$F = qvB\sin\theta$
$F = (1.6 \times 10^{-19} \text{ C}) \times (2 \times 10^7 \text{ m/s}) \times (1.5 \text{ T}) \times \sin(30^\circ)$
$F = 4.8 \times 10^{-12} \times 0.5 = 2.4 \times 10^{-12} \text{ N}$
A moving charge generates its own magnetic field. The external magnetic field interacts with this moving charge's magnetic field, resulting in a force. A stationary charge does not produce a magnetic field, so $v = 0 \implies F = q(0)B\sin\theta = 0$.
Radius $r = \frac{mv}{qB}$. For the same $v$ and $B$, $r \propto \frac{m}{q}$.
For $\alpha$-particle: $m_\alpha = 4m_p$, $q_\alpha = 2e$.
For proton: $m_p, e$.
Ratio $\frac{r_\alpha}{r_p} = \frac{4m_p / 2e}{m_p / e} = \frac{2}{1}$. Ratio is $2:1$.
Maximum force occurs when the charge moves perpendicular to the magnetic field ($\theta = 90^\circ$), because $\sin(90^\circ) = 1$. The max force is $F_{max} = qvB$.
The trajectory will be a straight line. If velocity is parallel to the field, $\theta = 0^\circ$ or $180^\circ$. Force $F = qvB\sin(0^\circ) = 0$. Since net force is zero, velocity remains unchanged.
Cyclotron frequency $\nu_c = \frac{qB}{2\pi m}$
$\nu_c = \frac{1.6 \times 10^{-19} \times 1}{2 \times 3.14 \times 1.67 \times 10^{-27}}$
$\nu_c \approx 1.52 \times 10^7 \text{ Hz}$ or $15.2 \text{ MHz}$.
Magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$. This force is always perpendicular to velocity $\vec{v}$. Work done $dW = \vec{F} \cdot d\vec{s} = \vec{F} \cdot (\vec{v}dt) = 0$. Since work is zero, kinetic energy remains constant (speed is unchanged).
Lorentz force is the total force exerted on a charged particle moving through a region containing both electric and magnetic fields.
$\vec{F} = \vec{F}_e + \vec{F}_m = q\vec{E} + q(\vec{v} \times \vec{B})$.
Electron moves in +x direction ($\vec{v} = v\hat{i}$). Field is in +y direction ($\vec{B} = B\hat{j}$). Charge $q = -e$.
$\vec{F} = -e(v\hat{i} \times B\hat{j}) = -ev(\hat{k}) = -ev\hat{k}$.
The force is directed along the negative z-axis.
The path becomes helical when the charge enters the magnetic field at an oblique angle (neither parallel, anti-parallel, nor exactly perpendicular). i.e., $0^\circ < \theta < 90^\circ$ or $90^\circ < \theta < 180^\circ$.
Biot-Savart Law states that the magnetic field $d\vec{B}$ due to a current element $Id\vec{l}$ is directly proportional to the current and length, and inversely proportional to the square of the distance.
Vector form: $d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \vec{r}}{r^3}$ or $\frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \hat{r}}{r^2}$.
$B = \frac{\mu_0 I}{2\pi r}$
$B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.20} = \frac{2 \times 10^{-7} \times 35}{0.20}$
$B = \frac{70 \times 10^{-7}}{0.20} = 350 \times 10^{-7} = 3.5 \times 10^{-5} \text{ T}$.
$B = \frac{\mu_0 N I}{2r}$
$B = \frac{4\pi \times 10^{-7} \times 100 \times 5}{2 \times 0.10}$
$B = \frac{2000\pi \times 10^{-7}}{0.20} = 10000\pi \times 10^{-7} \approx 3.14 \times 10^{-3} \text{ T}$.
Magnetic field at the center is $B = \frac{\mu_0 I}{2r}$. If $I$ becomes $2I$ and $r$ becomes $2r$, then $B' = \frac{\mu_0 (2I)}{2(2r)} = \frac{\mu_0 I}{2r} = B$. The field remains unchanged.
Similarity: Both laws follow the inverse square law with respect to distance ($\propto 1/r^2$).
Difference: Electric field is produced by a scalar source (charge $q$), while magnetic field is produced by a vector source (current element $Id\vec{l}$).
For a full circle, $B = \frac{\mu_0 I}{2R}$. For a semicircle, the field is exactly half.
$B = \frac{1}{2} \left(\frac{\mu_0 I}{2R}\right) = \frac{\mu_0 I}{4R}$.
Using the Right-Hand Thumb Rule, curling the fingers in the anti-clockwise direction points the thumb outwards. Therefore, the field is directed perpendicularly out of the plane of the paper.
Since $B \propto \frac{1}{r}$, the graph of $B$ versus $r$ is a rectangular hyperbola in the first quadrant, curving downwards as distance $r$ increases.
At the center, $B = \frac{\mu_0 I}{2R}$.
For loop P: $B_P = \frac{\mu_0 I}{2R}$.
For loop Q: $B_Q = \frac{\mu_0 (2I)}{2R} = 2 B_P$.
Ratio $B_P : B_Q = 1 : 2$.
If you hold the straight current-carrying conductor in your right hand such that the thumb points in the direction of the current, then the direction in which your fingers curl gives the direction of the magnetic field lines.
$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed}$
The line integral of magnetic field $\vec{B}$ around any closed path in a vacuum is equal to $\mu_0$ times the total steady current threading the surface bounded by the closed path.
Number of turns per unit length $n = \frac{N}{L} = \frac{500}{0.5} = 1000 \text{ turns/m}$.
$B = \mu_0 n I = (4\pi \times 10^{-7}) \times (1000) \times (5)$
$B = 20\pi \times 10^{-4} \approx 6.28 \times 10^{-3} \text{ T}$.
No. The formula for an ideal long solenoid is $B = \mu_0 n I$. This shows that the magnetic field is uniform inside and is independent of the radius of the solenoid (as long as it is long compared to its radius).
A toroid is a hollow circular ring on which a large number of turns of a wire are closely wound (like a solenoid bent into a circle).
Inside the core: $B = \mu_0 n I$ (where $n = \frac{N}{2\pi r}$).
If we draw an Amperian loop outside the toroid, the net current enclosed is zero (since every turn carrying current 'down' is canceled by the same current coming 'up'). By Ampere's Law, $\oint \vec{B} \cdot d\vec{l} = \mu_0 (0) \implies B = 0$.
$B = \mu_r \mu_0 n I$
$B = 400 \times (4\pi \times 10^{-7}) \times 1000 \times 2$
$B = 3200000\pi \times 10^{-7} = 0.32\pi \approx 1.0 \text{ T}$.
Inside the wire ($r < a$), $B \propto r$ (linear increase from zero at center). Outside the wire ($r > a$), $B \propto \frac{1}{r}$ (hyperbolic decrease). The graph is a straight line through the origin reaching a peak at $r=a$, then curving downwards.
Because for symmetric distributions, an Amperian loop can be chosen such that $\vec{B}$ is constant in magnitude and parallel to $d\vec{l}$ everywhere on the loop, simplifying the integral $\oint \vec{B} \cdot d\vec{l}$ to $B \oint dl = B(2\pi r)$.
At the exact ends of a long solenoid, the magnetic field is half of its value at the center.
$B_{ends} = \frac{1}{2} \mu_0 n I$.
The magnetic field in the empty space completely enclosed by the inner boundary of the toroid is zero, because an Amperian loop drawn there encloses no net current ($I_{enc} = 0$).
Force per unit length $f = \frac{\mu_0 I_1 I_2}{2\pi d}$
$f = \frac{4\pi \times 10^{-7} \times 5 \times 10}{2\pi \times 0.10} = \frac{100 \times 10^{-7}}{0.10} = 10^{-4} \text{ N/m}$.
Nature: Attractive (since currents are in the same direction).
One Ampere is the constant current which, if maintained in two straight parallel conductors of infinite length and negligible cross-section, placed 1 meter apart in vacuum, would produce a force of $2 \times 10^{-7}$ Newtons per meter of length between them.
Wire 1 creates a magnetic field at Wire 2. Using the right-hand grip rule, if currents are opposite, the field from Wire 1 causes a force on Wire 2 (via Fleming's Left-Hand rule) that points away from Wire 1. Thus, they repel.
It will experience a net force. The side of the loop closest to the wire experiences a stronger force (attractive or repulsive based on direction) than the far side (since $B \propto 1/r$). Thus, the opposing forces don't cancel, creating a net force.
$\vec{F} = I (\vec{l} \times \vec{B})$, where $\vec{l}$ is a vector whose magnitude is the length of the conductor and direction is the direction of the current.
Since the wire is perpendicular to the field, $\theta = 90^\circ$.
$F = I L B \sin(90^\circ) = I L B$.
The spring will contract (compress). This is because adjacent turns of the spring carry current in the same direction. Parallel currents in the same direction attract each other.
$f = \frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}$.
Fleming's Left-Hand Rule: Stretch thumb, forefinger, and middle finger of the left hand mutually perpendicular. If the forefinger points to the magnetic field and the middle finger to the current, then the thumb points to the direction of force.
When the conductor is placed exactly parallel or anti-parallel to the magnetic field ($\theta = 0^\circ$ or $180^\circ$). In this case, $\sin\theta = 0$, making the force $F = ILB\sin\theta = 0$.
$\vec{\tau} = \vec{m} \times \vec{B}$
$\tau = N I A B \sin\theta$
$A = \pi r^2 = \pi (0.08)^2 = 0.0064\pi \text{ m}^2$
$\tau = 30 \times 6 \times (0.0064\pi) \times 1.0 \times \sin(60^\circ)$
$\tau = 180 \times 0.0201 \times 0.866 \approx 3.13 \text{ N}\cdot\text{m}$.
A magnetic dipole is any system that produces a magnetic field similar to a bar magnet (e.g., a current loop). The dipole moment formula is $m = N I A$.
Bohr Magneton is the minimum magnetic dipole moment associated with an electron due to its orbital motion in the ground state of a hydrogen atom.
Value: $\mu_B = \frac{eh}{4\pi m_e} = 9.27 \times 10^{-24} \text{ A}\cdot\text{m}^2$.
Electric Dipole: $\vec{\tau} = \vec{p} \times \vec{E}$ (where $\vec{p}$ is electric dipole moment).
Magnetic Dipole: $\vec{\tau} = \vec{m} \times \vec{B}$ (where $\vec{m}$ is magnetic dipole moment). Both follow identical mathematical cross-product relationships.
$\mu_l = \frac{e}{2m_e} L$, where $e$ is electron charge, $m_e$ is its mass, and $L$ is the angular momentum. The ratio $\frac{\mu_l}{L} = \frac{e}{2m_e}$ is called the gyromagnetic ratio.
Stable equilibrium occurs when the magnetic moment vector $\vec{m}$ is exactly parallel to the external magnetic field $\vec{B}$ (i.e., angle $\theta = 0^\circ$).
Torque is maximum when the plane of the loop is parallel to the magnetic field. In this state, the normal vector to the loop is perpendicular to the field ($\theta = 90^\circ$). $\tau_{max} = NIAB$.
Initial moment $M = I \times A$.
New current $I' = I/2$, New area $A' = 2A$.
New moment $M' = I' \times A' = (I/2) \times (2A) = I \times A = M$.
The magnetic moment remains unchanged.
No. In a strictly uniform magnetic field, the net translational force on any closed current loop is exactly zero. It only experiences a torque (unless $\theta=0^\circ$ or $180^\circ$).
It is based on the principle that when a current-carrying coil is placed in a uniform magnetic field, it experiences a magnetic torque that tends to rotate it.
The soft iron core increases the strength of the magnetic field due to its high permeability, making the galvanometer more sensitive. It also helps to make the field radial.
Current sensitivity is the deflection produced per unit of current passing through the galvanometer. $I_s = \frac{\theta}{I} = \frac{NAB}{k}$.
SI Unit: rad/A or div/A.
1. Increasing the number of turns ($N$) in the coil.
2. Increasing the area ($A$) of the coil.
3. Using a stronger magnetic field ($B$).
4. Decreasing the restoring torque per unit twist ($k$) of the suspension wire.
Voltage sensitivity is the deflection produced per unit voltage applied ($V_s = \frac{\theta}{V} = \frac{NAB}{kR}$). No, increasing $N$ increases the length of the wire, which proportionally increases the resistance $R$. The ratio $N/R$ remains almost constant, so voltage sensitivity does not necessarily increase.
To convert to a voltmeter, connect a high resistance ($R_s$) in series.
$R_s = \frac{V}{I_g} - R_g = \frac{18}{3 \times 10^{-3}} - 12 = 6000 - 12 = 5988 \, \Omega$.
Connect $5988 \, \Omega$ in series.
To convert to an ammeter, connect a low resistance shunt ($S$) in parallel.
$S = \frac{I_g R_g}{I - I_g} = \frac{4 \times 10^{-3} \times 15}{6 - 0.004} = \frac{0.06}{5.996} \approx 0.010 \, \Omega$.
Connect $0.01 \, \Omega$ in parallel.
A radial magnetic field ensures that the plane of the coil is always parallel to the magnetic field regardless of its orientation. Thus, the torque $\tau = NIAB\sin(90^\circ) = NIAB$ remains maximum and directly proportional to current $I$.
The material should have a small restoring torque per unit twist ($k$) to ensure high sensitivity, and it should have high tensile strength and a steady elastic behavior.
An ammeter is connected in series because it must measure the total current flowing through the circuit component, and it has very low resistance to avoid altering that current. A voltmeter is connected in parallel to measure the potential difference across a component, and it has very high resistance so it doesn't draw significant current from the main circuit.
