OPhO 2020, Теория — Задача 10. Magnetostatics

Solution 10: For permeable media, two of Maxwell’s equations become

where $\mathbf H=\mathbf B/\mu$ and $\mathbf J_{free}=0$ because there is no free current. Note that $\mu=\mu_r\mu_0$. Thus, the equations become $\nabla\cdot\mathbf B=0$ and $\nabla\times\mathbf H=0$. Applying these to the boundary, we get the following boundary conditions:

where region 1 is $x<0$ and region 2 is $x>0$. We claim that if a magnetic monopole of magnetic charge $q$ is placed a distance $d$ away from the permeable medium instead of a dipole,

(1) in the region $x<0$, the magnetic field is equivalent to that of the original monopole and a magnetic monopole placed at $(d,0,0)$ with magnetic charge $q_1=-\frac{\mu_r-1}{\mu_r+1}q_0$ (2) in the region $x>0$, the magnetic field is equivalent to that of the original monopole at its current position but with charge $q_2=\frac{2\mu_r}{\mu_r+1}q_0$.

To show this, note that as long as we show that the boundary conditions are satisfied, we are done by the uniqueness theorem because the field in the rest of the space satisfies Laplace’s equation. For the region $x<0$, the boundary at the real magnetic monopole is already satisfied. Now the only boundary we need to consider is the interface $x=0$. Consider a point $P$ on the plane $x=0$ such that the line from $(-d,0,0)$ to $P$ forms an angle $\theta$ with the x-axis. Define $B_i$ to be the magnitude of the field at $P$ from $q_i$. According to the claim, the magnetic field near $P$ in region 1 has a magnetic field component perpendicular to the interface of $B_{1,\perp}=B_0\cos\theta-B_1\cos\theta$. This must be equal to the perpendicular component in region 2, so we have

Now, the parallel component of $\mathbf H$ in region 1 is $\frac{B_0}{\mu_0}\sin\theta+\frac{B_1}{\mu_0}\sin\theta$, which must be equal to the parallel component of $\mathbf H$ in region 2, so

Solving the two equations gives

Noting that $B_i\propto q_i$, we see that the claim is true.

We apply the claim to the problem via superposition. We treat the magnetic dipole as two magnetic monopoles of magnetic charges $q$ and $-q$ that are a small distance $s$ away from each other. Note that $m=qs$.

(a) If the dipole is at $(-d,0,0)$ and pointed towards the $+x$ direction, the field in region 1 is equivalent to that of the same dipole superposed with another dipole at $(d,0,0)$ pointing in the $+x$ direction with moment $m_1=q_1s=\frac{\mu_r-1}{\mu_r+1}m$. The field produced by the latter magnetic dipole at position of the original dipole is given by $\frac{\mu_0m_1}{2\pi r^3}$, where $r=2d$. The magnetic force on the original dipole is given by

Note that the force is attractive.

(b) We integrate the force from (a) to get the work done:

(c) Instead of rotating the dipole from the original orientation, we first bring the dipole to infinity, rotate it, and then bring it back to $(-d,0,0)$. From part (b), we have the work it takes to bring the dipole to infinity. Rotating it at infinity requires no work. Finally, to calculate the amount of work required to bring it back, we find the potential energy of the final system. From our claim, we can apply superposition to see that the setup is equivalent to one in which the permeable medium acts like a magnetic dipole at $(d,0,0)$ with moment $m_1=\frac{\mu_r-1}{\mu_r+1}$ and pointed in a direction that makes an angle $-\theta$ with the $+x$-axis. The general formula for dipole-dipole interaction is given by:

which equals

However, this is not the work we do to bring it back from infinity; rather, it is twice our work. To see why, imagine an external agent bringing in the imaginary dipole from $+\infty$ as we are bringing in the real dipole from $-\infty$. By Newton’s third law, the force they apply is equal and opposite to ours, and so they do the same amount of work as we do. Thus, the work it takes to bring in the dipole from $-\infty$ is

In total, the amount of work done is

(d) By similar reasoning from the previous parts, we can see that the image of the ring with current $I$ will be a ring centered at $x=d$ with current $\frac{\mu_r-1}{\mu_r+1}I$ in the same direction.

Claim 1: The mutual inductance between the real ring and the image dipole is given by $\frac{\mu_0R^2\cos\theta}{2(R^2+4d^2)^{\frac32}}*\frac{\mu_r-1}{\mu_r+1}A$, where $A$ is the area of the dipole.

Proof: By the mutual inductance reciprocity theorem $(M_{12}=M_{21})$, we just need to calculate the flux through the dipole over a current $I$ in the ring. Since the dipole is ideal, we can find the field at the dipole and do $\Phi=BA\cos\theta$. After plugging in well-known expression $B=\frac{\mu_0R^2}{2(R^2+x^2)^{\frac32}}$, where $x=2d$, we get the desired answer.

A natural corollary of this claim is that the mutual inductance between the real ring and the real dipole is $\frac{\mu_0\cos\theta}{2R}m$ Claim 2: The mutual inductance between the real ring and the image ring is given by $\mu_0R\ln\frac Rd$. Proof: Consider the field lines from the ring. Since $d\ll R$, the field lines near the edge of the ring are circular, so the flux through the image ring from the real ring is the same as the flux through a circle coplanar with the real ring with radius $R-d$. This is because of the fact that magnetic field has zero divergence. To find the flux through the circle of radius $R-d$, we use the fact

where $\mathbf A$ is the vector potential. By rotational symmetry, we only need to calculate the tangential component of $\mathbf A$ at a point $R-d$ from the center, and then multiply by $2\pi(R-d)$. We have

We simplify to get

Now, we realize that the integral diverges if $d=0$. Thus, since $d\ll R$, we see that the integral is dominated by the region where $\theta$ is close to zero. We can use the small angle approximation $\cos\theta\approx1-\frac{\theta^2}{2}$. We simplify to get

Finally, we multiply by $2\pi R$ and divide by $I$ to get the desired mutual inductance. Note: the result can also be achieved using Grover’s formula.

The flux through the superconducting ring must remain zero. The flux contributed by the ring itself is simply $LI$. The flux contributed by each dipole is the mutual inductance over the area times the dipole moment. The flux contributed by the image ring is the mutual inductance times the current in the image ring. Combining all the terms and setting it to zero gives:

Solving, we get

(e) The force on a dipole is given by $\mathbf F=\nabla(\mathbf p\cdot\mathbf B)$. We can break this force up into components. The x-component is given by $F_x=p_x\frac{\partial B_x}{\partial x}+p_y\frac{\partial B_x}{\partial y}$, where the dipole is in the x-y plane. Note that the ring contributes no gradient of $\mathbf B$ field at the dipole, so we only need to consider the field from the image ring and image dipole. First we find the force on the dipole due to the image ring. The field from the image ring is given by $B=\frac{\mu_0I_1R^2}{2(R^2+(2d)^2)^{\frac32}}$, so the gradient in the x direction is $\frac{3\mu_0I_1R^2d}{(R^2+(2d)^2)^{\frac52}}$. The x-component of the field from the image ring doesn’t change significantly in the y-direction, by symmetry, so we get $\frac{\partial B_x}{\partial y}=0$.

Now we consider the y-component $F_y=p_x\frac{\partial B_y}{\partial x}+p_y\frac{\partial B_y}{\partial y}$. The y-component of the field from the image ring along its axis is 0, so $\frac{\partial B_y}{\partial x}=0$. To find $\frac{\partial B_y}{\partial y}$, we use the fact that $\nabla\cdot\mathbf B=0$, so we get $\frac{\partial B_x}{\partial x}+2\frac{\partial B_y}{\partial y}=0$, where here we are only considering the field from the image ring. Thus, $\frac{\partial B_y}{\partial y}=-\frac{\partial B_x}{2\partial x}=-\frac{3\mu_0I_1R^2d}{2(R^2+(2d)^2)^{\frac52}}$.

It remains to find the force between the image dipole and the real dipole. The dipole-dipole interaction is given by equation (9). To find the x-component of the force $F_x$, we can calculate the derivative of the potential energy in the x direction. Moving the dipole in the x direction only changes $r$, so we have $F_x=\frac{dU}{dr}=-\frac{3\mu_0}{4\pi r^4}(mm_1\cos(2\theta)-3mm_1\cos^2\theta)=\frac{3\mu_0}{4\pi(2d)^4}mm_1(\cos^2\theta+1)=\frac{3\mu_0}{64\pi d^4}mm_1(\cos^2\theta+1)$.

To find the y-component of the force $F_y$, we can calculate the derivative of the potential energy in the y direction. Moving the dipole in the y direction does not change $r$ to first order, but does change $\hat{\mathbf r}$. Thus, $F_y=-\frac{dU}{dy}=-\frac{3\mu_0}{4\pi r^4}(2mm_1\cos\theta\sin\theta)=-\frac{3\mu_0}{32\pi d^4}mm_1\cos\theta\sin\theta$.

Finally, we add up all the forces to get

, where $I$ is given in part (d).