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: Consider a thin rigid wire-frame MNPP'N'M' in which MNN'M' and NPP'N' are two squares of side $L$ with resistance per unit-length $\lambda$ and their planes are perpendicular. The frame is rotated with a constant angular velocity $\omega$ around an axis passing through $\mathrm{NN}$ ' and put in a region with constant magnetic field $B$ pointing perpendicular to $\mathrm{NN}^{\prime}$. What is the total heat released on the frame per revolution (in Joules)? Use $L=1 \mathrm{~m}, \lambda=1 \Omega / \mathrm{m}, \omega=2 \pi \mathrm{rad} / \mathrm{s}$ and $B=1 \mathrm{~T}$.
<img_4311>

Solution:
In this setting, for every orientation during rotation the total magnetic flux passing though MNPP'N'M' is tthe same as through MPP'M', which has area $S=\sqrt{2} L^{2}$.
The magnetic flux is:

$$
\Phi(t)=B S \sin (\omega t)=\sqrt{2} B L^{2} \sin (\omega t) .
\tag{3}
$$

The emf running around the wire-frame is:

$$
E(t)=\frac{d}{d t} \Phi(t)=\sqrt{2} B L^{2} \omega \cos (\omega t)
\tag{4}
$$

The electrical current running around the wire-frame is:

$$
I(t)=\frac{E(t)}{6 \lambda L}=\frac{B L \omega \cos (\omega t)}{3 \sqrt{2} \lambda}
\tag{5}
$$

The heat released power is:

$$
\frac{d}{d t} Q(t)=I^{2}(t) \times 6 \lambda L=\frac{B^{2} L^{3} \omega^{2} \cos ^{2}(\omega t)}{3 \lambda}
\tag{6}
$$

Thus the total heat released per revolution is:

$$
Q=\int_{0}^{2 \pi / \omega} d t \frac{d}{d t} Q(t)=\frac{B^{2} L^{3} \omega^{2} \int_{0}^{2 \pi / \omega} d t \cos ^{2}(\omega t)}{3 \lambda}=\frac{\pi B^{2} L^{3} \omega}{3 \lambda} \approx 6.58 \mathrm{~J}
\tag{7}
$$

Final answer: $6.58$ J