Subfield: Electromagnetism

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: Two carts, each with a mass of $300 \mathrm{~g}$, are fixed to move on a horizontal track. As shown in the figure, the first cart has a strong, tiny permanent magnet of dipole moment $0.5 \mathrm{~A} \cdot \mathrm{m}^{2}$ attached to it, which is aligned along the axis of the track pointing toward the other cart. On the second cart, a copper tube of radius $7 \mathrm{~mm}$, thickness $0.5 \mathrm{~mm}$, resistivity $1.73 \cdot 10^{-8} \Omega$, and length $30 \mathrm{~cm}$ is attached. The masses of the magnet and coil are negligible compared to the mass of the carts. At the moment its magnet enters through the right end of the copper tube, the velocity of the first cart is $0.3 \mathrm{~m} / \mathrm{s}$ and the distance between the two ends of each cart is $50 \mathrm{~cm}$, find the minimum distance achieved between the two ends of the carts in centimeters. While on the track, the carts experience an effective coefficient of static friction (i.e., what it would be as if they did not have wheels) of 0.01 . Neglect the self-inductance of the copper tube.
<img_4305>

A picture of the two cart setup. The black rectangle represents the magnet while the gold rectangle represents the copper tube.

Hint 1: The magnetic field due to a dipole of moment $\vec{\mu}$, at a position $\vec{r}$ away from the dipole can be written as
$$
\vec{B}=\frac{\mu_{0}}{4 \pi} \frac{(3 \hat{r}(\vec{r} \cdot \vec{\mu})-\vec{\mu})}{r^{3}} \hat{r}
$$
where $\hat{r}$ is the unit vector in the direction of $\vec{r}$.

Hint 2: The following mathematical identity may be useful:
$$
\int_{-\infty}^{\infty} \frac{u^{2} d u}{\left(1+u^{2}\right)^{5}}=\frac{5 \pi}{128}
$$

Solution:
We analyze the thin $\mathrm{dz}$ of copper which center is at distance $\mathrm{z}$ away from the magnet. Notice that



the thickness of the copper tube $(w=0.5 \mathrm{~mm})$ is much smaller than its radius. Thus, we may model the copper tube as an infinitesimally thin current loop of radius $a_{0}=a+(w / 2)$ (note that $\frac{w}{2 a}$ is significant!) with resistance $\rho \frac{2 \pi a_{0}}{w d z}$ where $a=7 \mathrm{~mm}$. Using the formula for the magnetic field due to the dipole moment, it can be found that the B-field along the current loop is:
$$
B=\frac{\mu_{0} \mu}{4 \pi} \frac{\left(2 z^{2}-r^{2}\right) \hat{z}+3 r z \hat{r}}{\left(r^{2}+z^{2}\right)^{5 / 2}}
$$
where $\hat{z}$ is the unit vector parallel to the direction of motion of the magnet and $\hat{r}$ is the unit vector pointing radially outward from the center of the current loop. The flux through the loop is then:
$$
\Phi_{B}=\int_{0}^{a_{0}} B_{z}(2 \pi r) d r=\frac{1}{2} \mu_{0} \mu \frac{a_{0}^{2}}{\left(z^{2}+a_{0}^{2}\right)^{3 / 2}}
$$
in which Lenz law can be used to get the current of the loop, using the fact $v=-\frac{d z}{d t}$, where $v$ is the relative velocity of the right cart to the left.
$$
I=-\frac{d \Phi_{B}}{d t}\left(\rho \frac{2 \pi a_{0}}{w d z}\right)^{-1}=\frac{3 \mu_{0} \mu}{4 \pi} \frac{a_{0} w v z d z}{\rho\left(a_{0}^{2}+z^{2}\right)^{5 / 2}}
$$
The lorentz force on the current loop due to the magnet then obeys:
$$
F=I\left(2 \pi a_{0}\right) B_{r} \hat{z}=\frac{9}{8 \pi} \mu_{0}^{2} \mu^{2} \frac{a_{0}^{3} w v z^{2} d z}{\rho\left(a_{0}^{2}+z^{2}\right)^{5}} \hat{z}
$$
which is repulsive as expected. By Newton's third law, the force on the magnet due to the current in the copper tube is:
$$
F=-\frac{9}{8 \pi \rho} \mu_{0}^{2} \mu^{2} a_{0}^{3} w v \int_{-x}^{L-x} \frac{z^{2} d z}{\rho\left(a_{0}^{2}+z^{2}\right)^{5}} \hat{z}
$$
where $x$ is the depth of the magnet into the copper tube. However, when $x, L-x>>a_{0}$,
$$
\int_{-x}^{L-x} \frac{z^{2} d z}{\rho\left(a_{0}^{2}+z^{2}\right)^{5}} \hat{z} \approx \frac{a_{0}^{3}}{a_{0}^{10}} \int_{-\infty}^{\infty} \frac{u^{2} d u}{\left(1+u^{2}\right)^{5}}=\frac{5 \pi}{128 a_{0}^{7}}
$$
Therefore, for as long as the magnet is sufficiently within the tube, the force is instead:
$$
F=-\frac{45}{1024 \rho a_{0}^{4}} \mu_{0}^{2} \mu^{2} w v \hat{z}=-\gamma v \hat{z}
$$
where $\gamma=0.1815 \mathrm{~kg} / \mathrm{s}$.

The velocity of the second cart starts moving when $F>\mu_{s} M g$ with $\mu_{s}=0.01$ and $M=0.3 \mathrm{~kg}$. It turns out that $\frac{0.01 \mathrm{Mg}}{\gamma v_{0}} \approx 0.5$, so this static friction is overcome very quickly. Once both carts are in motion, and since there is no kinetic friction, momentum is in fact conserved! As we'll see soon, this results in the carts performing an (infinitely-long) perfectly inelastic collision. In addition, it allows us to use a reduced mass of $M_{\text {red }}=M / 2$ to calculate the closest distance achieved between the two carts. While $x>>a$ is obeyed:
$$
-\gamma v_{r e l}=\frac{1}{2} M \frac{d v_{r e l}}{d t}
$$
giving:
$$
v_{r e l}=v_{0} e^{-\frac{2 \gamma}{M} t}
$$

in which $v_{r e l}$ goes to 0 as $t$ goes to $\infty$, resulting in a perfectly inelastic collision. Since $v_{r e l}=-\frac{d x}{d t}$
$$
x_{\min }=x_{0}-\int_{0}^{\infty} v_{r e l} d t=x_{0}-\frac{m v_{0}}{2 \gamma}=0.252 m
$$

Final answer: 25.2 cm