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: At Hanoi-Amsterdam High School in Vietnam, every subject has its own flag (see Figure A, taken by Tung X. Tran). While the flags differ in color, they share the same central figure. Consider a planar conducting frame of that figure rotating at a constant angular velocity in a uniform magnetic field (see Figure B). The frame is made of thin rigid wires with same uniform curvature and same resistance per unit length. What fraction of the total heat released is released by the outermost wires?<img_4327>

Solution:
We call the current looping in the wires $I_{1}, I_{2}, I_{3}, I_{4}$ as shown in the attached Fig., and define the resistance of every quarter-circular section (radius $R$ ) of the wires to be $\rho$, then considering the



EMF on every loop gives:

$$
\begin{aligned}
& \left(2 I_{1}+\left(I_{1}-I_{2}\right)-I_{4}\right) \rho=\frac{d B_{\perp}}{d t} \times \frac{\pi+2}{2} R^{2} \\
& \left(2 I_{2}-I_{3}-\left(I_{1}-I_{2}\right)\right) \rho=\frac{d B_{\perp}}{d t} \times \frac{\pi+2}{2} R^{2} \\
& \left(2 I_{4}+2\left(I_{1}-I_{2}-I_{3}+I_{4}\right)\right) \rho=\frac{d B_{\perp}}{d t} \times 2 R^{2} \\
& \left(2 I_{3}-2\left(I_{1}-I_{2}-I_{3}+I_{4}\right)\right) \rho=\frac{d B_{\perp}}{d t} \times 2 R^{2}
\end{aligned}
\tag{9}
$$

in which $B_{\perp}$ is the perpendicular component of the magnetic field.

<img_4300>

The set of equations in Eq. (9) can be solved to get the relation between currents:

$$
I_{1}=I_{2}, I_{3}=I_{4}, \frac{I_{1}}{I_{3}}=\frac{\pi+4}{4}
\tag{10}
$$

The fraction of heat released on the outermost wires can be calculated:

$$
\frac{Q_{\text {outermost }}}{Q_{\text {all }}}=\frac{\left(8 I_{1}^{2}+8 I_{2}^{2}\right) \rho}{\left(8 I_{1}^{2}+8 I_{2}^{2}+4 I_{3}^{2}+4 I_{4}^{2}\right) \rho}=\frac{(\pi+4)^{2}}{(\pi+4)^{2}+8} \approx 0.864 \mathrm{~J}
\tag{11}
$$

Final answer: $0.864$ J