Return to Answer

Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$

The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$

So the current in the ring (in an anticlockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.

[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.]

Flux will be escaping out of the sides of the core, all along its length, so $\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$) $$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$

in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.

The upward motor effect (Laplace) force on the ring will be $$F_z=-B_r I 2 \pi r=\frac{\partial \Phi}{\partial z} \times I=\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$

Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is

$$F_z=-\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$ That is $$F_z=-\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$

Since the magnitude of $\phi$ diminishes with $z$, $\frac{d}{d z} [\phi(z)]^2$ is negative so $F_z$ is positive; the ring jumps rather than dives! This is irrespective of the direction of the flux.

$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution.

It's easy enough, though, to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search\ peak}=n \omega \phi(z)$$

The mean relative permeability, $\mu_r$, of the core over the a.c. cycle will no doubt be in the order of $10^3$. Therefore the magnetic field affecting the ring will be almost entirely that due to the core, rather than that due directly to the coil. We shall assume a sinusoidal variation of flux in the core, ignoring the variation of $\mu_r$ over the cycle.

Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$

The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$

So the current in the ring (in an anticlockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.

[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.]

Flux will be escaping out of the sides of the core, all along its length, so $\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$) $$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$

in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.

The upward motor effect (Laplace) force on the ring will be $$F_z=-B_r I 2 \pi r=\frac{\partial \Phi}{\partial z} \times I=\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$

Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is

$$\langle F_z \rangle =-\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$ That is $$\langle F_z \rangle =-\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$

Since the magnitude of $\phi$ diminishes with $z$, $\frac{d}{d z} [\phi(z)]^2$ is negative so $F_z$ is positive; the ring jumps rather than dives!

$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution.

It's easy enough, though, to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search\ peak}=n \omega \phi(z)$$

Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$

The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$

So the current in the ring (in a clockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.

[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.]

Flux will be escaping out of the sides of the core, all along its length, so $\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$) $$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$

in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.

The upward motor effect (Laplace) force on the ring will be $$F_z=B_r I 2 \pi r=-\frac{\partial \Phi}{\partial z} \times I=-\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$

Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is

$$F_z=\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$ That is $$F_z=\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$

$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution.

It's easy enough, though, to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search\ peak}=n \omega \phi(z)$$

Almost inevitably there is a sign adrift in the main argument; I will investigate and correct the mistake.

Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$

The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$

So the current in the ring (in an anticlockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.

[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.]

Flux will be escaping out of the sides of the core, all along its length, so $\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$) $$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$

in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.

The upward motor effect (Laplace) force on the ring will be $$F_z=-B_r I 2 \pi r=\frac{\partial \Phi}{\partial z} \times I=\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$

Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is

$$F_z=-\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$ That is $$F_z=-\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$

Since the magnitude of $\phi$ diminishes with $z$, $\frac{d}{d z} [\phi(z)]^2$ is negative so $F_z$ is positive; the ring jumps rather than dives! This is irrespective of the direction of the flux.

$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution.

It's easy enough, though, to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search\ peak}=n \omega \phi(z)$$

Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$

The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$

So the current in the ring (in a clockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.

[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.]

Flux will be escaping out of the sides of the core, all along its length, so $\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$) $$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$

in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.

The upward motor effect (Laplace) force on the ring will be $$F_z=B_r I 2 \pi r=-\frac{\partial \Phi}{\partial z} \times I=-\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$

Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is

$$F_z=\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$ That is $$F_z=\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$

$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution.

It's easy enough, though to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search}=n \omega \phi(z)$$

Almost inevitably there is a sign adrift in the main argument; I will investigate and correct the mistake.

Suppose that at height $z$ up the core, the flux in the $z$ direction through the core is $$\Phi=\phi(z)\ \cos(\omega t).$$

The emf induced in the ring (if it is at height $z$) is $$\mathscr E =-\frac{\partial \Phi}{\partial t}=\phi(z)\ \omega\ \sin(\omega t).$$

So the current in the ring (in a clockwise sense seen from above) will be $$I=\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}$$ in which $R$ is the ring's resistance, as a turn of a coil, and $L$ is the ring’s inductance.

[We assume the ring has thickness $h$ in the z direction, an inner radius $r$ only just bigger than that of the core, and a very small radial thickness, $b$.]

Flux will be escaping out of the sides of the core, all along its length, so $\frac{\partial \Phi}{\partial z}$ will be negative. But (using $\text{div}\vec B =0$) $$-\frac{\partial \Phi}{\partial z}=2 \pi r B_r$$

in which $B_r$ is the radial flux density just outside the core at height z, due to flux escaping from the core.

The upward motor effect (Laplace) force on the ring will be $$F_z=B_r I 2 \pi r=-\frac{\partial \Phi}{\partial z} \times I=-\frac{d \phi}{d z}\ \cos(\omega t) \times\left[\frac{\phi(z)\ \omega R\ \sin (\omega t)}{R^2 + \omega^2 L^2} -\frac{\phi(z)\ \omega^2 L\ \cos (\omega t)}{R^2 + \omega^2 L^2}\right]$$

Over a number of cycles the mean of $\sin(\omega t)\ \cos(\omega t)$ is zero and that of $\cos^{2}(\omega t)$ is $\frac12$, so the mean value of $F_z$ is

$$F_z=\frac12 \frac{d \phi(z)}{d z} \times\frac{\phi(z)\ \omega^2 L}{R^2 + \omega^2 L^2}$$ That is $$F_z=\frac14 \frac{d}{d z} [\phi(z)]^2 \times\frac{\omega^2 L}{R^2 + \omega^2 L^2}$$

$L$ and $R$ can also be evaluated or at least estimated. The force problem has therefore been shown to be equivalent to the simpler-sounding one of finding $\phi(z)$ for a steel rod with a coil wound round one end. This is essentially a magnetostatics problem, which may well already have a standard solution.

It's easy enough, though, to determine $\phi(z)$ experimentally. We run a.c. through the coil at the bottom of the rod and measure the emf induced in a 'flat' search coil placed around the rod at various values of $z$. If the search coil has $n$ turns then $$\mathscr E_{search\ peak}=n \omega \phi(z)$$

Almost inevitably there is a sign adrift in the main argument; I will investigate and correct the mistake.