Answer the kibble balance problem

_psets/4.md

@@ -250,6 +250,64 @@ relates to the current through it and the vertical gradient of the magnetic fiel
 definition of the force on an infinitesimal current, Green’s theorem, and the formula for the
 divergence of a magnetic field).
 
+Instead of a coil, I'll just consider one current loop. A coil can then be modeled as a stack of
+such loops. Assume that our loop is centered at the origin and lies in the xy plane. The
+differential force generated on a differential current by a magnetic field is $$dF = I (dl \times
+B)$$. In a uniform magnetic field the forces cancel out, so let's consider a linearly nonuniform
+magnetic field
+
+$$
+B(x) = B_0 + D[B] x
+$$
+
+where $$B(x)$$, $$x$$, and $$B_0$$ are 3-vectors and $$D[B]$$ is the (3x3) Jacobian matrix of
+$$B(x)$$.
+
+Then the $$z$$ component of $$I (dl \times B)$$ at any location $$(x, y, 0)$$ on our loop is
+
+$$
+I dl_x \left( B_{0x} + \frac{\partial B_y}{\partial x} x + \frac{\partial B_y}{\partial y} y \right)
+- I dl_y \left( B_{0y} + \frac{\partial B_x}{\partial x} x + \frac{\partial B_x}{\partial y} y \right)
+$$
+
+Let's move to radial coordinates to make the integration easier. Let's say our loop has radius
+$$r$$. For a point on our loop at angular position $$\theta$$,
+
+$$
+dl = (-r \mathrm{d} \theta \sin{\theta}, r \mathrm{d} \theta \cos{\theta}, 0)
+$$
+
+Thus the $$z$$ component of $$I (dl \times B)$$ is
+
+$$
+-I r \mathrm{d} \theta \sin{\theta} \left( B_{0x} + \frac{\partial B_y}{\partial x} \cos{\theta} +
+    \frac{\partial B_y}{\partial y} \sin{\theta} \right) \\
+- I r \mathrm{d} \theta \cos{\theta} \left( B_{0y} + \frac{\partial B_x}{\partial x} \cos{\theta} +
+    \frac{\partial B_x}{\partial y} \sin{\theta} \right)
+$$
+
+We need to integrate this over $$\theta$$ from 0 to $$2 \pi$$. Though the expression above looks
+complicated, the integral ends up being easy. To spare all the algebra, any term that only involves
+$$\cos{\theta}$$ or $$\sin{\theta}$$ integrates to zero since we're integrating over one complete
+period. The same happens for both terms that have $$\cos{\theta} \sin{\theta}$$. So there are only
+two terms left, and they reduce to $$\cos^2{\theta}$$ and $$\sin^2{\theta}$$, both of which
+integrate to $$\pi$$. So in the end the $$z$$ component of the force on the loop is
+
+$$
+-I \pi r^2 \left( \frac{\partial B_y}{\partial y} + \frac{\partial B_x}{\partial x} \right)
+= I \pi r^2 \frac{\partial B_z}{\partial z}
+$$
+
+since $$\nabla \cdot B = 0$$.
+
+The current is adjusted until this force exactly balances the weight of the mass, so we have
+
+$$
+mg = I \pi r^2 \frac{\partial B_z}{\partial z}
+$$
+
+where $$g = 9.8 \si{m/s^2}$$ is the local acceleration due to gravity.
+
 ### (b)
 
 {:.question}
@@ -257,17 +315,109 @@ In the dynamic phase, the coil is moved at a constant vertical velocity and the
 measured. Calculate how the voltage depends on the velocity and vertical gradient of the magnetic
 field.
 
+The induced voltage is proportional to the time rate of change of the magnetic flux through the
+current loop. The flux is $$\pi r^2 B_z$$. If we move the loop along the z axis with a velocity
+$$v$$, the rate of change of the flux is
+
+$$
+\pi r^2 v \frac{\partial B_z}{\partial z}
+$$
+
+So the voltage is
+
+$$
+V = - \pi r^2 v \frac{\partial B_z}{\partial z}
+$$
+
 ### (c)
 
 {:.question}
 Combine these results to express the weight in terms of the voltage and current.
 
+From part (a) we have
+
+$$
+\pi r^2 \frac{\partial B_z}{\partial z} = \frac{mg}{I}
+$$
+
+and from part (b)
+
+$$
+\pi r^2 \frac{\partial B_z}{\partial z} = -\frac{V}{v}
+$$
+
+Thus
+
+$$
+m = -\frac{I V}{g v}
+$$
+
+If we use a coil with $$n$$ windings instead of a single loop then we just need to multiply by
+$$n$$.
+
+### (a), (b), and (c) without the assumption of linearity
+
+Let's not assume that the magnetic field varies linearly with distance. Instead, let's assume that
+the field is radially symmetric. This is a reasonable assumption since if the field isn't at least
+periodically radially symmetric then there will be a torque on the loop. Call the radial component
+of the field experienced by our loop $$B_r(z)$$. (There must also be some z component, but we won't
+end up needing it so we won't give it a name.)
+
+Now calculating the force on the loop is easy. The z component of $$B$$ doesn't matter since it just
+pulls the loop outward. And the radial component produces a force along z. So the force is $$-2 \pi
+r I B_r(z)$$.
+
+But how can we calculate the flux? Consider a z-aligned cylinder of radius $$r$$ and height $$z$$,
+with the center of its base at the origin. The magnetic flux through the bottom is some constant
+amount $$\Phi_\text{bottom}$$. The flux through the walls is
+
+$$
+\Phi_\text{walls}(z) = 2 \pi r \int_0^z B_r(z') \mathrm{d}z'
+$$
+
+Then since the divergence of $$B$$ is zero everywhere, Green's theorem tells us that
+
+$$
+\Phi_\text{top}(z) = -\Phi_\text{walls}(z) - \Phi_\text{bottom}
+$$
+
+So by the
+[fundamental theorem of calculus](https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)
+
+$$
+\frac{\partial}{\partial z} \Phi_\text{top}(z) = -2 \pi r B_r(z)
+$$
+
+Thus when our loop travels at a fixed velocity $$v$$ the induced voltage is
+
+$$
+2 \pi r v B_r(z)
+$$
+
+Solving these equations for $$m$$ yields
+
+$$
+m = -\frac{I V}{g v}
+$$
+
+as before.
+
 ### (d)
 
 {:.question}
 Why are the voltage and current measured separately, rather than by doing a measurement of both at
 the same time?
 
+Well the voltage is only generated when the coil is moving, but the current is defined as that which
+keeps the coil still. So you can't do both at the same time. You could start the coil moving and
+then measure the current that keeps it moving at a constant velocity, simultaneously measuring the
+resulting voltage. But then the voltage is a combination of the applied voltage and the induced
+voltage, and ditto for the current. The field is much harder to analyze.
+
+At the end of the day the point is to have the effects of the magnetic field cancel out, so it
+doesn't need to be perfect as it did in the previous definition. In other words, we just want to
+measure a voltage and a current and not worry about the specifics of the magnetic field.
+
 
 ## (6.6)