A constant current in a wire creates a static magnetic field. So at the start of the problem, you have $B_1$ from the top loop, and $B_2$ from the bottom loop. They are both in the same direction. The question also asks you to use Lenz's Law which states when an induced current flows, it always acts to oppose the change which produced it. In other words, Lenz's Law states that induced current will act to try and maintain the status quo of the original setup.

If you take the current in the top wire to zero, this is equivalent to saying that we are turning $B_1$ off. Because $B_1$ was created by a clockwise current, by Lenz's Law loop 2 will get an induced current in the clockwise direction to create a magnetic field in the same direction $B_1$ was pointing. Thus the current in the bottom loop actually increases to compensate for the lost current in the top loop.

Let's call the circuit in the origin circuit one and it's line element $\mathrm{d}l_1=(0,\mathrm{d}y_1,0)$ and the one to it's right $r_2=(d,y_2,0)$ then the force between them is $\mathrm{d}F_{12}=i \mathrm{d}l_2 \times B_1$ where

$$B_1=\frac{\mu_0i}{4\pi}\int_{l_1} \frac{\mathrm{d}l_1\times \Delta r}{(\Delta r)^3}$$

and $\Delta r=(d,(y_2-y_1),0)$ so we have that

$$\mathrm{d}l_1\times \Delta r=(0,0,-\mathrm{d}y_1 d)$$

so we get as you wrote:

$$B_1=-\frac{\mu_0 i d}{4 \pi}\int_{0}^{d} \frac{\mathrm{d}y_1}{(d^2+(y_2-y_1)^2)^{\frac{3}{2}}}$$

Ok now let's call $y_2-y_1=t$ so $\mathrm{d}t=-\mathrm{d}y_1$ then we can write

$$B_1=\frac{\mu_0 i d}{4\pi}\int_{y_2}^{y_2-d}\frac{\mathrm{d}t}{(d^2+t^2)^{\frac{3}{2}}}$$

we now make the substitution

$$t=d\cdot \sinh(u)$$

and we obtain

$$\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$

and then

$$B_1=\frac{\mu_0 i d}{4\pi}\int \mathrm{d}u \frac{d \cosh(u)}{d^3 \cosh(u)^3}$$

in which we used

$$\cosh(u)^2-\sinh(u)^2=1$$
$$B_1=\frac{\mu_0 i }{4\pi d}\int \frac{\mathrm{d}u}{\cosh(u)^2}$$

now $\frac{1}{\cosh^2(u)}$ is the derivative of $\tanh(u)$ so

$$\int \frac{\mathrm{d}u}{\cosh(u)^2}=\tanh(u)$$

we get then

$$B_1=\frac{\mu_0 i }{4\pi d}\tanh\left(a\sinh\left(\frac{y_2-y_1}{d}\right)\right)+\text{const}$$

where we have substituted back all parameters

$$u=a\sinh\left(\frac{t}{d}\right) \\ t=y_2-y_1$$

so knowing that (where $a\sinh(x)$ is the inverse function of $\sinh(x)$):

$$\tanh(a\sinh(x))=\frac{x}{\sqrt{x^2+1}}$$

finally

$$B_1=\frac{\mu_0 i }{4\pi d} \frac{\frac{y_2-y_1}{d}}{\sqrt{(\frac{y_2-y_1}{d})^2+1}}+\text{const}=\frac{\mu_0 i }{4\pi} \frac{1}{\sqrt{(y_2-y_1)^2+d^2}}+\text{const}$$

now we calculate it between $y_1=0$ and $y_1=d$ which yields

$$B_1=\frac{\mu_0 i }{4\pi} \left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]$$

to calculate the force we take $B_1=(0,0,B_1 \hat{z})$ and we operate the following:

$$\mathrm{d}F_{12}=i\mathrm{d}l_2 \times B_1=i(B_1\mathrm{d}y_2,0,0)$$

now we have to integrate on the circuit two:

$$F_{12}=\frac{\mu_0 i^2 }{4\pi} \int_{0}^{d}\mathrm{d}y_2\left[ \frac{1}{\sqrt{(y_2-d)^2+d^2}}-\frac{1}{\sqrt{y_2^2+d^2}}\right]=\frac{\mu_0 i^2 }{4\pi} \left[I(y_2-d)-I(y_2)\right]$$

and know we do the same trick as before

$$t=y_2-d \ \text{or}\ t=y_2\ \text{for the second piece}$$
$$t=d\cdot \sinh(u)$$
$$\mathrm{d}y_2=\mathrm{d}t=d\cdot \cosh(u)\mathrm{d}u$$

then:

$$I=\int \mathrm{d}u\cdot d \cdot \cosh(u) \frac{1}{\sqrt{d^2\cosh^2(u)}}=u=a\sinh\left(\frac{t}{d}\right)$$

we finally get

$$F_{12}=\frac{\mu_0 i^2 }{4\pi}\left[a\sinh\left(\frac{y_2-d}{d}\right)-a\sinh\left(\frac{y_2}{d}\right)\right]_0^d$$

which curiously enough is zero for this choice of parameters! I hope that helped!

## Best Answer

(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'

squareloops 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 $b^{-4}$.