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: An open electrical circuit contains a wire loop in the shape of a semi-circle, that contains a resistor of resistance $R=0.2 \Omega$. The circuit is completed by a conducting pendulum in the form of a uniform rod with length $\ell=0.1 \mathrm{~m}$ and mass $m=0.05 \mathrm{~kg}$, has no resistance, and stays in contact with the other wires at all times. All electrical components are oriented in the $y z$ plane, and gravity acts in the $z$ direction. A constant magnetic field of strength $B=2 \mathrm{~T}$ is applied in the $+x$ direction.
<img_4318>

Ignoring self inductance and assuming that $\alpha \ll 1$, the general equation of motion is in the form of $\theta(t)=A(t) \cos (\omega t+\varphi)$, where $A(t) \geq 0$. Find $\omega^{2}$.

Solution:
The area enclosed by the wire loop is
$$
A=\frac{1}{2} \ell^{2} \alpha+A_{0}
$$

for small angles $\alpha$, and $A_{0}$ is a constant number (which gets ignored since we really care about how this angle is changing). The flux is $\Phi=B A$ and from Lenz's Law, we have,

$$
\varepsilon=-\frac{d \Phi}{d t}=-\frac{1}{2} B \ell^{2} \dot{\alpha}
$$

One can verify that if $\alpha$ is increasing, the current will flow in the clockwise direction, so we set the counterclockwise direction as positive. The current through the wire is thus,

$$
i=\frac{\varepsilon}{R}=-\frac{B \ell^{2}}{2 R} \dot{\alpha}
$$

The magnetic force acting on it is $F_{B}=i B \ell$ and the resulting torque is

$$
\tau_{B}=F_{B} \frac{\ell}{2}=-\frac{B^{2} \ell^{4}}{4 R} \dot{\alpha}
$$

Please verify that the sign is correct. The gravitational torque is $\tau_{g}=-m g \frac{\ell}{2} \alpha$, so the torque equation gives us

$$
\begin{aligned}
0 & =\frac{1}{3} m \ell^{2} \ddot{\alpha}+\frac{B^{2} \ell^{4}}{4 R} \dot{\alpha}+m g \frac{\ell}{2} \alpha \\
0 & =\ddot{\alpha}+\frac{3}{4} \frac{B^{2} \ell^{2}}{m R} \dot{\alpha}+\frac{3}{2} \frac{g}{\ell} \alpha
\end{aligned}
$$

Recall that for a damped harmonic oscillator in the form of $\ddot{\alpha}+\gamma \dot{\alpha}+\omega_{0}^{2} \alpha=0$, the frequency of oscillations is $\omega^{2}=\omega_{0}^{2}-\gamma^{2} / 4$, so in our case, we have

$$
\omega^{2}=\frac{3}{2} \frac{g}{\ell}-\frac{9}{64}\left(\frac{B^{2} \ell^{2}}{m R} \dot{\alpha}\right)^{2}=145 \mathrm{~s}^{-1}
$$

Final answer: $145$ $\mathrm{~s}^{-1}$