EOM: $m\frac{d\vec{v}}{dt} = q(\vec{E} + \vec{v} \times \vec{B})$. $\implies m\ddot{x} = qB\dot{y}$ and $m\ddot{y} = qE - qB\dot{x}$.
Let $\omega = qB/m$. $\ddot{y} + \omega^2 y = \frac{qE}{m}$. Solving with initial conditions: $x(t) = \frac{E}{\omega B}(\omega t - \sin\omega t) + \frac{v_0}{\omega}\sin\omega t$, and $y(t) = \frac{E}{\omega B}(1 - \cos\omega t) + \frac{v_0}{\omega}(1 - \cos\omega t)$.
Condition for cycloid: Initial velocity must be $v_0 = 0$ or fields balanced such that it drifts. If $v_0 = 0$, $x(t) = R(\omega t - \sin\omega t)$ and $y(t) = R(1 - \cos\omega t)$ where $R = E/\omega B$.
The magnetic force provides transverse acceleration. For the proton to barely reach the center (grazing it), the radius of its circular trajectory $r_c$ must equal $R/2$.
Since $r_c = \frac{mv}{qB} = \frac{R}{2} \implies v_{min} = \frac{qBR}{2m}$.
Motion in z-axis is uniform acceleration: $v_z(t) = \frac{qE_0}{m}t$.
Pitch $P(t) = \int_{t}^{t+T} v_z dt$, where $T = \frac{2\pi m}{qB_0}$.
$P(t) = \frac{qE_0}{m} \int_{t}^{t+T} t' dt' = \frac{qE_0}{2m} [(t+T)^2 - t^2] = \frac{qE_0}{2m} (2tT + T^2)$.
$P(t) = \frac{2\pi E_0}{B_0} t + \frac{2\pi^2 m E_0}{q B_0^2}$.
Both charges follow circular paths of radius $R = \frac{mv_0}{qB_0}$. Because their initial velocities are perpendicular, their centers of revolution $C_1$ and $C_2$ are separated by $\sqrt{R^2 + R^2} = R\sqrt{2}$.
Maximum separation occurs when they are at diametrically opposite ends relative to their respective centers. Max separation $= R\sqrt{2} + R + R = R(2 + \sqrt{2}) = \frac{mv_0}{qB_0}(2 + \sqrt{2})$.
The particle undergoes circular motion. To cross without reflecting, its radius of curvature must be strictly greater than the width of the region. $r > d \implies \frac{mv}{qB} > d \implies v > \frac{qBd}{m}$.
Consider an elemental ring of radius $r$, $dr$. Charge $dq = \sigma 2\pi r dr$. Current $dI = \frac{dq}{2\pi/\omega} = \sigma \omega r dr$.
Force on ring $dF = dI (2\pi r) B_0$. The direction of $d\vec{l} \times \vec{B}$ is along the z-axis (axial force).
Total force $F = \int_0^R 2\pi B_0 \sigma \omega r^2 dr = 2\pi B_0 \sigma \omega \left[ \frac{r^3}{3} \right]_0^R = \frac{2}{3} \pi \sigma \omega B_0 R^3$.
Electric field of a moving charge includes a Lorentz factor $\gamma = 1/\sqrt{1 - v^2/c^2}$. The transverse electric field is $E_y = \gamma \frac{1}{4\pi\epsilon_0} \frac{q}{r^2}$.
Using Ampere-Maxwell law or relativistic transformation, $B = \frac{v}{c^2} \times E$.
$B_z = \gamma \frac{v}{c^2} \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} = \gamma \frac{\mu_0}{4\pi} \frac{qv}{r^2}$.
Conservation of energy and magnetic moment $\mu = \frac{mv_{\perp}^2}{2B}$.
Let field at center be $B_{min}$ and at ends be $B_{max}$.
$\frac{v_{\perp 0}^2}{B_{min}} = \frac{v_{\perp}^2}{B_{max}}$. For mirroring, $v_{\parallel}$ goes to 0 at the ends, so $v_{\perp} = v_0$.
$\frac{v_0^2 \sin^2\theta_c}{B_{min}} = \frac{v_0^2}{B_{max}} \implies \sin^2\theta_c = \frac{B_{min}}{B_{max}}$. Particles with $\theta > \theta_c$ are trapped.
Force $\vec{F} = q(v_x\hat{i} + v_y\hat{j}) \times (\alpha y \hat{k}) = q\alpha y (v_y\hat{i} - v_x\hat{j})$.
$m\frac{dv_x}{dt} = q\alpha y \frac{dy}{dt} \implies m v_x = q\alpha \frac{y^2}{2} + C$. Initially $y=0, v_x=v_0 \implies C = mv_0$.
$v_x = v_0 + \frac{q\alpha}{2m}y^2$. Using $v_x^2 + v_y^2 = v_0^2$ (energy cons.), $\left(v_0 + \frac{q\alpha}{2m}y^2\right)^2 + \dot{y}^2 = v_0^2$. This defines the trajectory.
Kinetic energy $K = qV \implies p = \sqrt{2mqV}$. Radius $R = \frac{p}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$.
Diameter $D = 2R = \frac{2}{B}\sqrt{\frac{2V}{q}} \sqrt{m}$.
Spatial separation $\Delta D = D_2 - D_1 = \frac{2}{B}\sqrt{\frac{2V}{q}} (\sqrt{m_2} - \sqrt{m_1})$.
A regular polygon of $n$ sides has apothem $r = R \cos(\pi/n)$. Each side subtends angle $2\pi/n$.
Field of one side $B_1 = \frac{\mu_0 I}{4\pi r} 2\sin(\pi/n)$. Total $B = n B_1 = \frac{\mu_0 n I}{2\pi R \cos(\pi/n)} \sin(\pi/n) = \frac{\mu_0 n I}{2\pi R} \tan(\pi/n)$.
As $n \to \infty$, $n \tan(\pi/n) \to n(\pi/n) = \pi$. Thus $B \to \frac{\mu_0 \pi I}{2\pi R} = \frac{\mu_0 I}{2R}$ (circular loop).
Charge density $\sigma = Q / \pi R^2$. Current of elemental ring $dI = \frac{\sigma \omega r dr}{\pi} \implies dI = \frac{Q\omega}{\pi R^2} r dr$.
$dB = \frac{\mu_0 dI r^2}{2(r^2 + z^2)^{3/2}} = \frac{\mu_0 Q \omega}{2\pi R^2} \frac{r^3 dr}{(r^2 + z^2)^{3/2}}$.
Integrating from $0$ to $R$, $B = \frac{\mu_0 Q \omega}{2\pi R^2} \left[ \frac{R^2 + 2z^2}{\sqrt{R^2 + z^2}} - 2z \right]$.
Biot-Savart Law in polar coordinates: $dB = \frac{\mu_0 I dl \sin\alpha}{4\pi r^2}$. Noting $dl \sin\alpha = r d\theta$.
$B = \int_{0}^{2\pi} \frac{\mu_0 I r d\theta}{4\pi r^2} = \frac{\mu_0 I}{4\pi} \int_{0}^{2\pi} \frac{d\theta}{a + b\theta}$.
$B = \frac{\mu_0 I}{4\pi b} [\ln(a + b\theta)]_0^{2\pi} = \frac{\mu_0 I}{4\pi b} \ln\left(1 + \frac{2\pi b}{a}\right)$.
The point $(a,a,0)$ lies in the xy-plane. Wire 1 (x-axis) creates field at distance $y=a$. Angle is $45^\circ$ to $180^\circ$. $B_1 = \frac{\mu_0 I}{4\pi a} (\sin 90^\circ + \sin 45^\circ) = \frac{\mu_0 I}{4\pi a} (1 + 1/\sqrt{2}) \hat{k}$.
Wire 2 (y-axis) symmetric, distance $x=a$. $B_2 = \frac{\mu_0 I}{4\pi a} (1 + 1/\sqrt{2}) \hat{k}$.
Total $\vec{B} = B_1 + B_2 = \frac{\mu_0 I}{2\pi a} \left(1 + \frac{1}{\sqrt{2}}\right) \hat{k}$.
Consider a volume element in spherical coordinates. Alternatively, view it as nested rotating shells. Field of a shell of radius $r$ at center is $dB = \frac{2}{3} \mu_0 \sigma \omega r = \frac{2}{3} \mu_0 (\rho dr) \omega r$.
$B = \int_0^R \frac{2}{3} \mu_0 \rho \omega r dr = \frac{1}{3} \mu_0 \rho \omega R^2$.
Field $B(x) = \frac{\mu_0 N I R^2}{2} \left[ (R^2 + (x - R/2)^2)^{-3/2} + (R^2 + (x + R/2)^2)^{-3/2} \right]$.
Evaluating $\frac{dB}{dx}$ and $\frac{d^2B}{dx^2}$ at $x=0$, terms naturally cancel due to symmetry for the first derivative. Setting the second derivative equal to zero yields the exact separation distance $d = R$. This provides highly uniform fields.
The focus is at $(a,0)$. Using Biot-Savart, $dB = \frac{\mu_0 I (\vec{dl} \times \vec{r})}{4\pi r^3}$. Converting to polar coordinates originating at the focus, a parabola is $r = \frac{2a}{1-\cos\theta}$.
Integrating $dB = \frac{\mu_0 I}{4\pi} \frac{d\theta}{r}$ from $0$ to $2\pi$ (excluding origin path).
$B = \frac{\mu_0 I}{4\pi (2a)} \int_0^{2\pi} (1-\cos\theta) d\theta = \frac{\mu_0 I}{8\pi a} (2\pi - 0) = \frac{\mu_0 I}{4a}$.
Parametric equations: $x = a\cos\phi$, $y = b\sin\phi$. $dl = \sqrt{dx^2 + dy^2}$. $r^2 = a^2\cos^2\phi + b^2\sin^2\phi$.
$B = \frac{\mu_0 I}{4\pi} \oint \frac{\vec{dl} \times \hat{r}}{r^2} = \frac{\mu_0 I}{4\pi} \int_0^{2\pi} \frac{ab}{(a^2\cos^2\phi + b^2\sin^2\phi)^{3/2}} d\phi$.
This results in a complete elliptic integral of the second kind. $B = \frac{\mu_0 I}{2\pi} \frac{1}{ab} E(e)$, where $e$ is eccentricity.
Turn density $n = \frac{N}{b-a}$. In radial thickness $dr$, there are $dN = n dr$ turns.
$dB = \frac{\mu_0 (I dN)}{2r} = \frac{\mu_0 I N}{2(b-a)} \frac{dr}{r}$.
Integrating from $a$ to $b$: $B = \frac{\mu_0 I N}{2(b-a)} \ln(b/a)$.
Wire along x: $B_x = \frac{\mu_0 I}{2\pi} \frac{a\hat{j} - a\hat{k}}{a^2+a^2} = \frac{\mu_0 I}{4\pi a} (\hat{j} - \hat{k})$.
Wire along y: $B_y = \frac{\mu_0 I}{4\pi a} (\hat{k} - \hat{i})$. Wire along z: $B_z = \frac{\mu_0 I}{4\pi a} (\hat{i} - \hat{j})$.
Total $\vec{B} = B_x + B_y + B_z = \vec{0}$. The fields cancel exactly due to cyclic symmetry.
$I_{enc} = \int_0^r J(r') 2\pi r' dr' = \int_0^r J_0 \frac{r'}{R} 2\pi r' dr' = \frac{2\pi J_0}{R} [\frac{r'^3}{3}]_0^r = \frac{2\pi J_0 r^3}{3R}$.
For $r < R$: $B(2\pi r) = \mu_0 \frac{2\pi J_0 r^3}{3R} \implies \vec{B} = \frac{\mu_0 J_0 r^2}{3R} \hat{\theta}$.
For $r > R$: $I_{total} = \frac{2\pi J_0 R^2}{3}$. $B = \frac{\mu_0 I_{total}}{2\pi r} = \frac{\mu_0 J_0 R^2}{3r} \hat{\theta}$.
Superposition: Solid cylinder $+J$ and smaller cylinder $-J$.
Field of solid at $\vec{r}$: $\vec{B}_1 = \frac{\mu_0 J}{2} (\vec{k} \times \vec{r})$. Field of cavity at offset $\vec{r} - \vec{d}$: $\vec{B}_2 = -\frac{\mu_0 J}{2} (\vec{k} \times (\vec{r} - \vec{d}))$.
$\vec{B}_{net} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 J}{2} (\vec{k} \times \vec{d})$. Since $\vec{d}$ is constant, $\vec{B}_{net}$ is purely uniform. Magnitude $B = \frac{\mu_0 J d}{2}$.
Field of a single infinite sheet with $\vec{K}$ is $\vec{B} = \frac{\mu_0}{2} (\vec{K} \times \hat{n})$.
Upper sheet ($z=a$): Below it, $\hat{n} = -\hat{k}$. $\vec{B}_1 = \frac{\mu_0}{2} (K_0\hat{i} \times -\hat{k}) = \frac{\mu_0 K_0}{2} \hat{j}$.
Lower sheet ($z=-a$): Above it, $\hat{n} = +\hat{k}$. $\vec{B}_2 = \frac{\mu_0}{2} (-K_0\hat{i} \times \hat{k}) = \frac{\mu_0 K_0}{2} \hat{j}$.
Total $\vec{B} = \vec{B}_1 + \vec{B}_2 = \mu_0 K_0 \hat{j}$.
Magnetic field inside toroid at distance $r$ is $B = \frac{\mu_0 N I}{2\pi r}$.
Flux $d\Phi = B dA = \left( \frac{\mu_0 N I}{2\pi r} \right) (h dr)$.
Total Flux $\Phi_B = \int_a^b \frac{\mu_0 N I h}{2\pi r} dr = \frac{\mu_0 N I h}{2\pi} \ln\left(\frac{b}{a}\right)$.
Apply Ampere's loop symmetrically about $y=0$. $\oint B \cdot dl = \mu_0 I_{enc}$. $\vec{B}$ is along $\pm\hat{k}$.
Inside ($-d < y < d$): $B(2L) = \mu_0 (J_0 2y L) \implies B = \mu_0 J_0 y$. Vectorially: $\vec{B} = -\mu_0 J_0 y \hat{k}$.
Outside ($|y| > d$): $B(2L) = \mu_0 (J_0 2d L) \implies B = \mu_0 J_0 d$. $\vec{B} = \mp\mu_0 J_0 d \hat{k}$.
Total return current must be $I_0$. $\int_b^c (\alpha/r) 2\pi r dr = I_0 \implies 2\pi\alpha (c-b) = I_0 \implies \alpha = \frac{I_0}{2\pi(c-b)}$.
For $b < r < c$: $I_{enc} = I_0 - \int_b^r \frac{\alpha}{r'} 2\pi r' dr' = I_0 - 2\pi\alpha(r-b) = I_0 \left(1 - \frac{r-b}{c-b}\right) = I_0 \frac{c-r}{c-b}$.
$B = \frac{\mu_0 I_{enc}}{2\pi r} = \frac{\mu_0 I_0}{2\pi r} \left(\frac{c-r}{c-b}\right)$.
Consider an infinite solenoid as two semi-infinite solenoids joined at $x=0$. By superposition, $\vec{B}_{infinite} = \vec{B}_{left\_semi} + \vec{B}_{right\_semi}$.
At the junction $x=0$, by symmetry, $\vec{B}_{left} = \vec{B}_{right}$.
Since $\vec{B}_{infinite} = \mu_0 n I \hat{i}$, we have $2\vec{B}_{semi} = \mu_0 n I \hat{i} \implies \vec{B}_{semi} = \frac{1}{2}\mu_0 n I \hat{i}$.
Mathematically, Ampere's law ($\oint \vec{B} \cdot d\vec{l} = \mu_0 I$) is always valid for any closed loop containing steady currents. However, a square loop lacks the continuous rotational symmetry of $\vec{B}$. The dot product $\vec{B} \cdot d\vec{l}$ varies along the square's edge, preventing us from factoring $B$ out of the integral. Thus, it is practically useless for algebraically solving for $B$.
Work done $W = \oint \vec{F}_m \cdot d\vec{l}$. For a monopole $q_m$, $\vec{F}_m = q_m \vec{B}$.
$W = q_m \oint \vec{B} \cdot d\vec{l}$. By Ampere's Law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$.
Therefore, $W = q_m \mu_0 I = (1) (\mu_0) (1) = \mu_0 \text{ Joules}$. This non-conservative work implies magnetic scalar potential is multi-valued.
Angular velocity $\omega(t) = \alpha t$. Surface current density $K(t) = \sigma v = \sigma (R \omega(t)) = \sigma R \alpha t$.
The cylinder acts as an infinite solenoid with current per unit length $K$.
Thus, $\vec{B}(t) = \mu_0 K(t) \hat{k} = \mu_0 \sigma R \alpha t \hat{k}$.
Field of the infinite sheet is strictly uniform: $B = \frac{\mu_0 K}{2}$, parallel to the sheet and perpendicular to current flow.
Force on wire $F = I_1 L B = I_1 L \left( \frac{\mu_0 K}{2} \right)$.
Force per unit length $f = \frac{F}{L} = \frac{\mu_0 I_1 K}{2}$. (Direction depends on relative current directions, either pulling wire to sheet or pushing away).
$dF = I_2 (\vec{dl} \times \vec{B})$. $B = \frac{\mu_0 I_1}{2\pi(d - R\cos\theta)}$.
Integrating over semicircle: $F = \int_{-\pi/2}^{\pi/2} \frac{\mu_0 I_1 I_2 R \cos\theta d\theta}{2\pi(d - R\cos\theta)}$.
This exact integral evaluates to $F = \frac{\mu_0 I_1 I_2}{\pi} \left[ \frac{d}{\sqrt{d^2-R^2}} - 1 \right]$, directed towards/away from the straight wire.
Take the top vertex wire. Force from bottom-left wire is $F_1 = \frac{\mu_0 I^2}{2\pi a}$ directed along the vector connecting them (attractive). Force from bottom-right is $F_2 = \frac{\mu_0 I^2}{2\pi a}$ attractive.
Angle between these two force vectors is exactly $60^\circ$.
Resultant $F_{net} = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos 60^\circ} = \sqrt{3} F_1 = \frac{\sqrt{3} \mu_0 I^2}{2\pi a}$, directed towards the center of the base.
Divide the ribbon into infinite wires of width $dx$ at distance $x$. $dI = \frac{I}{w} dx$.
Force $dF = \frac{\mu_0 I_0 dI}{2\pi x} = \frac{\mu_0 I_0 I}{2\pi w} \frac{dx}{x}$.
Integrating from $x=a$ to $x=a+w$: $F_{net} = \frac{\mu_0 I_0 I}{2\pi w} \ln\left(\frac{a+w}{a}\right)$.
Consider a small arc element subtending $d\theta$. Magnetic outward force $dF_m = I (R d\theta) B$.
This is balanced by the inward radial component of tension: $2 T \sin(d\theta/2) \approx T d\theta$.
$T d\theta = I R B d\theta \implies T = I R B$.
The ring snaps if $I R B > T_{max} \implies B > \frac{T_{max}}{IR}$.
Field of wire 1 at wire 2 position $(0, y, d)$ is $\vec{B} = \frac{\mu_0 I_1}{2\pi \sqrt{y^2+d^2}} (-\sin\theta \hat{i} + \cos\theta \hat{k})$. Since it's at $z=d$, $\vec{B} = \frac{\mu_0 I_1}{2\pi (y^2+d^2)} (-d\hat{i} + y\hat{k})$.
Force on element $dy$ is $d\vec{F} = I_2(dy\hat{j}) \times \vec{B} = \frac{\mu_0 I_1 I_2}{2\pi (y^2+d^2)} (y\hat{i} + d\hat{k}) dy$.
Torque $d\vec{\tau} = \vec{r} \times d\vec{F} = (y\hat{j} + d\hat{k}) \times d\vec{F}$. Integrating evaluates to $\tau = \frac{\mu_0 I_1 I_2 L}{\pi}$.
Let infinite wire be x-axis. Finite wire starts at $y=a$, aligned at angle $\theta$. Its distance is $r = a + l\sin\theta$.
$dF = I_2 dl B \sin(90^\circ) = I_2 dl \frac{\mu_0 I_1}{2\pi (a + l\sin\theta)}$.
$F = \int_0^L \frac{\mu_0 I_1 I_2}{2\pi (a + l\sin\theta)} dl = \frac{\mu_0 I_1 I_2}{2\pi \sin\theta} \ln\left(1 + \frac{L\sin\theta}{a}\right)$.
At distance $r$, $F_{near} = \frac{\mu_0 I_1 I_2 a}{2\pi r}$ (attractive) and $F_{far} = \frac{\mu_0 I_1 I_2 a}{2\pi (r+a)}$ (repulsive).
Net force $F_{net} = \frac{\mu_0 I_1 I_2 a^2}{2\pi r(r+a)}$. This force opposes the velocity vector as determined by Lenz's Law equivalent (work done against this pulls energy for induced EMF if it were a closed loop, though here it's purely Ampere force).
Current $I$ flowing parallel creates an azimuthal magnetic field $\vec{B}$ inside. $B(r) = \frac{\mu_0 I r}{2\pi R^2}$.
Force on a volume element $dV$ carrying current density $J = I/\pi R^2$ is radially inward. $dP = J \times B dr = \left(\frac{I}{\pi R^2}\right) \left(\frac{\mu_0 I r}{2\pi R^2}\right) dr$.
Total pinch pressure $P = \int_0^R \frac{\mu_0 I^2 r}{2\pi^2 R^4} dr = \frac{\mu_0 I^2}{4\pi^2 R^2}$.
Field of dipole 1 along its axis: $\vec{B}_1 = \frac{\mu_0}{4\pi} \frac{2\vec{m}_1}{r^3}$.
Force on dipole 2 is $\vec{F} = \nabla(\vec{m}_2 \cdot \vec{B}_1)$. Since they are parallel, $\vec{m}_2 \cdot \vec{B}_1 = \frac{\mu_0}{4\pi} \frac{2 m_1 m_2}{r^3}$.
$\vec{F} = \frac{\partial}{\partial r} \left( \frac{\mu_0}{4\pi} \frac{2 m_1 m_2}{r^3} \right) \hat{r} = -\frac{3\mu_0 m_1 m_2}{2\pi r^4} \hat{r}$. (Attractive and drops as $1/r^4$).
Divide sphere into discs. For any rotating charge distribution with uniform mass/charge ratio, $dq/dm = Q/M$.
$d\mu = dI \cdot A = \frac{dq}{T} \pi r^2 = \frac{dq \omega}{2\pi} \pi r^2 = \frac{1}{2} dq \omega r^2$.
$dL = dm \omega r^2$. Thus $d\mu = \frac{1}{2} \frac{Q}{M} dm \omega r^2 = \frac{Q}{2M} dL$.
Integrating, $\vec{\mu} = \frac{Q}{2M}\vec{L}$. Yes, it holds universally for uniform distributions (discs, spheres, shells).
The loop can be viewed as three triangles in the xy, yz, and zx planes. Vector area $\vec{A} = \frac{1}{2}a^2(\hat{i} + \hat{j} + \hat{k})$.
Magnetic moment $\vec{m} = I\vec{A} = \frac{1}{2} I a^2 (\hat{i} + \hat{j} + \hat{k})$.
Torque $\vec{\tau} = \vec{m} \times \vec{B} = \left[ \frac{1}{2} I a^2 (\hat{i} + \hat{j} + \hat{k}) \right] \times \left[ B_0(\hat{i} + \hat{j} + \hat{k}) \right]$.
Since cross product of a vector with itself is 0, $\vec{\tau} = \vec{0}$.
Initial state: $\vec{m}_2$ is along $+\hat{i}$. $B_1$ at $\vec{m}_2$ is $\frac{\mu_0}{4\pi} \frac{2m_0}{r^3} \hat{i}$.
$U_i = -\vec{m}_2 \cdot \vec{B}_1 = -m_0 \left( \frac{\mu_0 2m_0}{4\pi r^3} \right) = -\frac{\mu_0 m_0^2}{2\pi r^3}$.
Final state: $\vec{m}_2$ is along $+\hat{j}$. $U_f = -(m_0\hat{j}) \cdot (\frac{\mu_0 2m_0}{4\pi r^3}\hat{i}) = 0$.
Work done $W = U_f - U_i = 0 - \left( -\frac{\mu_0 m_0^2}{2\pi r^3} \right) = \frac{\mu_0 m_0^2}{2\pi r^3}$.
$\nabla \cdot \vec{B} = 0 \implies \alpha + \beta + \gamma = 0 \implies \gamma = -(\alpha + \beta)$.
Force $d\vec{F} = I(dx\hat{i} + dy\hat{j}) \times (\alpha x\hat{i} + \beta y\hat{j} + \gamma z\hat{k})$. On xy-plane, $z=0$, so $\vec{B} = \alpha x\hat{i} + \beta y\hat{j}$.
$d\vec{F} = I [ (\alpha x dy - \beta y dx)\hat{k} ]$. Integrating over closed loop using Green's Theorem: $\oint (\alpha x dy - \beta y dx) = \iint (\alpha - (-\beta)) dx dy = (\alpha + \beta) A$.
$\vec{F} = I A (\alpha + \beta) \hat{k} = -I A \gamma \hat{k}$.
Restoring torque $\tau = -\mu B \sin\theta$. For small angles, $\sin\theta \approx \theta$, so $\tau \approx -\mu B \theta$.
Equation of motion: $I_{cm} \frac{d^2\theta}{dt^2} = -\mu B \theta \implies \frac{d^2\theta}{dt^2} + \left( \frac{\mu B}{I_{cm}} \right) \theta = 0$.
This is SHM with $\omega^2 = \frac{\mu B}{I_{cm}}$. Time period $T = 2\pi \sqrt{\frac{I_{cm}}{\mu B}}$.
As it rotates, it constitutes current $I = (\lambda 2\pi R) \frac{\omega}{2\pi} = \lambda R \omega$.
It doesn't experience torque from static $B$, but friction causes deceleration. Wait, $B$ is vertical, no $m \times B$ torque.
Only frictional torque $\tau_f = \mu (mg) R$ acts. $I_{cm}\alpha = -\mu mg R \implies (mR^2)\alpha = -\mu mg R \implies \alpha = -\frac{\mu g}{R}$.
Time to stop $t = \frac{\omega_0}{|\alpha|} = \frac{\omega_0 R}{\mu g}$. ($B$ field doesn't affect motion unless it's changing).
Torque $\vec{\tau} = \vec{\mu} \times \vec{B}_0$. Also, $\vec{\tau} = \frac{d\vec{L}}{dt}$. Since $\vec{\mu} = \gamma \vec{L}$, we have $\frac{d\vec{L}}{dt} = \gamma \vec{L} \times \vec{B}_0$.
This equation represents a vector $\vec{L}$ rotating around $\vec{B}_0$. The magnitude of rate of change is $dL = L \sin\theta d\phi$.
$\frac{L \sin\theta d\phi}{dt} = \gamma L B_0 \sin\theta \implies \omega_L = \frac{d\phi}{dt} = \gamma B_0$.
Charge element $dq = \sigma(2\pi R\sin\theta)(R d\theta) = \sigma_0\cos\theta (2\pi R^2\sin\theta) d\theta$.
Current $dI = dq \frac{\omega}{2\pi} = \sigma_0 \omega R^2 \sin\theta \cos\theta d\theta$.
$dm = dI (\pi (R\sin\theta)^2) = \pi \sigma_0 \omega R^4 \sin^3\theta \cos\theta d\theta$.
Integrate $0$ to $\pi$: $m = \pi \sigma_0 \omega R^4 \int_0^\pi \sin^3\theta \cos\theta d\theta$. By symmetry, integral evaluates to exactly **Zero**. No net dipole moment.
Regardless of the central force $F(r)$, the electron's motion remains uniform circular. $\mu_l = I \cdot A = (\frac{ev}{2\pi r})(\pi r^2) = \frac{evr}{2}$.
Angular momentum $L = m_e v r$.
Ratio $\frac{\mu_l}{L} = \frac{evr/2}{m_e v r} = \frac{e}{2m_e}$. This gyromagnetic ratio is a universal constant for orbital motion and is identical to the Bohr orbit.
The wire creates field $B = \frac{\mu_0 I_0}{2\pi r}$ with $\frac{dB}{dr} = -\frac{\mu_0 I_0}{2\pi r^2}$.
Force on a dipole in non-uniform field is $F_z = m \frac{dB}{dr}$ (assuming alignment).
$m = NIA$. Change in weight $\Delta W = F_z = (NIA) \left( -\frac{\mu_0 I_0}{2\pi h^2} \right)$ where $h$ is distance from wire. Balance reading reduces.
Torque impulse: $\int \tau dt = \int (NAB I) dt = NAB \int I dt = NAB Q$.
This equals change in angular momentum: $I_{mom} \omega_0 = NAB Q$.
Kinetic energy $\frac{1}{2}I_{mom}\omega_0^2$ converts to spring potential $\frac{1}{2}k\theta_0^2$. $\implies \omega_0 = \theta_0\sqrt{k/I_{mom}}$.
Substitute: $I_{mom} \theta_0 \sqrt{k/I_{mom}} = NAB Q \implies Q = \frac{\sqrt{k I_{mom}}}{NAB} \theta_0$. Thus $Q \propto \theta_0$.
As coil rotates with $\omega$, flux changes. Induced EMF $e = NAB\omega$.
With shunt $S$ and resistance $R_g$, total circuit resistance is essentially $S$ (if $S \ll R_g$). Induced current $i_{ind} = \frac{NAB\omega}{R_{total}}$.
Damping torque $T_D = N i_{ind} A B = \frac{(NAB)^2}{R_{total}} \omega$. This opposes motion, heavily damping it to prevent oscillation (dead-beat).
Current sensitivity $I_s = \frac{NAB}{C}$. So $I_s \propto \frac{1}{C} \propto \frac{l}{r^4}$.
Fractional change $\frac{dI_s}{I_s} = \frac{dl}{l} - 4\frac{dr}{r}$.
Given $dl/l = -0.10$ and $dr/r = -0.05$.
$\frac{dI_s}{I_s} = -0.10 - 4(-0.05) = -0.10 + 0.20 = +0.10$. Current sensitivity **increases by 10%**.
Range $V_1$: $V_1 = I_g(R_g + R_1) \implies R_1 = \frac{V_1}{I_g} - R_g$.
Range $V_2$ (tapped after $R_2$): $V_2 = I_g(R_g + R_1 + R_2) \implies R_2 = \frac{V_2 - V_1}{I_g}$.
Range $V_3$ (tapped after $R_3$): $V_3 = I_g(R_g + R_1 + R_2 + R_3) \implies R_3 = \frac{V_3 - V_2}{I_g}$.
Original measured $I = I_g(1 + R_g/S)$. Due to temp $\Delta T$, $R_g' = R_g(1 + \alpha\Delta T)$.
True full scale current becomes $I' = I_g(1 + R_g'/S) = I_g \left(1 + \frac{R_g}{S}(1 + \alpha\Delta T)\right)$.
Error $\Delta I = I' - I = I_g \left(\frac{R_g}{S}\right) \alpha\Delta T$.
Fractional error $\frac{\Delta I}{I} = \frac{R_g/S}{1 + R_g/S} \alpha\Delta T \approx \alpha\Delta T$ (since $R_g \gg S$).
In a uniform field $B_0$, the torque $\tau = N I A B_0 \cos\theta$ (where $\theta$ is deflection from equilibrium aligned with $B_0$ field lines).
Equating to spring torque: $k\theta = N I A B_0 \cos\theta \implies I = \frac{k}{NAB_0} \frac{\theta}{\cos\theta} = \frac{k}{NAB_0} \theta \sec\theta$. The scale compresses at larger $\theta$.
Initial: $I_1 = \frac{E}{R+G} = k\theta$.
Final: Eq resistance $R_{eq} = R' + \frac{GS}{G+S}$. Main current $I_{main} = \frac{E}{R_{eq}}$.
Current through G is $I_2 = I_{main} \frac{S}{G+S} = k(\theta/2) = \frac{I_1}{2}$.
Substituting and equating yields the standard exact formula: $G = \frac{R S}{R - S}$ (assuming constant $E$).
Characteristic equation roots: $\lambda = \frac{-b \pm \sqrt{b^2 - 4I_{mom}k}}{2I_{mom}}$.
Critical damping occurs when roots are real and equal, i.e., discriminant vanishes: $b^2 - 4I_{mom}k = 0 \implies b = 2\sqrt{I_{mom}k}$.
Here $b = \frac{(NAB)^2}{R}$. So Critical Resistance $R_c = \frac{(NAB)^2}{2\sqrt{I_{mom}k}}$.
Single MCG: $V_s = \frac{\theta}{V} = \frac{NAB}{kR_g}$.
Two in series: Total deflection $\theta_{total} = \theta_1 + \theta_2$. Current is $I = \frac{V}{2R_g}$.
$\theta_1 = \theta_2 = \frac{NAB}{k} I = \frac{NAB}{k} \frac{V}{2R_g}$.
$\theta_{total} = 2 \left( \frac{NAB}{k} \frac{V}{2R_g} \right) = \frac{NAB}{k R_g} V$.
Combined $V_{s,eq} = \frac{\theta_{total}}{V} = \frac{NAB}{kR_g}$. The combined voltage sensitivity is **exactly the same**.
Deflection of coil is $\theta = \frac{NAB}{k} I$.
By laws of reflection, the reflected beam rotates by $2\theta$.
Displacement on scale $y = D \tan(2\theta)$.
Small-angle approximation: $\tan(2\theta) \approx 2\theta \implies y \approx 2D\theta = \frac{2NDAB}{k} I$. This holds valid for $2\theta \le 10^\circ$ (or $\sim 0.17$ rad).
