Unit 03: Magnetic Effects of Current • Chapter 04[cite: 12, 13]
Lorentz Force & Particle Trajectories in E and B Fields
Cross-product dynamics, circular orbits, helical pitch, zero-work theorem, and velocity selector crossed fields.
Electric field ek accelerator pedal jaisa hai: chahe particle ruka ho ya chal raha ho, yeh uspar force lagakar uski kinetic energy badha deta hai. Lekin Magnetic Field ek pure steering wheel hai! Ruke hue charge par yeh zero force lagata hai ($v = 0 \implies F_B = 0$)[cite: 13]. Aur jab charge chalta hai, toh yeh force hamesha velocity ke strictly perpendicular ($90^\circ$) lagata hai[cite: 13]. Iska matlab magnetic force kabhi particle ki speed ya kinetic energy change nahi kar sakta ($W = 0$)—yeh sirf uska path bend karta hai. Agar injection perpendicular ho toh circular orbit banta hai, aur agar kisi angle $\theta$ par ho toh corkscrew spring jaisi Helical Motion ban jati hai!
1. Lorentz Force & Work Theorem
Total electromagnetic force acting on charge $q$ moving with velocity $\vec{v}$[cite: 13]:
Zero-work identity: Since $\vec{F}_B \perp \vec{v}$, instantaneous power $P = \vec{F}_B \cdot \vec{v} = 0 \implies W = \Delta K = 0$ (speed is constant).
2. Circular Orbit & Helical Parameters
Centripetal equilibrium ($qv_\perp B = \frac{mv_\perp^2}{r}$) yields radius, period, and pitch:
Crucial fact: Time period $T$ and cyclotron frequency $f_c = \frac{qB}{2\pi m}$ are strictly independent of speed $v$ and radius $r$!
Circular Trajectory & Centripetal Balance ($\theta = 90^\circ$)[cite: 13]
Fig 4.3A: Uniform circular motion of a positive charge with centripetal magnetic force $\vec{F}_B \perp \vec{v}$[cite: 13].
1. A charge $q$ enters a uniform magnetic field $\vec{B}$ perpendicular to its velocity ($\vec{v} \perp \vec{B}, \theta = 90^\circ$)[cite: 13].
2. Magnetic force magnitude is $F_B = q v B \sin 90^\circ = q v B$. By right-hand cross product, $\vec{F}_B$ is perpendicular to $\vec{v}$ at every instant, providing the required centripetal acceleration:
3. Time period $T$ to complete one revolution:
Core Invariant: Period $T$ and cyclotron frequency $f_c$ are completely independent of speed $v$ and radius $r$!
Helical Trajectory & Pitch Geometry ($0 < \theta < 90^\circ$)[cite: 13]
Fig 4.3B: Resolution of velocity into $v_\parallel$ (translation) and $v_\perp$ (circular rotation) forming a corkscrew helix of pitch $p$[cite: 13].
1. When velocity $\vec{v}$ makes an angle $\theta$ with $\vec{B}$, resolve into two orthogonal components:
2. Magnetic force on parallel component: $\vec{F}_\parallel = q(\vec{v}_\parallel \times \vec{B}) = 0$ ($\because \sin 0^\circ = 0$). Hence $v_\parallel$ remains constant, producing uniform linear translation along the field lines.
3. Magnetic force on normal component: $\vec{F}_\perp = q(\vec{v}_\perp \times \vec{B})$ provides centripetal acceleration, producing circular motion of radius:
4. Superposition of uniform translation along the field and circular rotation in the transverse plane forms a helix.
5. The Pitch $p$ is the linear distance advanced along the field in one full revolution ($t = T = \frac{2\pi m}{qB}$):
Velocity Selector in Mutually Perpendicular Crossed Fields ($\vec{E} \perp \vec{B}$)[cite: 13]
1. Set up mutually perpendicular electric and magnetic fields ($\vec{E} = E\hat{j}$ along $y$-axis, $\vec{B} = B\hat{k}$ along $z$-axis). A positively charged particle $q$ enters with velocity $\vec{v} = v\hat{i}$ along $x$-axis[cite: 13].
2. Electric force acts along $+y$: $\vec{F}_E = q\vec{E} = qE\hat{j}$[cite: 13].
3. Magnetic force acts along $-y$: $\vec{F}_B = q(\vec{v}\times\vec{B}) = q(v\hat{i}\times B\hat{k}) = -qvB\hat{j}$[cite: 13].
4. For the particle to pass straight through without any angular deflection, the two opposing forces must balance out ($\vec{F}_{\text{net}} = 0$)[cite: 13]:
Filter Mechanism: Particles with $v > E/B$ deflect towards magnetic force; particles with $v < E/B$ deflect towards electric force. Only particles with exactly $v = E/B$ emerge undeflected, completely independent of their charge $q$ or mass $m$[cite: 13]!
Trap 1: The "Work Done by Magnetic Field" Myth
*Board Question:* A proton enters a magnetic field with velocity $\vec{v}$ and completes a semicircle. What is the work done by the magnetic field?
*Trap Answer:* $\frac{1}{2}mv^2$ or non-zero work.
*Correct Answer:* ZERO! Because $\vec{F}_B = q(\vec{v}\times\vec{B})$, force is always strictly perpendicular to instantaneous displacement $d\vec{r}$. Therefore, $dW = \vec{F}_B \cdot d\vec{r} = F_B ds \cos 90^\circ = 0$. Magnetic field changes direction of momentum, but kinetic energy remains strictly constant.
Trap 2: Radius Comparison for Different Particles
Be extremely alert to what is kept constant when comparing paths of Proton ($p$), Deuteron ($d$), and Alpha particle ($\alpha$):
• Same velocity $v$: $r = \frac{mv}{qB} \implies r \propto \frac{m}{q}$ ($r_p : r_d : r_\alpha = 1 : 2 : 2$).
• Same kinetic energy $K$: $r = \frac{\sqrt{2mK}}{qB} \implies r \propto \frac{\sqrt{m}}{q}$ ($r_p : r_d : r_\alpha = 1 : \sqrt{2} : 1$).
• Same accelerating voltage $V$: $K = qV \implies r = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}} \implies r \propto \sqrt{\frac{m}{q}}$ ($r_p : r_d : r_\alpha = 1 : \sqrt{2} : \sqrt{2}$).
Trap 3: Sign of Charge in Velocity Selector
Does the velocity selector condition $v = E/B$ depend on whether the particle is positive or negative[cite: 13]?
NO! If charge is negative (e.g. electron), $\vec{F}_E$ flips direction downward, but $\vec{F}_B$ also flips direction upward by cross-product. The two forces remain exactly opposite. Both electrons and protons pass straight through if $v = E/B$[cite: 13]!
Trap 4: Helical Motion at $\theta = 0^\circ$ or $180^\circ$
If a charged particle moves parallel or antiparallel to $\vec{B}$ ($\theta = 0^\circ$ or $180^\circ$):
$\vec{F}_B = q v B \sin(0^\circ) = 0$.
The particle experiences zero magnetic deflection and continues moving along a straight line with unaffected velocity.
Problem: An electron projected with speed $v = 4.0 \times 10^6\text{ m/s}$ enters a uniform magnetic field $B = 0.5\text{ T}$ at an angle of $30^\circ$ to the field lines. Find (i) the radius of the helical path, (ii) the time period of one revolution, and (iii) the pitch of the helix. (Take $m_e = 9.1 \times 10^{-31}\text{ kg}, e = 1.6 \times 10^{-19}\text{ C}$).
Solution:
1. Velocity components:
$$v_\perp = v\sin 30^\circ = 4.0 \times 10^6 \times 0.5 = 2.0 \times 10^6\text{ m/s}$$
$$v_\parallel = v\cos 30^\circ = 4.0 \times 10^6 \times \frac{\sqrt{3}}{2} \approx 3.464 \times 10^6\text{ m/s}$$
2. Radius of the helix ($r$):
3. Time period ($T$):
4. Pitch of the helix ($p$):
Problem: A proton and an alpha particle having equal kinetic energies enter perpendicularly into a uniform magnetic field. What is the ratio of their orbit radii? If they enter mutually perpendicular crossed fields $\vec{E}$ and $\vec{B}$ with the same initial speeds, will both emerge undeflected simultaneously[cite: 13]?
Solution:
1. Radius in terms of kinetic energy $K$: $r = \frac{\sqrt{2mK}}{qB}$.
For proton: $m_p = m, q_p = e$. For alpha particle: $m_\alpha = 4m, q_\alpha = 2e$.
2. In crossed fields, undeflected condition is strictly $v = E/B$[cite: 13]. Since both particles have equal speeds $v$, both satisfy $v = E/B$ and emerge undeflected together[cite: 13]!
Lorentz Force
$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$[cite: 13]
Orbit Radius
$r = \frac{m v_\perp}{q B}$
Helical Pitch
$p = \frac{2\pi m v\cos\theta}{q B}$
Crossed Fields Filter
$v = \frac{E}{B}$[cite: 13]