Radius $r = \frac{mv}{qB} = \frac{p}{qB}$. Since momentum $p$ and field $B$ are constant, $r \propto \frac{1}{q}$.
For proton, $q_p = e$. For alpha particle, $q_\alpha = 2e$.
Therefore, $\frac{r_p}{r_\alpha} = \frac{q_\alpha}{q_p} = \frac{2e}{e} = 2:1$.
$\vec{F} = q(\vec{v} \times \vec{B}) = (-1.6 \times 10^{-19}) \times 10^6 \left[ (2\hat{i} + 3\hat{j}) \times (0.2\hat{i} - 0.3\hat{j} + 0.1\hat{k}) \right]$
Cross product: $(2\hat{i} \times -0.3\hat{j} \to -0.6\hat{k})$, $(2\hat{i} \times 0.1\hat{k} \to -0.2\hat{j})$, $(3\hat{j} \times 0.2\hat{i} \to -0.6\hat{k})$, $(3\hat{j} \times 0.1\hat{k} \to 0.3\hat{i})$
$\vec{v} \times \vec{B} = 0.3\hat{i} - 0.2\hat{j} - 1.2\hat{k}$
$\vec{F} = -1.6 \times 10^{-13} (0.3\hat{i} - 0.2\hat{j} - 1.2\hat{k})$ N.
Kinetic energy gained $K = qV$. Also, $K = \frac{p^2}{2m} \implies p = \sqrt{2mqV}$.
Radius of circular path in magnetic field $r = \frac{p}{qB}$.
Substituting $p$: $r = \frac{\sqrt{2mqV}}{qB} = \sqrt{\frac{2mV}{qB^2}}$.
Pitch $P = v_{\parallel} \times T = (v \cos\theta) \times \left( \frac{2\pi m}{qB} \right)$.
$P = (3 \times 10^7 \times \cos 30^\circ) \times \frac{2 \times 3.14 \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 1.5 \times 10^{-3}}$
$P \approx (2.598 \times 10^7) \times (23.8 \times 10^{-9}) \approx 0.618 \text{ m} = 61.8 \text{ cm}$.
For an undeflected particle, the electric and magnetic forces must exactly balance: $qE = qvB \implies v = \frac{E}{B}$.
$v = \frac{3 \times 10^4}{0.015} = 2 \times 10^6 \text{ m/s}$.
$r = \frac{\sqrt{2mK}}{qB} \implies K = \frac{q^2 B^2 r^2}{2m}$.
$K = \frac{(1.6 \times 10^{-19})^2 (0.8)^2 (0.05)^2}{2 \times 1.67 \times 10^{-27}} = 1.22 \times 10^{-13} \text{ Joules}$.
In MeV: $\frac{1.22 \times 10^{-13}}{1.6 \times 10^{-13}} \approx 0.76 \text{ MeV}$.
The magnetic field will force the particle into a circular motion in the transverse plane. Simultaneously, the parallel electric field will constantly accelerate the particle along the axis. The resulting trajectory is a helix with continuously increasing pitch.
If $\vec{E}$ is switched off, only the magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ remains. Since $\vec{v}$ is perpendicular to $\vec{B}$ (from the initial cross-field condition), the beam will immediately adopt a circular trajectory.
From Q3, $r = \sqrt{\frac{2mV}{qB^2}} \implies r \propto \sqrt{\frac{m}{q}}$.
For deuteron ($d$): $m_d = 2m_p$, $q_d = e$. For alpha ($\alpha$): $m_\alpha = 4m_p$, $q_\alpha = 2e$.
Ratio $\frac{r_d}{r_\alpha} = \frac{\sqrt{2/1}}{\sqrt{4/2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1:1$.
Lorentz magnetic force is $\vec{F} = q(\vec{v} \times \vec{B})$. By definition of the cross product, $\vec{F}$ is always perpendicular to $\vec{v}$. Work done $W = \vec{F} \cdot d\vec{s} = \vec{F} \cdot (\vec{v}dt) = 0$. By the work-energy theorem, zero work means $\Delta K = 0$, so kinetic energy is strictly constant.
Magnetic field from wire: $B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 0.1} = 10^{-4} \text{ T}$.
Velocity is parallel to wire, so $\vec{v} \perp \vec{B}$. $F = qvB = (1.6 \times 10^{-19}) \times 10^7 \times 10^{-4} = 1.6 \times 10^{-16} \text{ N}$.
Direction: Since it moves opposite to current, by Fleming's Left Hand rule (considering $-e$ charge), the force is repulsive (away from the wire).
Field of X: $B_X = \frac{\mu_0 N_1 I_1}{2 R_1} = \frac{4\pi \times 10^{-7} \times 20 \times 16}{2 \times 0.16} = 4\pi \times 10^{-4} \text{ T}$.
Field of Y: $B_Y = \frac{\mu_0 N_2 I_2}{2 R_2} = \frac{4\pi \times 10^{-7} \times 25 \times 18}{2 \times 0.10} = 9\pi \times 10^{-4} \text{ T}$.
Since currents are opposite, fields subtract: $B_{net} = B_Y - B_X = 5\pi \times 10^{-4} \text{ T} \approx 1.57 \times 10^{-3} \text{ T}$.
Using $dB = \frac{\mu_0 I dl \sin 90^\circ}{4\pi (R^2+x^2)}$. Resolving components, the perpendicular components cancel. The axial components add: $dB_{axial} = dB \sin\phi = dB \frac{R}{\sqrt{R^2+x^2}}$.
Integrating over the loop ($2\pi R$): $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
For a full circle, $B = \frac{\mu_0 I}{2R}$. For a semi-circle, $B = \frac{1}{2} \left( \frac{\mu_0 I}{2R} \right) = \frac{\mu_0 I}{4R}$.
$B = \frac{4\pi \times 10^{-7} \times 12}{4 \times 0.02} = \frac{48\pi \times 10^{-7}}{0.08} = 600\pi \times 10^{-7} \approx 1.88 \times 10^{-4} \text{ T}$.
$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 8}{2\pi \times 0.04} = \frac{16 \times 10^{-7}}{0.04} = 4 \times 10^{-5} \text{ T}$.
By Right-Hand Thumb rule, wire on y-axis, point on +x axis $\implies$ field is in $-z$ direction ($-\hat{k}$).
The square has 4 finite wires of length $a$. Distance to center is $a/2$. Angles are $45^\circ$.
For one side: $B_1 = \frac{\mu_0 I}{4\pi (a/2)} (\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi a} (\sqrt{2})$.
Total $B = 4 \times B_1 = \frac{2\sqrt{2} \mu_0 I}{\pi a}$.
$B = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$. Here $R=0.1$ m, $x=0.1$ m. $R^2+x^2 = 0.02$.
$B = \frac{4\pi \times 10^{-7} \times 200 \times 2 \times 0.01}{2(0.02)^{3/2}} = \frac{16\pi \times 10^{-5}}{2 \times 0.00282} \approx 8.9 \times 10^{-4} \text{ T}$.
Point $(a,a,0)$ is distance $a$ from both wires. Wire 1 (x-axis) creates $B_1 = \frac{\mu_0 I}{2\pi a}$ in $+\hat{k}$.
Wire 2 (y-axis) creates $B_2 = \frac{\mu_0 (2I)}{2\pi a}$ in $-\hat{k}$.
Net $B = B_2 - B_1 = \frac{\mu_0 I}{2\pi a}$ in the $-\hat{k}$ direction.
Length $L$ is constant. Circle: $2\pi R = L \implies R = L/2\pi$. $B_c = \frac{\mu_0 I}{2R} = \frac{\pi \mu_0 I}{L}$.
Square: $4a = L \implies a = L/4$. $B_s = \frac{2\sqrt{2} \mu_0 I}{\pi (L/4)} = \frac{8\sqrt{2} \mu_0 I}{\pi L}$.
Ratio $\frac{B_c}{B_s} = \frac{\pi}{L} \times \frac{\pi L}{8\sqrt{2}} = \frac{\pi^2}{8\sqrt{2}}$.
The incoming/outgoing straight wires point at the center, so their $\vec{dl} \times \vec{r} = 0$, producing no field. The loop splits into parallel resistances proportional to lengths. Current inversely splits. $I_1 l_1 = I_2 l_2$. Field $B_1 \propto I_1 l_1/R^2$ and $B_2 \propto I_2 l_2/R^2$. Thus magnitudes are equal but opposite in direction. Net $B = 0$.
Outside ($r > a$): $\oint B \cdot dl = B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$.
Inside ($r < a$): Current enclosed $I' = I \frac{\pi r^2}{\pi a^2}$. $B(2\pi r) = \mu_0 I \frac{r^2}{a^2} \implies B = \frac{\mu_0 I r}{2\pi a^2}$.
Total turns $N = 3 \times 300 = 900$. Length $L = 0.6$ m. $n = \frac{N}{L} = 1500$ turns/m.
$B = \mu_0 n I = (4\pi \times 10^{-7}) \times 1500 \times 2 = 12\pi \times 10^{-4} \approx 3.77 \times 10^{-3} \text{ T}$.
(i) Outside: Amperian loop encloses net zero current. $B_{out} = 0$.
(ii) Inside core: Mean radius $r = \frac{25+26}{2} = 25.5 \text{ cm} = 0.255 \text{ m}$.
$B = \frac{\mu_0 N I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 3500 \times 11}{2\pi \times 0.255} = \frac{77000 \times 10^{-7}}{0.255} \approx 0.03 \text{ T}$.
Take an Amperian loop of radius $r$ ($R_1 < r < R_2$). The current enclosed is only the inner conductor's current $I$. The outer shell's current is outside this loop.
$B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$.
No. For non-steady currents (charging capacitor), the conduction current is discontinuous between the plates. Ampere's original law fails because it yields different fields depending on the chosen bounding surface. Maxwell corrected this by adding "displacement current".
$n = 1000/1 = 1000$. The field at the exact end of a long solenoid is half the field at its center.
$B_{end} = \frac{1}{2} \mu_0 n I = \frac{1}{2} (4\pi \times 10^{-7}) (1000) (5) = 10\pi \times 10^{-4} \approx 3.14 \times 10^{-3} \text{ T}$.
Draw a rectangular Amperian loop with one side far outside parallel to the axis. If $B_{out}$ existed, $\oint B \cdot dl$ would be non-zero. But for an ideal infinite solenoid, the field lines cannot spread out and return; they must remain parallel to the axis infinitely. Thus $B_{out}$ must vanish to satisfy boundary conditions.
Since $r = 2$ mm $< a = 5$ mm, use internal field formula: $B = \frac{\mu_0 I r}{2\pi a^2}$.
$B = \frac{4\pi \times 10^{-7} \times 10 \times 0.002}{2\pi \times (0.005)^2} = \frac{4 \times 10^{-8}}{0.000025} = 1.6 \times 10^{-3} \text{ T}$.
The plot must show a straight line starting from origin $(0,0)$ with a positive slope until $r=R$ (where it peaks at $B = \frac{\mu_0 I}{2\pi R}$). Beyond $r=R$, it must show a smoothly decaying hyperbolic curve ($B \propto 1/r$), halving its peak value at $r=2R$ and further decaying at $r=3R$.
Formula is $B = \frac{\mu_0 N I}{2\pi r}$.
If $N \to 2N$ and $I \to I/2$, the new field $B' = \frac{\mu_0 (2N) (I/2)}{2\pi r} = \frac{\mu_0 N I}{2\pi r} = B$.
There is 0% change in the magnetic field.
Force per unit length $f = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 8 \times 5}{2\pi \times 0.04} = 2 \times 10^{-4} \text{ N/m}$.
For a 10 cm ($0.1$ m) section, Force $F = f \times L = 2 \times 10^{-4} \times 0.1 = 2 \times 10^{-5} \text{ N}$ (Attractive).
Attractive force on near side (2 cm): $F_1 = \frac{\mu_0 I_1 I_2 L}{2\pi d_1} = \frac{4\pi \times 10^{-7} \times 25 \times 15 \times 0.25}{2\pi \times 0.02} = 9.375 \times 10^{-4} \text{ N}$.
Repulsive force on far side (12 cm): $F_2 = \frac{\mu_0 I_1 I_2 L}{2\pi d_2} = \frac{4\pi \times 10^{-7} \times 25 \times 15 \times 0.25}{2\pi \times 0.12} = 1.5625 \times 10^{-4} \text{ N}$.
Net Force $= F_1 - F_2 = 7.81 \times 10^{-4} \text{ N}$ (Towards the wire).
Q is attracted by P (left) and R (right).
Force by P: $f_{QP} = \frac{\mu_0 (2)(4)}{2\pi (0.02)} = \frac{8 \times 2 \times 10^{-7}}{0.02} = 8 \times 10^{-5} \text{ N/m}$ (Left).
Force by R: $f_{QR} = \frac{\mu_0 (4)(6)}{2\pi (0.02)} = \frac{24 \times 2 \times 10^{-7}}{0.02} = 24 \times 10^{-5} \text{ N/m}$ (Right).
Net force $= 24 - 8 = 16 \times 10^{-5} \text{ N/m}$ towards wire R.
Forces on the two perpendicular sides exactly cancel. Net force is the difference between forces on the parallel sides.
$F_{net} = F_{near} - F_{far} = \frac{\mu_0 I_1 I_2 a}{2\pi x} - \frac{\mu_0 I_1 I_2 a}{2\pi (x+a)}$.
$F_{net} = \frac{\mu_0 I_1 I_2 a}{2\pi} \left( \frac{1}{x} - \frac{1}{x+a} \right) = \frac{\mu_0 I_1 I_2 a^2}{2\pi x (x+a)}$.
No net translational force. The magnetic field of wire 1 at wire 2 is perpendicular to wire 2. Integrating the force $Id\vec{l} \times \vec{B}$ from $-\infty$ to $+\infty$ yields zero because the force components on symmetrically opposite sides of the origin cancel out perfectly. However, they do experience a twisting torque.
Magnetic force must balance gravity: $ILB = mg$.
$m = \frac{ILB}{g} = \frac{2 \times 1.5 \times 0.5}{10} = \frac{1.5}{10} = 0.15 \text{ kg}$ (or 150 grams).
Wire 1 carrying current $I_1$ upwards creates a magnetic field $\vec{B}$ into the page at the location of Wire 2. Wire 2 carries current $I_2$ downwards. Applying $\vec{F} = I_2 (\vec{l} \times \vec{B})$, $\vec{l}$ is down, $\vec{B}$ is in. The cross product points outwards (away from Wire 1), indicating repulsion.
$F = ILB\sin\theta = mg$. For minimum field, $\theta = 90^\circ$ (perpendicular).
$B = \frac{mg}{IL} = \frac{3 \times 10^{-3} \times 9.8}{5 \times 0.1} = \frac{0.0294}{0.5} \approx 0.0588 \text{ T}$.
By Fleming's Left Hand Rule, if current is East, field must be North to yield an upward (supportive) force.
Let third wire carry current $I_3$ in the same plane. It must be placed between them so the forces from $I$ and $2I$ oppose each other. Let it be distance $x$ from wire $I$.
$\frac{\mu_0 I I_3 L}{2\pi x} = \frac{\mu_0 (2I) I_3 L}{2\pi (d-x)} \implies \frac{1}{x} = \frac{2}{d-x} \implies d-x = 2x \implies 3x = d \implies x = d/3$.
It should be placed at $d/3$ from the wire carrying current $I$.
1 Ampere is defined as the current which, flowing in two parallel infinite conductors 1m apart in vacuum, produces a force of $2 \times 10^{-7} \text{ N/m}$. It is preferred over charge because measuring current mechanically via macroscopic magnetic forces (using a current balance) is practically much more accurate than isolating and measuring individual static charges.
Total Force on any closed loop in a uniform field is always Zero.
Torque $\tau = NIAB\sin\theta$. Here normal to plane is parallel to field, so $\theta = 0^\circ$.
Torque $\tau = 20 \times 5 \times (\pi \times 0.1^2) \times 0.1 \times \sin(0^\circ) = \mathbf{0}$.
Consider a loop of sides $a$ and $b$. Forces on horizontal sides cancel. Forces on vertical sides $b$ are $F = I b B$. They form a couple separated by perpendicular distance $a \sin\theta$.
Torque $\tau = \text{Force} \times \text{Perpendicular distance} = (IbB)(a\sin\theta)$.
Since Area $A = a \times b$, $\tau = I A B \sin\theta = \vec{m} \times \vec{B}$.
Area $A = (0.1)^2 = 0.01 \text{ m}^2$. Angle with normal $\theta = 30^\circ$.
$\tau = NIAB\sin\theta = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^\circ$
$\tau = 2.4 \times 0.80 \times 0.5 = 0.96 \text{ N}\cdot\text{m}$.
A current loop in a magnetic field experiences torque $\vec{\tau} = I\vec{A} \times \vec{B}$, exactly mirroring an electric dipole in an electric field ($\vec{\tau} = \vec{p} \times \vec{E}$). By direct mathematical analogy, it behaves as a magnetic dipole with moment $\vec{m} = I\vec{A}$ (for $N$ turns, $\vec{m} = NI\vec{A}$).
Current $I = \frac{e}{T} = \frac{ev}{2\pi r}$. Area $A = \pi r^2$.
$M = I \times A = \left( \frac{ev}{2\pi r} \right) (\pi r^2) = \frac{evr}{2}$.
$M = \frac{1.6 \times 10^{-19} \times 2.2 \times 10^6 \times 0.53 \times 10^{-10}}{2} = 9.3 \times 10^{-24} \text{ A}\cdot\text{m}^2$.
Initial angle $\theta_1 = 0^\circ$ (perpendicular implies normal is parallel to field). Final angle $\theta_2 = 180^\circ$.
$M = NIA = 500 \times 1 \times 0.05 = 25 \text{ A}\cdot\text{m}^2$.
Work done $W = MB(\cos\theta_1 - \cos\theta_2) = 25 \times (4 \times 10^{-5}) \times (1 - (-1)) = 10^{-3} \times 2 = 2 \times 10^{-3} \text{ J}$.
Circle: $2\pi R = L \implies R = L/2\pi \implies Area_c = \pi(L/2\pi)^2 = L^2/4\pi$.
Square: $4a = L \implies a = L/4 \implies Area_s = (L/4)^2 = L^2/16$.
Since $4\pi \approx 12.56$, $Area_c > Area_s$. Since $M = IA$, the circular loop has a greater magnetic dipole moment.
Stable Equilibrium: When magnetic moment $\vec{m}$ is strictly parallel to $\vec{B}$ ($\theta = 0^\circ$). Potential energy is minimum ($-MB$).
Unstable Equilibrium: When magnetic moment $\vec{m}$ is strictly anti-parallel to $\vec{B}$ ($\theta = 180^\circ$). Potential energy is maximum ($+MB$).
Potential energy $U = -\vec{m} \cdot \vec{B} = -MB\cos\theta$.
$U = -(2.5) \times (0.2) \times \cos 120^\circ$.
Since $\cos 120^\circ = -0.5$, $U = -0.5 \times -0.5 = \mathbf{+0.25 \text{ Joules}}$.
In a non-uniform field, the magnitudes of the $\vec{B}$ vectors on opposite sides of the loop differ. Therefore, the opposing forces $Id\vec{l} \times \vec{B}$ do not cancel out completely, yielding a net translational force. Additionally, since the lines of action of these unequal forces usually do not intersect at the center of mass, they generate a net rotational torque.
Principle: A current loop in a magnetic field experiences torque. Mechanism: A coil suspended in a radial field rotates until magnetic torque ($NIAB$) equals restoring spring torque ($k\theta$). The radial magnetic field ensures the plane of the coil is always parallel to $\vec{B}$, keeping $\sin\theta=1$, making deflection $\theta$ linearly proportional to current $I$.
Connect a shunt resistance $S$ in parallel.
$S = \frac{I_g R_g}{I - I_g} = \frac{2.5 \times 10^{-3} \times 12}{7.5 - 0.0025} \approx \frac{0.03}{7.4975} \approx 0.004 \, \Omega$.
Net resistance $R_{net} = \frac{R_g S}{R_g + S} \approx S = 0.004 \, \Omega$.
Connect a resistance $R$ in series.
$V = I_g(R_g + R) \implies R = \frac{V}{I_g} - R_g$
$R = \frac{20}{5 \times 10^{-3}} - 50 = 4000 - 50 = 3950 \, \Omega$.
Figure of merit ($k$) is the current required to produce a deflection of one division. $k = \frac{I}{\theta}$.
Current sensitivity is deflection per unit current ($I_s = \frac{\theta}{I}$). Therefore, Figure of Merit is the exact mathematical reciprocal of current sensitivity ($k = \frac{1}{I_s}$).
Current Sensitivity Ratio: $\frac{I_{s2}}{I_{s1}} = \frac{N_2 A_2 B_2}{N_1 A_1 B_1} = \frac{42 \times 1.8 \times 0.50}{30 \times 3.6 \times 0.25} = \frac{37.8}{27} = 1.4$ (or $7:5$).
Voltage Sensitivity Ratio: $\frac{V_{s2}}{V_{s1}} = \frac{I_{s2} / R_2}{I_{s1} / R_1} = 1.4 \times \frac{10}{14} = 1.4 \times 0.714 = 1:1$.
1. It is a ferromagnetic material, so its high permeability concentrates the magnetic field lines, significantly increasing the field strength $B$ and instrument sensitivity.
2. Combined with concave pole pieces, it actively helps to bend the magnetic field lines to make the field strictly radial.
Here $I_g = 1.0$ A (the ammeter acts as a pseudo-galvanometer), $R_g = 0.80 \, \Omega$, new $I = 5.0$ A.
$S = \frac{I_g R_g}{I - I_g} = \frac{1.0 \times 0.80}{5.0 - 1.0} = \frac{0.80}{4.0} = 0.20 \, \Omega$.
Initial state: Full deflection current $I_g = \frac{V}{G}$.
Final state: For range $nV$, $I_g = \frac{nV}{G + R}$.
Equating $I_g$: $\frac{V}{G} = \frac{nV}{G+R} \implies G+R = nG \implies R = (n-1)G$.
A voltmeter is connected in parallel across a component to measure voltage drop. If it had low resistance, it would draw a massive amount of current from the main circuit, entirely altering the potential difference it was supposed to measure. High resistance ensures it draws negligible current.
No. An AC current reverses direction rapidly (e.g., 50 times a second for 50Hz). The magnetic torque would reverse direction identically. Due to the mechanical inertia of the coil assembly, it cannot respond fast enough to these rapid oscillations, resulting in a net zero (null) deflection.
