Physics

Table of contents

- Electric Field Equations
- Magnetic Field Equations
- More Magnetic Field Equations

Electric Field Equations

$$\vec{\mathit{F}}{}_0=q_0\vec{\mathit{E}}$$ Force exerted on a point charge by and electric field.

Magnetic Field Equations

(1)

$$F=\left|q\right|\upsilon {}_{\perp }B=\left|q\right|\mathit{\upsilon B}\text{sin}\varphi$$ This is the magnitude of the magnetic force on a moving charge. Some things we can observe about the magnetic force from this equation:

1. its magnitude is proportional to the magnitude of the electric charge,
2. its magnitude is proportional to the magnitude of the magnetic field,
3. it is dependent on the moving charge's velocity,
4. by experimentation, its magnitude is always perpendicular to the magnetic field and the velocity vector,
5. it is proportional to the magnitude of the velocity vector's component perpendicular to the magnetic field (i.e. the y-component of the velocity vector if the magnetic field is traveling uniformly along the x-axis, and vice versa). In other words, the magnitude of the magnetic force on the charged particle is greatest when the velocity vector of the particle is perpendicular to the magnetic field and weakest, or nil, when the velocity vector is parallel to the magnetic field.

(2)

$$\overrightarrow{\mathit{F}}=q\overrightarrow{\mathit{v}}\times \overrightarrow{\mathit{B}}$$ This is the magnetic force experienced by a moving charged particle in a magnetic field. Observe that the direction of the magnetic force on a charged particle is dependent on the charge and direction of the particle.

(3)

$$F=\left|q\right|\upsilon B_{\perp }$$ This is the magnitude of the magnetic force felt by a moving particle when the magnetic field is perpendicular to the velocity vector of the particle. This equation allows you to observe the effect of the charged particle's velocity vector on the magnetic force vector. From here, we should point out that the SI units for a magnetic field is the following:

$$1\text{tesla}=1\text{T}=1\text{N}/\text{A}\cdot \text{m}$$

(4)

$$\vec{\mathit{F}}=q(\vec{\mathit{E}}+\vec{\mathit{\upsilon }}\times \vec{\mathit{B}})$$ This is the total force a charged particle experiences when it travels through a region where both electric and magnetic fields are present.

(5)

$$d{\Phi }_B=B_{\perp }dA=B\text{cos}\varphi dA=\vec{\mathit{B}}\cdot d\vec{\mathit{A}}$$ This is the magnetic flux through a small, uniform area of a surface.

(6)

$$\begin{array}{rl}{\mathrm{\Phi }}_B=\int B\cos \varphi dA=\int B_{\perp }dA& =\int \overrightarrow{B}\cdot d\overrightarrow{A}\\ =\int \overrightarrow{B}\cdot d\overrightarrow{A}\end{array}$$ This is the total magnetic flux through a surface. The component of the magnetic field vector normal to the surface may vary across the surface if the surface is not uniform.

(7)

$${\Phi }_B=B_{\perp }A=BA\text{cos}\varphi$$ This is the result of equation (6) when the component of the magnetic field vectors perpendicular to the surface and the angle between the magnetic field vectors and the normal vector to the surface are the same everywhere on a plane surface. If the magnetic field is also perpendicular to the surface, then the equation will be reduced to simply the product of the magnitude of the magnetic field and the the surface area.

$$1\text{Wb}=1{\text{T⋅m}}^2=1\text{N}\cdot m/A$$ This is the SI unit of magnetic flux.

(8)

$$\text{∮}\overrightarrow{\mathit{B}}\cdot d\overrightarrow{\mathit{A}}=0$$ This is Gauss's law for magnetism. It states that the total magnetic flux through any closed surface equals zero. If we were to observe the magnetic field lines of a bar magnet, you would notice that the field lines that enter one side of the magnet (the south magnetic pole) come out of the other end (the north magnetic pole) resulting in a net flux equal to zero. Magnetic field lines have no endpoints unlike electric field lines; and that is why magnetic monopoles do not exist. In the case when there is a magnetic field on an open surface, the direction of the area vector can be arbitrary but should be consistent throughout the calculations.

(9)

$$B={\displaystyle \frac{d{\text{Φ}}_B}{d{A}_{\perp }}}$$ This represents the magnitude of the magnetic field when the area of an element is entirely at a right angle to the field lines.

(10)

$$F=\left|q\right|\upsilon \text{B}=m{\displaystyle \frac{\upsilon {}^2}{R}}$$ This equation arised from the observation that charged particles moving through a magnetic field must move in a circular motion because of the constant magnitudes of the magnetic force and the velocity vector on the charged particle. Because the magnitude of the velocity vector is constant, and the magnitude of the magnetic force is proportional to it, we can describe the magnitude of the magnetic force on a charged particle as the product of the mass of the particle and the centripetal acceleration (the magnitude of acceleration of an object in uniform circular motion).

(11)

$$R=\frac{mv}{|q|B}$$ This is the equation for the radius of a charged particle's circular orbit in a magnetic field derived from equation (10).

(12)

$$\omega ={\displaystyle \frac{\upsilon }{R}}=\upsilon {\displaystyle \frac{\left|q\right|B}{m\upsilon }}={\displaystyle \frac{\left|q\right|B}{m}}$$ From mechanical physics, we know that the angular speed of a rigid body is the magnitude of the body's angular velocity. Note: speed is a magnitude and does not have direction; thus it is always positive.

(13)

$$\upsilon ={\displaystyle \frac{E}{B}}$$ When a charged particle moves at this speed through an electromagnetic field where the electric and magnetic fields are perpendicular to each other, the electric and magnetic forces cancel each other out. This is the equation often used by ionic velocity selectors known as Wien filters. These are used in mass spectronomy.

(14)

$${\displaystyle \frac{1}{2}}m\upsilon {}^2=eV$$

(15)

$${\displaystyle \frac{e}{m}}={\displaystyle \frac{E^2}{2V{B}^2}}$$

$$e/m\approx 1.759\times {10}^{11}\text{C}/kg$$

(16)

$$F=\left(nAl\right)(q\upsilon {}_{\text{d}}B)=(nq\upsilon {}_{\text{d}}A)\left(lB\right)$$

(17)

$$F=IlB$$ Use this equation to determine the magnetic force exerted on a current carrying rod/wire of a specific length in a magnetic field where the angle between the magnetic field lines and the current carrying rod/wire is 90 degrees.

(18)

$$F=Il{B}_{\perp }=IlB\text{sin}\varphi$$ This is similar to equation (17) except it describes how the angle between the magnetic field lines and the current carrying rod/wire has an effect on the magnetic force experienced by the rod/wire. Observe that if the angle between the magnetic field lines and the rod/wire is 0, the magnetic force is 0.

(19)

$$\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}$$

(20)

$$d\overrightarrow{\mathit{F}}=Id\overrightarrow{\mathit{l}}\times \overrightarrow{\mathit{B}}$$ This is the magnetic force experienced by an infinitesimal section of current carrying wire inside a magnetic field. Note: the length vector points in the direction of the current flow, and it is tangent to a curve on the current carrying wire (hence the derivative).

(21)

$$F=IaB$$ This describes the force on a side of length 'a' of a closed, current-carrying loop caused by a magnetic field perpendicular to the normal of the loop's plane. This same force would be felt on both sides of a square current-carrying loop positioned inside a magnetic field, except the forces on each side would be opposite in opposite directions. This produces torque on the loop. Summing up the forces on a closed current-carrying loop in a magnetic field, the net force would be zero. This is similar to equation (17).

(22)

$$\tau =2F(b/2)\text{sin}\varphi =\left(IBa\right)(b\text{sin}\varphi )$$ This describes the magnitude of the net torque on a square, current-carrying loop in a magnetic field. The torque is equal to the force experienced by two straight edges of the square, current-carrying loop about the y-axis times the distance from each of these edges to the center of axis of rotation (i.e. the y-axis, in this case). The force depends on the angle between the square plane and the magnetic field where the force is greatest when the normal vector is perpendicular to the magnetic field.

(23)

$$\begin{array}{rl}{\mathrm{\Phi }}_B=\int B\cos \varphi dA=\int B_{\perp }dA& =\int \overrightarrow{B}\cdot d\overrightarrow{A}\\ =\int \overrightarrow{B}\cdot d\overrightarrow{A}\end{array}$$ This is the same as equation (6).

(24)

$$\mu =IA$$ This is the magnetic dipole moment of the loop described in (22).

(25)

$$\tau =\mu B\text{sin}\varphi$$

(26)

$$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$

(27)

$$U=-\overrightarrow{\mathit{\mu }}\cdot \overrightarrow{\mathit{B}}=-\mu B\cos \varphi$$

(28)

$$\tau =NIAB\text{sin}\varphi$$

(29)

$$V_{ab}=\epsilon +Ir$$

(30)

$$nq=\frac{-J_x{B}_y}{E_z}$$

More Magnetic Field Equations

(1)

$$B={\displaystyle \frac{{\mu }_0}{4\pi }}{\displaystyle \frac{\left|q\right|\upsilon \text{sin}\varphi }{r^2}}$$ This describes the magnitude of a magnetic field produced by a point charge moving at constant velocity. The magnetic field lines produced by a moving point charge can be thought of as forming circles around the velocity vector of the moving point charge.

(2)

$$\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{q\overrightarrow{\mathit{v}}\times \hat{\mathit{r}}}{r_{}^2}$$ This describes the magnetic field produced by a moving charge at some distance.

(3)

$$\begin{array}{lll}{\mu }_0\hfill & =\hfill & 1.25663706\times {10}^{-6}\text{N}\cdot {\text{s}}^2{\text{C}}^2=1.25663706\times {10}^{-6}\text{ Wb}/\text{A}\cdot \text{m}\hfill \\ \hfill & \cong \hfill & 4\pi \times {10}^{-7}\text{ T}\cdot \text{m}/\text{A}\hfill \end{array}$$

(4)

$$c={\displaystyle \frac{1}{\sqrt{{\in }_0{\mu }_0}}}=2.998\times {10}^8\text{m}/s$$ This is the value of the speed of light [in a vacuum].

(6)

$$d\overrightarrow{B}=\frac{{\mu }_0}{4\pi }\frac{Id\overrightarrow{l}\times \hat{r}}{r^2}$$

(9)

$$B={\displaystyle \frac{{\mu }_{0}I}{2\pi r}}$$ This describes the magnitude of the magnetic field at a distance from a long, straight, current-carrying conductor.

(11)

$$\frac{F}{L}=\frac{{\mu }_{0}I{I}^{\mathrm{\prime }}}{2\pi r}$$

(15)

$$B_x=\frac{{\mu }_{0}I{a}^2}{2{(x^2+a^2)}^{3/2}}$$

(17)

$$B_x=\frac{{\mu }_{0}NI}{2a}$$

(20)

$$\text{∮}\overrightarrow{\mathit{B}}\cdot d\overrightarrow{\mathit{l}}={\mu }_0{I}_{\text{encl}}$$