Kuliah Fisika 3 2025 pertemuan 3

Force on a Charged Particle in Magnetic Fields (Example 1: Proton)
📌 A proton with a kinetic energy ($K$) of $5.3 \text{ MeV}$ is shot northward into a uniform magnetic field ($\mathbf{B}$) of $1.2 \text{ mT}$ pointing out of the screen.
📐 Using the right-hand rule (thumb for velocity $\mathbf{V}$, fingers for $\mathbf{B}$ field direction), the magnetic force ($\mathbf{F}_B$) on the proton is directed to the right.
🧮 The magnitude of the magnetic force is calculated as $F_B = QVB \sin 90^\circ$, requiring the velocity ($V$) derived from the kinetic energy ($K = \frac{1}{2}mV^2$).
⚡ The calculated velocity is approximately $3.2 \times 10^7 \text{ m/s}$, resulting in a magnetic force magnitude of about $6.1 \times 10^{-15} \text{ N}$.

Cathode Ray Tube (CRT) and Electron Deflection
📺 The Cathode Ray Tube (CRT) setup is analogous to old television tubes, where a heated filament emits electrons accelerated by a potential difference.
⬆️ In the presence of an electric field ($\mathbf{E}$) directed downward, an electron (negative charge) experiences an upward force ($F_E = qE$), causing it to deflect upwards ($\text{acceleration } a_y = qE/m$).
⬇️ In the presence of a magnetic field ($\mathbf{B}$) directed into the screen, the electron experiences a downward force (due to the Lorentz force rule applied to a negative charge), causing deflection downward ($F_B = qVB$).

Velocity Selector Principle ($E$ and $B$ Fields)
🎯 When both $\mathbf{E}$ (downward) and $\mathbf{B}$ (inward) fields are applied such that the electron beam passes undeflected (straight through), the upward electric force balances the downward magnetic force: $F_E = F_B$.
⚖️ This balance yields the relationship for the particle's velocity ($V$): $V = E/B$ (since $\sin \theta = 1$ for perpendicular fields).
💨 In an example with $E = 4 \times 10^3 \text{ V/m}$ and $B = 2 \times 10^{-3} \text{ T}$, an undeflected electron beam implies a velocity of $V = 2 \times 10^6 \text{ m/s}$.

Determining Charge-to-Mass Ratio ($m/q$)
📏 The transverse displacement ($Y$) of an electron beam due to the electric field alone (over a plate length $L$) is given by $Y = \frac{qEL^2}{2mV^2}$.
🔄 By combining the deflection equation with the velocity selector condition ($V^2 = (E/B)^2$), the charge-to-mass ratio can be determined from the measured deflection ($Y$):
$$\frac{m}{q} = \frac{B^2 L^2}{2YE}$$
🔬 Applying this to a "mysterious particle" with known $E, B, L,$ and deflection $Y = 5 \times 10^{-3} \text{ m}$, the calculated $m/q$ ratio was $3.1 \times 10^{-9} \text{ kg/C}$, which was significantly larger than the known $m_e/q_e$ ratio ($5.7 \times 10^{-12} \text{ kg/C}$).

Key Points & Insights
➡️ The direction of the magnetic force ($\mathbf{F}_B$) on a moving charge is determined by the right-hand rule (for positive charges) or the opposite for negative charges (like electrons).
➡️ In a velocity selector setup, zero deflection occurs when the electric force ($qE$) exactly cancels the magnetic force ($qVB$), allowing for the direct calculation of velocity $V = E/B$.
➡️ The Thomson apparatus setup allows the experimental determination of the charge-to-mass ratio ($m/q$) of particles by balancing or measuring deflection caused by known electromagnetic fields.
➡️ The calculated $m/q$ ratio for the mystery particle was three orders of magnitude larger than that of an electron, suggesting it was likely a positive ion rather than an electron.

📸 Video summarized with SummaryTube.com on Nov 24, 2025, 11:09 UTC