Question

Two circular coil of radii $r$ and $3R (R>>r)$ carrying current $I$ and $2I$ respectively are placed coaxially and parallel to each other at separation $4R$.

The force between two coils is $\frac{y\mu_0\pi}{(5)^5} (\frac{rI}{R})^2$. Find $\frac{y}{10}$.

Answer 1

270

Answer 2

The force between two coaxial circular coils can be approximated when one coil is much smaller than the other and the separation is significant. Let coil 1 have radius $r_1=r$ and current $I_1=I$, and coil 2 have radius $r_2=3R$ and current $I_2=2I$. The separation is $d=4R$. Given $R>>r$, coil 1 is much smaller than coil 2 and the separation.

We can treat coil 1 as a magnetic dipole with magnetic moment $\vec{m}_1 = I_1 A_1 \hat{k} = I (\pi r^2) \hat{k}$. The force on this dipole in a magnetic field $\vec{B}_2$ produced by coil 2 is given by $\vec{F} = \nabla(\vec{m}_1 \cdot \vec{B}_2)$. Since the field $B_2$ is along the z-axis, $F_z = m_1 \frac{\partial B_{2z}}{\partial z}$.

The magnetic field on the axis of coil 2 (radius $a=3R$, current $i=2I$) at a distance $z$ from its center is: $B_{2z}(z) = \frac{\mu_0 i a^2}{2(a^2 + z^2)^{3/2}} = \frac{\mu_0 (2I) (3R)^2}{2((3R)^2 + z^2)^{3/2}} = \frac{9\mu_0 I R^2}{(9R^2 + z^2)^{3/2}}$

Let coil 2 be at $z=0$ and coil 1 be at $z=d=4R$. We need to find the gradient of $B_{2z}$ at $z=d$: $\frac{\partial B_{2z}}{\partial z} = \frac{\partial}{\partial z} \left[ 9\mu_0 I R^2 (9R^2 + z^2)^{-3/2} \right] = 9\mu_0 I R^2 (-\frac{3}{2}) (9R^2 + z^2)^{-5/2} (2z) = -27\mu_0 I R^2 z (9R^2 + z^2)^{-5/2}$

Evaluating at $z=d=4R$: $\frac{\partial B_{2z}}{\partial z}|_{z=4R} = -27\mu_0 I R^2 (4R) (9R^2 + (4R)^2)^{-5/2} = -108\mu_0 I R^3 (25R^2)^{-5/2} = -108\mu_0 I R^3 (125R^5)^{-1} = -\frac{108\mu_0 I}{125R^2}$

The magnitude of the force on coil 1 is: $F = |m_1 \frac{\partial B_{2z}}{\partial z}| = |(I \pi r^2) (-\frac{108\mu_0 I}{125R^2})| = \frac{108 \mu_0 \pi I^2 r^2}{125R^2}$

The problem states the force is $F = \frac{y\mu_0\pi}{(5)^5} (\frac{rI}{R})^2$. $F = \frac{y\mu_0\pi}{3125} \frac{r^2 I^2}{R^2}$

Comparing our result with the given form: $\frac{108}{125} = \frac{y}{3125}$ $y = \frac{108}{125} \times 3125 = 108 \times 25 = 2700$

The question asks for $\frac{y}{10}$. $\frac{y}{10} = \frac{2700}{10} = 270$