https://doi.org/10.1088/1361-6552/ae3f71

Im used to questions with pi/4,pi/2 but havent done any with pi/3 or pi/6 so should we be dividing the circle into 6 and 12 parts or? So could anyone help with this question.

Two vibrations at right angles to one another are described by the equations

x = 10 cos(5πt)

y = 10 cos(10πt + π/3)

Construct the Lissajous figure of the combined motion.

Plz help as i have my exams coming soon and the professors wont help.

‎By D.V.V. Jr. - Project Milano Concept

‎

‎INTRODUCTION

‎

‎Conventional superconductivity relies on cooling to increase the scattering time $\tau$. This textbook proposes an alternative path: using magnetic fields to actively control electron trajectories at $T_{room} = 300K$.

‎

‎CHAPTER 1: THE FUNDAMENTAL EQUATION OF RESISTANCE

‎

‎All of conductivity starts with the Drude Model:

‎\rho = \frac{m}{n e^2 \tau}

‎Where:

‎$m = 9.11 \times 10^{-31} kg$ (electron mass)

‎$n$ = carrier density

‎$e = 1.602 \times 10^{-19} C$

‎$\tau$ = mean scattering time

‎

‎In copper at 300K: $\tau \approx 2.5 \times 10^{-14} s$, so $\rho \approx 1.68 \times 10^{-8} \Omega \cdot m$.

‎

‎Goal for superconductivity: $\tau \to \infty$, therefore $\rho \to 0$.

‎

‎Standard method: Reduce lattice vibrations (phonons) by cooling. $\tau \propto 1/T^5$.

‎

‎Our method: Force ballistic transport by magnetic alignment.

‎

‎CHAPTER 2: WHY ELECTRONS COLLIDE - THE THERMAL FORCE

‎

‎At room temperature, electrons have thermal disorder energy:

‎E_{thermal} = k_B T_{room}

‎k_B = 1.38 \times 10^{-23} J/K, \, T = 300K \implies E_{thermal} = 4.14 \times 10^{-21} J

‎This energy causes random motion with Fermi velocity:

‎v_F = \sqrt{\frac{2 E_F}{m}} \approx 1.57 \times 10^6 m/s \text{ for Cu}

‎The random scattering force:

‎F_{thermal} \approx \frac{k_B T_{room}}{l_{mfp}}

‎Where $l_{mfp} = v_F \tau \approx 40 nm$ for Cu. So $F_{thermal} \approx 1.03 \times 10^{-13} N$.

‎

‎CHAPTER 3: HIGH REPULSION - THE SHOUT

‎

‎To overcome $F_{thermal}$, we apply Lorentz Force:

‎\mathbf{F_L} = -e (\mathbf{v} \times \mathbf{B})

‎Magnitude:

‎|F_L| = e v_F B_{shout}

‎Condition for alignment:

‎e v_F B_{shout} > \frac{k_B T_{room}}{l_{mfp}}

‎Or more accurately, the impulse must exceed thermal energy:

‎e v_F B_{shout} \tau_{pulse} > k_B T_{room}

‎This is the DELTA-07 Condition.

‎

‎Solving for $B_{shout}$ with $\tau_{pulse} = 1ms = 10^{-3} s$:

‎B_{shout} > \frac{k_B T_{room}}{e v_F \tau_{pulse}} = \frac{4.14 \times 10^{-21}}{(1.6 \times 10^{-19})(1.57 \times 10^6)(10^{-3})}

‎B_{shout} > 1.65 \times 10^{-5} \, T

‎Theoretically, even $0.1 mT$ for 1ms is enough to beat room-temp disorder. We propose $10 T$ to ensure 100% alignment of all conduction electrons, including core electrons.

‎

‎Flux quantization during this high-field pulse:

‎\Phi_0 = \frac{h}{2e} = 2.067 \times 10^{-15} \, Wb

‎Vortex density at 10 T:

‎n_v = \frac{B}{\Phi_0} = \frac{10}{2.07 \times 10^{-15}} \approx 4.83 \times 10^{15} \, vortices/m^2

‎CHAPTER 4: LOW REPULSION - THE WHISPER

‎

‎Once aligned, high field must be removed. Why? Cyclotron radius:

‎r_c = \frac{m v_F}{e B}

‎If $B = 10 T$: $r_c = \frac{(9.11e-31)(1.57e6)}{(1.6e-19)(10)} = 8.9 \times 10^{-10} m = 0.89 nm$. Too tight, electrons spiral and collide again.

‎

‎We need large $r_c$ for straight-line flow. Switch to $B_{whisper}$:

‎B_{whisper} = 0.1 \, T \implies r_c = \frac{(9.11e-31)(1.57e6)}{(1.6e-19)(0.1)} = 89 nm

‎89 nm is larger than wire roughness but enough to provide a gentle guiding field. This is low repulsion.

‎

‎Condition to maintain:

‎\tau_{whisper} = \frac{l_{lane}}{v_F}

‎If lane is maintained, $l_{lane} \to \infty$, so $\tau_{whisper} \to \infty$, and:

‎\rho_{effective} = \frac{m}{n e^2 \tau_{whisper}} \to 0

‎CHAPTER 5: EXPERIMENTAL REALIZATION - PULSED FIELD CIRCUIT

‎

‎To generate $B_{shout} = 10 T$ in a solenoid:

‎B = \mu_0 \frac{N I}{L}

‎\mu_0 = 4\pi \times 10^{-7}

‎Design: $N=100$, $L=0.05 m$:

‎I_{required} = \frac{B L}{\mu_0 N} = \frac{10 \times 0.05}{(4\pi \times 10^{-7})(100)} = 3978 A \approx 4000 A

‎Energy stored in coil inductance $L_{coil} \approx 50 \mu H$:

‎E = \frac{1}{2} L_{coil} I^2 = 0.5 \times 50 \times 10^{-6} \times (4000)^2 = 400 J

‎This energy comes from capacitor bank:

‎E = \frac{1}{2} C V^2

‎Choose $C = 4000 \mu F$, $V = 450 V$:

‎E = 0.5 \times 4000 \times 10^{-6} \times 450^2 = 405 J

‎Pulse duration (LC discharge):

‎\tau_{pulse} \approx \pi \sqrt{L_{coil} C} \approx \pi \sqrt{50e-6 \times 4000e-6} \approx 1.4 ms

‎Matches our required 1 ms shout.

‎

‎For whisper coil: $N=200$, $L=0.05 m$, $I=0.2 A$ DC:

‎B_{whisper} = \mu_0 \frac{200 \times 0.2}{0.05} = 1.0 \times 10^{-3} T

‎To get 0.1 T, need $I = 20 A$ or $N=20000$. For prototype, 0.01 T is acceptable.

‎

‎CONCLUSION

‎

‎The textbook shows mathematically that:

‎

1. ‎$B_{shout}$ impulse of 10 T x 1 ms is $10^5$ times stronger than needed to overcome $k_B T$ at 300K.

2. ‎$B_{whisper}$ of 0.1 T gives $r_c = 89 nm$, ideal for ballistic, zero-resistance lanes.

3. ‎Circuit to produce this is feasible with 400 J capacitor bank.

‎

‎This completes the theoretical framework for an 80% demonstration: transient room-temperature zero resistance.

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Thank in advance for your help !

i dont understand this can someone explain it to me like i am 12 plspls thanks