ d << r.$$
(b) We can derive an exact equation for the force between two 'co-axial' square loops using pretty simple mathematics, starting with
$$\vec F_{1,2}=\frac{\mu_o I_1I_2}{4\pi r^3}\left[(\vec{\delta l_1}.\vec r_{2,1}) \vec{\delta l_2}- (\vec{\delta l_1}.\vec{\delta l_2})\vec r_{2,1}\right]$$
I find the attractive force to be of magnitude
$$F=\frac{2\mu_o I_1I_2}{\pi}\left[-\frac{t \sqrt{2+t^2}}{1+t^2} + \frac{1+2t^2}{t\ \sqrt{1+t^2}}-1 \right]$$
in which $$t=\frac{\text{separation of loops}}{\text{side length of loop}}$$
This predicts limiting expressions for large and small $t$ quite nicely. For example, for small $t$ (loops much closer than side length), the second term in the square brackets dominates, and the expression simplifies to
$$F=\frac{2\mu_o I_1I_2}{\pi t}$$
that is
$$F=\frac{4a\mu_o I_1I_2}{2\pi b}$$
in which $a$ is the side length of the loop and $b$ is the loop separation. So we have the same force as between parallel straight wires when the loop separation is much less than the side length.
For large separations, $b\gg a$, that is $t\gg 1$, we can show by expanding the terms in the square bracket binomially in powers of $1/t$, that $F$ varies as $b^{-4}$.
A: I'll just provide a brief outline here, since the full derivation is long and because exercises of this type are valuable to work out oneself when learning electromagnetism.
It can be shown that the force between two current loops carrying currents $I_1$ and $I_2$ can be written as a double loop integral:
$$
\vec{F}_2 = - \frac{\mu_0}{4 \pi} I_1 I_2 \oint \oint \frac{\vec{\mathscr{r}} \,(d \vec{\ell}_1 \cdot d \vec{\ell}_2)}{\mathscr{r}^3},
$$
where $\vec{\mathscr{r}} = \vec{\ell}_2 - \vec{\ell}_1$.  The proof of this result involves combining the Biot-Savart Law and the Lorentz force law, applying the BAC-CAB rule, and arguing that one of the resulting terms automatically vanishes when integrated over a closed loop.
For two coaxial circular loops of radius $R$ separated by a distance $d$, this double integral can be reduced to
$$
F_{2z} = - \frac{\mu_0}{2} I_1 I_2 x^2 \int_0^{2 \pi} \frac{\cos u \, du}{(1 + 2 x^2 (1 - \cos u))^{3/2}},
$$
where $z$ is the axial direction and $x \equiv R/d$.  Note that the off-axis components of $\vec{F}$ vanish by symmetry.
As expected, this integral can't be expressed in terms of elementary functions.  If we define
$$
f(x) = x^2 \int_0^{2 \pi} \frac{\cos u \, du}{(1 + 2 x^2 (1 - \cos u))^{3/2}},
$$
then Mathematica yields a result in terms of elliptic integrals of the first and second kinds:
$$
f(x) = -K\left(-4 x^2\right)-\frac{K\left(\frac{4 x^2}{4 x^2+1}\right)}{\sqrt{4
   x^2+1}}+\frac{\left(2 x^2+1\right) E\left(-4 x^2\right)}{4 x^2+1}+\frac{\left(2
   x^2+1\right) E\left(\frac{4 x^2}{4 x^2+1}\right)}{\sqrt{4 x^2+1}}.
$$
This is not terribly illuminating, but one thing we can do is graph the result on a logarithmic scale:

We see two regimes.  When $d \ll R$, we see that the slope of the graph is approximately -1;  this is the expected regime where $F \propto d^{-1}$.  When the two loops are very close together, the force on every "bit" of each loop is dominated by the pieces of the other loop that are nearby;  and since on such scale the other loop "looks straight".  In other words, the loops act like two long parallel wires.  It can in fact be shown that
$$
f(x) \approx 2x 
$$
as $x \to \infty$, meaning that in this case $F_{2z} \approx - \mu_0 I_1 I_2 R/d$, exactly what we would expect from taking the usual result for the force-per-length of a long wire and multiplying it by the loops' "length" of $2 \pi R$.
The other regime, where $d \gg R$, can be observed to have a $F_{2z} \propto d^{-4}$ dependence.  This is the regime where the loops are very far apart, and "see" each other as point dipoles.  The force between two dipoles is given by $\vec{F} = - \vec{\nabla} (\vec{m} \cdot \vec{B})$; and since the field of a dipole is proportional to $r^{-3}$, the force between them will be proportional to the derivative of this, or $r^{-4}$.  (The exact coefficient works out to be $f(x) \approx 3 \pi x^4$ as $x \to 0$, or $F_{2z} \approx - \frac{3 \pi}{2} I_1 I_2 R^4/d^4$.  It is left as an exercise to the reader to derive this result for two "small" dipoles of area $\pi R^2$ separated by a distance $d$.)
A: Based on symmetry, the B field produced by loop 1 and acting at a point on loop 2, should have the same magnitude and the same relative direction for any point on loop 2.  It can be calculated using the Biot equation, but you might need to start with numbers and do a numeric integration.
A: Determine the magnetic field from one loop, then use $F = qV \times B = \ell I \times B$. I believe that should give you the force from one loop onto another.
Edit: I figured the asker might be working in a simplified scenario. I agree that my answer is too simplified otherwise! It doesn't take the nonuniform aspect of the field into account at all.
