702 lines
27 KiB
Markdown
702 lines
27 KiB
Markdown
# ECE 106: Electricity and Magnetism
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## MATH 117 review
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!!! definition
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A definite integral is composed of:
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- the **upper limit**, $b$,
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- the **lower limit**, $a$,
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- the **integrand**, $f(x)$, and
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- the **differential element**, $dx$.
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$$\int^b_a f(x)\ dx$$
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The original function **cannot be recovered** from the result of a definite integral unless it is known that $f(x)$ is a constant.
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## N-dimensional integrals
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Much like how $dx$ represents an infinitely small line, $dx\cdot dy$ represents an infinitely small rectangle. This means that the surface area of an object can be expressed as:
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$$dS=dx\cdot dy$$
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Therefore, the area of a function can be expressed as:
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$$S=\int^x_0\int^y_0 dy\ dx$$
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where $y$ is usually equal to $f(x)$, changing on each iteration.
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!!! example
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The area of a circle can be expressed as $y=\pm\sqrt{r^2-x^2}$. This can be reduced to $y=2\sqrt{r^2-x^2}$ because of the symmetry of the equation.
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$$
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\begin{align*}
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A&=\int^r_0\int^{\sqrt{r^2-x^2}}_0 dy\ dx \\
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&=\int^r_0\sqrt{r^2-x^2}\ dx
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\end{align*}
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$$
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!!! warning
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Similar to parentheses, the correct integral squiggly must be paired with the correct differential element.
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These rules also apply for a system in three dimensions:
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| Vector | Length | Area | Volume |
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| --- | --- | --- | --- |
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| $x$ | $dx$ | $dx\cdot dy$ | $dx\cdot dy\cdot dz$ |
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| $y$ | $dy$ | $dy\cdot dz$ | |
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| $z$ | $dz$ | $dx\cdot dz$ | |
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Although differential elements can be blindly used inside and outside an object (e.g., area), the rules break down as the **boundary** of an object is approached (e.g., perimeter). Applying these rules to determine an object's perimeter will result in the incorrect deduction that $\pi=4$.
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Therefore, further approximations can be made using the Pythagorean theorem to represent the perimeter.
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$$dl=\sqrt{(dx^2) + (dy)^2}$$
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### Polar coordinates
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Please see [MATH 115: Linear Algebra#Polar form](/1a/math115/#polar-form) for more information.
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In polar form, the difference in each "rectangle" side length is slightly different.
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| Vector | Length difference |
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| --- | --- |
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| $\hat r$ | $dr$ |
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| $\hat\phi$ | $rd\phi$ |
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Therefore, the change in surface area can be approximated to be a rectangle and is equal to:
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$$dS=(dr)(rd\phi)$$
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!!! example
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The area of a circle can be expressed as $A=\int^{2\pi}_0\int^R_0 r\ dr\ d\phi$.
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$$
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\begin{align*}
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A&=\int^{2\pi}_0\frac{1}{2}R^2\ d\phi \\
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&=\pi R^2
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\end{align*}
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$$
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If $r$ does not depend on $d\phi$, part of the integral can be pre-evaluated:
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$$
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\begin{align*}
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dS&=\int^{2\pi}_{\phi=0} r\ dr\ d\phi \\
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dS^\text{ring}&=2\pi r\ dr
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\end{align*}
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$$
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So long as the variables are independent of each other, their order does not matter. Otherwise, the dependent variable must be calculated first.
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!!! tip
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There is a shortcut for integrals of cosine and sine squared, **so long as $a=0$ and $b$ is a multiple of $\frac\pi 2$**:
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$$
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\int^b_a\cos^2\phi=\frac{b-a}{2} \\
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\int^b_a\sin^2\phi=\frac{b-a}{2}
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$$
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The side length of a curve is as follows:
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$$dl=\sqrt{(dr^2+(rd\phi)^2}$$
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!!! example
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The side length of the curve $r=e^\phi$ (Archimedes' spiral) from $0$ to $2\pi$:
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\begin{align*}
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dl &=d\phi\sqrt{\left(\frac{dr}{d\phi}\right)^2 + r^2} \\
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\tag{$\frac{dr}{d\phi}=e^\phi$}&=d\phi\sqrt{e^{2\phi}+r^2} \\
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&=????????
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\end{align*}
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Polar **volume** is the same as Cartesian volume:
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$$dV=A\ dr$$
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!!! example
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For a cylinder of radius $R$ and height $h$:
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$$
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\begin{align*}
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dV&=\pi R^2\ dr \\
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V&=\int^h_0 \pi R^2\ dr \\
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&=\pi R^2 h
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\end{align*}
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$$
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### Moment of inertia
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The **mass distribution** of an object varies depending on its surface density $\rho_s$. In objects with uniformly distributed mass, the surface density is equal to the total mass over the total area.
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$$dm=\rho_s\ dS$$
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The formula for the **moment of inertia** of an object is as follows, where $r_\perp$ is the distance from the axis of rotation:
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$$dI=(r_\perp)^2dm$$
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If the axis of rotation is perpendicular to the plane of the object, $r_\perp=r$. If the axis is parallel, $r_\perp$ is the shortest distance to the axis. Setting an axis along the axis of rotation is easier.
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!!! example
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In a uniformly distributed disk rotating about the origin like a CD with mass $M$ and radius $R$:
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$$
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\begin{align*}
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\rho_s &= \frac{M}{\pi R^2} \\
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dm &= \rho_s\ r\ dr\ d\phi \\
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dI &=r^2\ dm \\
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&= r^2\rho_s r\ dr\ d\phi \\
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&= \rho_s r^3dr\ d\phi \\
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I &=\rho_s\int^{2\pi}_{\phi=0}\int^R_{r=0} r^3dr\ d\phi \\
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&= \rho_s\int^{2\pi}_{\phi=0}\frac{1}{4}R^4d\phi \\
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&= \rho_s\frac{1}{2}\pi R^4 \\
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&= \frac 1 2 MR^2
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\end{align*}
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$$
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## Electrostatics
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!!! definition
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- The **polarity** of a particle is whether it is positive or negative.
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The law of **conservation of charge** states that electrons and charges cannot be created nor destroyed, such that the **net charge in a closed system stays the same**.
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The law of **charge quantisation** states that charge is discrete — electrons have the lowest possible quantity.
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Please see [SL Physics 1#Charge](/sph3u7/#charge) for more information.
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**Coulomb's law** states that for point charges $Q_1, Q_2$ with distance from the first to the second $\vec R_{12}$:
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$$\vec F_{12}=k\frac{Q_1Q_2}{||R_{12}||^2}\hat{R_{12}}$$
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!!! warning
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Because Coulomb's law is an experimental law, it does not quite cover all of the nuances of electrostatics. Notably:
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- $Q_1$ and $Q_2$ must be point charges, making distributed charges inefficient to calculate, and
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- the formula breaks down once charges begin to move (e.g., if a charge moves a lightyear away from another, Coulomb's law says the force changes instantly. In reality, it takes a year before the other charge observes a difference.)
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### Dipoles
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An **electric dipole** is composed of two equal but opposite charges $Q$ separated by a distance $d$. The dipole moment is the product of the two, $Qd$.
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The charge experienced by a positive test charge along the dipole line can be reduced to as the ratio between the two charges decreases to the point that they are basically zero:
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$$\vec F_q=\hat x\frac{2kQdq}{||\vec x||^3}$$
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## Maxwell's theorems
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Compared to Coulomb's law, $Q_1$ creates an electric field around itself — each point in space is assigned a vector that depends on the distance away from the charge. $Q_2$ *interacts* with the field. According to Maxwell, as a charge moves, it emits a wave that carries information to other charges.
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The **electric field strength** $\vec E$ is the force per unit *positive* charge at a specific point $p$:
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$$\vec E_p=\lim_{q\to 0}\frac{\vec{F}}{q}$$
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Please see [SL Physics 1#Electric potential](/sph3u7/#electric-potential) for more information.
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### Electric field calculations
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If charge is distributed over a three-dimensional object, integration similar to moment of inertia can be used. Where $dQ$ is an infinitely small point charge at point $P$, $d\vec E$ is the electric field at that point, and $r$ is the vector representing the distance from any arbitrary point:
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$$d\vec E = \frac{kdQ}{r^2}\hat r$$
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!!! warning
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As the arbitrary point moves, both the direction and the magnitude of the distance from the desired point $P$ change (both $\hat r$ and $r$).
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Generally, if a decomposing the vector into Cartesian forms $d\vec E_x$, $d\vec E_y$, and $d\vec E_z$ is helpful even if it is easily calculated in polar form because of the significantly easier ability to detect symmetry in the shape. Symmetry about the axis allows deductions such as $\int d\vec E_y=0$, which makes calculations easier.
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In a **one-dimensional** charge distribution (a line), the charge density is used in a similar way as moment of inertia's surface density:
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$$dQ=\rho_\ell d\ell$$
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**Two-dimensional** charge distributions are more or less the same, but polar or Cartesian forms of the surface area work depending on the shape.
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$$dQ=\rho_s dS$$
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!!! example
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A rod of uniform charge density and length $L$ has a charge density of $p_\ell=\frac{Q}{L}$.
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1. Determine the formula for the charge density $\rho$
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2. Choose an origin and coordinate system (along the axes of the object when possible)
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3. Choose an arbitrary point $A$ on the charge
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4. Create a right-angle triangle with $A$, the desired point, and usually the origin
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5. Attempt to find symmetry
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6. Solve
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## Gauss's law
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!!! definition
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- A **closed surface** is any closed three-dimensional object.
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- **Electric flux** represents the number of electric field lines going through a surface.
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At an arbitrary surface, the **normal** to the plane is its vector form:
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$$\vec{dS}=\vec n\cdot dS$$
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The **electric flux density** $\vec D$ is an alternate representation of electric field strength. In a vacuum:
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$$\vec D = \epsilon_0\vec E$$
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**Electric flux** is the electric flux density multiplied by the surface area at every point of an object.
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$$\phi_e=\epsilon_0\int_s\vec E\bullet\vec{dS}$$
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The flux from charges outside a closed surface will **always be zero at the surface**. A point charge in the centre of a closed space has a flux equal to its charge. Regardless of the charge distribution or shape, the **total flux** through a closed surface is equal to the **total charge within** the closed surface.
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$$\oint \vec D\bullet\vec{dS}=Q_\text{enclosed}$$
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This implies $\phi_e>0$ is a net positive charge enclosed.
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!!! warning
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Gauss's law only applies when $\vec E$ is from all charges in the system
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### Charge distributed over a line/cylinder
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!!! warning "Limitations"
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To apply this strategy, the following conditions must hold:
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- $Q$ must not vary with the length of the cylinder or $\phi$
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- The charge must be distributed over either a cylindrical surface or the volume of the cylinder.
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- In the real world, $r$ must be significantly smaller than $L$ as an approximation.
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- The strategy is more accurate for points closer to the centre of the wire.
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Please see [Maxwell's integral equations#Gauss's law](https://en.wikiversity.org/wiki/MyOpenMath/Solutions/Maxwell%27s_integral_equations) for more information.
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**Outside** the radius $R$ of the cylinder of the Gaussian surface, the enclosed charge is, where $L$ is the length of the cylinder:
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$$Q_{enc}=\pi R^2\rho_0L$
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such that the field at any radius $r>R$ is equal to:
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$$\vec E(r)=\frac{\rho_0\pi R^2}{2\pi\epsilon_0r}\hat r$$
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**Inside** the radius $R$ of the cylinder, the enclosed charge depends on $r$. For a uniform charge density:
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$$Q_{enc}=\pi r^2\rho_0L$$
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such that the field at any radius $r< R$ is equal to:
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$$\vec E(r)=\frac{\rho_0}{2\epsilon_0}r\hat r$$
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The direction of $\vec E$ should always be equal to that of $\vec r$. Generally, where $lim$ is $r$ if $r$ is *inside* the cylinder or $R$ otherwise, $\rho_v$ is the function for charge density based on radius, and $r_1$ is hell if I know:
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$$\epsilon_0 E2\pi rL=\int^{lim}_0\rho_v(r_1)2\pi r_1L\ dr_1$$
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### Charge distributed over a plane
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!!! warning
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To apply this strategy, the following conditions must hold:
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- $Q$ must not vary with the lengths of the plane
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- The charge must be distributed over a plane or slab
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- In the real world, the thickness $z$ must be significantly smaller than the lengths as an approximation
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Where $\rho_v$ is an **even** surface density function and $lim$ is from $0$ to $z$ if the desired field is outside of the charge, or $0$ to field height $h$ if it is inside the charge:
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$$\epsilon_0 E=\int_{lim}\rho_v\ dh_1$$
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Any two points have equal electric fields regardless of distance due to the construction of a uniform electric field.
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Where $\rho_v$ is not an even surface density function, $d$ is the thickness of the slab, and $E$ is the electric field **outside** the slab:
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$$2\epsilon_0E = \int^d_0\rho_v(A)dh_1$$
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Where $E$ is the electric field **inside** the slab at some height $z$:
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$$E=\frac{\rho_0}{4\epsilon_0}(2z^2-d^2),0\leq z\leq d$$
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If $E$ is negative, it must point opposite the original direction ($\hat z$).
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Generally:
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1. Determine $\vec E$ outside the slab.
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2. Set one outside surface and one inside surface as a pillbox and apply rules.
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## Electrostatic potential
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At a point $P$, the electrostatic potential $V_p$ or voltage is the work done per unit positive test charge from infinity to bring it to point $P$ by an external agent.
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$$
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V_p=\lim_{q\to 0^+}\frac{W_i}{q} \\
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W_I=\int^p_\infty\vec F_I\bullet \vec{dl}=\Delta U=QV_p
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$$
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Because the desired force acts opposite to the force from the electric field, as long as $\vec E$ is known at each point:
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$$
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V_p=-\int^p_\infty\vec E\bullet\vec{dl} \\
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V_p=-\int^p_\infty E\ dr
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$$
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The work done only depends on initial and final positions — it is conservative, thus implying Kirchoff's voltage law.
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Where $\vec dl$ is the path of the test charge from infinity to the point, and $\vec dr$ is the direct path from the origin through the point to the charge, because $dr=-dl$:
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$$\vec E\bullet\vec{dl}=Edr$$
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Therefore, the potential due to a point charge is equal to (the latter is true only if distance from charge is always constant, regardless of distribution):
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$$V_p=-\int^p_\infty\frac{kQ}{r^2}dr=\frac{kQ}{r}$$
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**Positive** charges naturally move to **lower** potentials ($V$ decreases) while negative charges do the opposite. Potential energy always decreases.
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In order to calculate the voltage for charge distributions:
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- If $\vec E$ is easy to find via Gauss law:
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$$V_p=-\int^p_\infty\vec E\bullet\vec{dl}$$
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- If the charge is asymmetric:
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$$V_p=\int_\text{charge dist}\frac{kdQ}{r}$$
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The electric field always points in the direction of **lower** potential, and is equal to the **negative gradient** of potential.
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$$\vec E=-\nabla V$$
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If $\vec E$ is constant:
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$$\vec E=\frac{Q_{enc\ net}}{\epsilon_0\oint dS}$$
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The **superposition** principle allows potential due to different charges to be calculated separately and summed together to achieve the same result.
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## Conductors
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An **ideal conductor** has electrons loosely bound to atoms such that an electric field causes them to freely move by $F=Q_e E$. However, this assumes that there are infinite electrons in the conductor, and that the electrons will move with **zero resistance** to the surface of the conductor but **not leave it**.
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A conductor placed in an external electric field will cause electrons to hop from atom to atom to reach the surface, charging one surface negatively and the other positively. The **induced electric field** from this imbalance opposes the external field force, slowing down electron movement until equilibrium is reached.
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$$\text{equilibrium}\iff \vec E_{ext}+\vec E_{ind}=\vec 0$$
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At equilibrium, **every point in the conductor is equipotential**. Gauss's law implies that there is no volume charge inside a conductor.
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At its surface, $\vec E$ tangent to the surface must be zero. Normal to the surface:
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$$|\vec E_N|=\frac{|\rho_0|}{\epsilon_0}$$
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- $\rho_0$ is negative if field lines **enter** the conductor.
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- $\rho_0$ is positive if field lines exit the conductor.
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### Conductor cavities
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A cavity surface must have **zero surface charge**. This creates a Faraday cage — outside fields cannot affect the cavity, but fields from the cavity can affect the outside world.
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If there is a fixed/non-moving charge $Q$ in the cavity:
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- $\vec E=0$ inside the conductor, so the boundary surface charge must be $-Q$.
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- Electrons are taken from the surface, so the surface charge outside the conductor must be $Q$, propagating the effect of the charge to the outside world.
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### Ground
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A **ground** is a reservoir or sink of charges that never changes, regardless of the quantity added or removed from it. At the connection point, $V=0$ is always guaranteed.
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Grounding a conductor means that it takes charges from the ground to balance an internal charge, neutralising it.
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A charge released into a conductor (e.g., battery into wire) will always go to the outside surface, regardless of the point of insertion. Two charged objects connected by a thin conductor will redistribute their charge such that:
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- their potentials are equal
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- conservation of charge is followed.
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This implies that a larger object has more charge, but a smaller object has a denser charge and thus stronger electric field.
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$$Q_1=\frac {R_1} {R_2}Q_2$$
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!!! example
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For two spheres, as $\rho=\frac{Q_1}{4\pi R^2}$:
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$$\rho_1=\frac {R_2} {R_1}\rho_2$$
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A non-uniform object, such as a cube, will have larger charge density / stronger electric field at sharper points in its shape. Symmetrical surfaces always have uniform charge density.
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!!! warning
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An off-centre charge in a cavity will require a non-uniform induced charge to cancel out the internal field, but the external surface charge will be uniform (or non-uniform if the surface is odd).
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### Nutshell
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**Inside** a conductor:
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- $\vec E=0$
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- $\Delta V=0$
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- $\rho_v=0$
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Inside a cavity, if there exists an external field:
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- $\vec E=0$
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- $\rho_s=-Q$
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- $\rho_{s\ outer}=Q$
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The inner surface charge distribution matches that of the inner charge, but the outer surface charge distribution is dependent only on the shape of the conductor.
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On conductor surfaces, the only $\vec E$ is **normal** to the surface and dependents on the shape of the surface.
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$$|\vec E_N|=\frac{|\rho_s|}{\epsilon_0}$$
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Grounding a conductor neutralises any free charges.
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In slabs, as $A>>d$, assume $Q$ is uniformly distributed.
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To solve systems:
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- Assigning charge **density** is easier with sheets
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- Assigning **charges** is easier with cylinders/spheres
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## Dielectrics
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!!! definition
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- An **insulator** has electrons tightly bound to atoms.
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### Polarisation
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Polarisation is the act of inducing a dipole to a lesser extent than conductors. The induced field cannot reduce $\vec E$ inside the insulator to zero, but it will reduce its effects. The **polarisation vector** $\vec P$ is an average of the effects of all induced fields on a certain point inside a volume.
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$$\vec P=\lim_{\Delta V\to 0}\frac{\sum^{N\Delta v}\vec p_i}{\Delta v}$$
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where:
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- $\Delta v\approx dv$ is the volume of the insulator
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- $p_i$ is the dipole moment at a point
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- $N$ is the total number of atoms in the volume
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Polarisation is proportional to electric field and the **electric susceptibility** $X_e$ of a material to external fields.
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$$\boxed{\vec P=\epsilon_0X_e\vec E}$$
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The **relative permittivity** $\epsilon_r$ of a material is the ratio of decreasing $\vec E$ inside a medium relative to free space.
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$$\epsilon_r=1+X_e$$
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The new **flux density** formula includes polarised charges, so now $Q_{enc}$ includes **only free charges** (i.e., not polarised charges).
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$$\boxed{\vec D=\epsilon_0\vec E+\vec P=\epsilon_0\epsilon_r\vec E}$$
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$$\boxed{\oint\vec D\bullet\vec{dS}=Q_{enc,free}}$$
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In uniform charge distributions, the surface charge density is related to its polarisation. Where $\hat n$ is the unit normal of the surface:
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$$\rho_s=\vec P\bullet\hat n$$
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### Boundary conditions
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Regardless of permittivity, the $\vec E$ **tangential to the boundary** between two materials must be equal.
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## Capacitors
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!!! definition
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- A **capacitor** is a device that uses the capacitance of materials to store energy in electric fields. It is usually composed of two conductors separated by a dielectric.
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**Capacitance** is a measurement of the charge that can be stored per unit difference in potential.
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$$\boxed{Q=C\Delta V}$$
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To determine $C$:
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1. Place a positive and a negative charge on conductors
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2. Determine charge distribution
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3. Determine $\vec E$ between the conductors
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4. Find a path from the negative to the positive conductor and determine voltage
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??? example
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For two plates separated by distance $d$, with charges of $+Q$ and $-Q$, and a dielectric in between with permittivity $\epsilon_0\epsilon_r$:
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- Clearly $\rho_0=\frac Q A$ as sheets must have uniform distribution. $-\rho_0$ is on the negative plate.
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- From Gauss' law, creating a Gaussian surface outside the capacitor to between the plates gives $DA=\rho_0A$.
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- $D=\epsilon_0\epsilon_rE$ gives $E=\frac{\rho_0}{\epsilon_0\epsilon_r}$
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- Sheets have uniform fields, thus $\Delta V=Ed$
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- Finally, $C=\epsilon_0\epsilon_r\frac A d$
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!!! warning
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If three dielectrics with different permittivities are allowed to touch each other, they will create **fringe fields** at their intersection that destroy the boundary condition.
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### Capacitors and energy
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The stored energy inside capacitors is the same as any other energy.
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$$\boxed{U_e=\frac 1 2CV^2}$$
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Much like VIR, it's usually easier to work with the form of the equation that has squared constants.
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$$U_e=\frac 1 2 \frac {Q^2}{C}=\frac 1 2 QV$$
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Adding dielectrics increases capacitance but decrease stored energy.
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## Magnetism
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All magnetic field lines are closed, i.e., they all return to the same magnetic object, much like a dipole. All lines must be perpendicular to the surface:
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$$\oint\vec B\bullet\vec{dS}=0$$
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Per **Biot-Savart's law**, magnets are complicated.
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$$\boxed{d\vec B_p=\frac{\mu_0}{4\pi}I\frac{\vec {dl}\times\hat r}{|r|^2}}$$
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where:
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- $\mu_0$ is the magnetic permeability of free space
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- $\hat r$ is the unit vector pointing from an arbitrary point of a wire to the desired point
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- $I$ is current
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- $dl$ follows the direction of current
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The final direction can be determined in advance with the **right-hand rule**. Therefore, magnitude can be reduced to:
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$$|dl\times\hat r|=|dl||\hat r|\sin\theta=|dl|\sin\theta$$
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### Calculations
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1. Define coordinate system
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2. Go to some arbitrary point $A$ on a coordinate axis such that $r=AP$
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3. Determine magnitude of the cross product
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4. Determine final magnetic field direction (should be constant)
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5. Rewrite equation in terms of one variable (usually $\theta$)
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6. Integrate
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### Selenoids
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It's easiest to place the origin at the target point.
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A selenoid with $N$ turns around a coil of length $L$ has density $n$, and has parallel electric fields inside.
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$$n=\frac N L$$
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The effective current of a selenoid for magnetic purposes is the sum of all currents.
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$$\boxed{I_{eff}=ndzI}$$
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where:
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- $dz$ is the axis in the direction of current
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- $I$ is current
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This can be substituted directly into Biot-Savart's law, although definite integration should be done **in the direction of the axis** (from the desired point to the farthest point of the selenoid).
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### Velocity and current
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Biot-Savart's law can be applied to moving charges:
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$$I\cdot \vec{dl}=\frac{dq\cdot dl}{dt}=dq\cdot \vec v$$
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### Ampere's law
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!!! definition
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- **Drift velocity** is the average speed of electrons through a material.
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The **current density** $\vec J$ is the amount of charge per unit time that flows through a unit area of a cross section.
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$$\boxed{\vec J=nq\vec u=\rho_v\vec u}$$
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where:
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- $\vec u$ is drift velocity
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- $n$ is the charge per unit volume
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- $q$ is the total charge
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Ohmic resistors have current density proportional to electric field by a material's **conductivity** $\sigma$.
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$$\vec J=\sigma\vec E$$
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Resistivity is related to conductivity: $\rho=\frac 1\sigma$
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Integrating over a cross section returns current:
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$$\boxed{I=\oint\vec J\bullet\vec{dS}}$$
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**Ampere's law** asserts that magnetic flux due to all currents is equal to current enclosed inside a closed boundary/loop.
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$$
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\boxed{\begin{align*}
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\oint\vec B\bullet\vec{dl}&=\mu_0I_{enc} \\
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&=\mu_0\oint\vec J\bullet\vec{dS}
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\end{align*}}
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$$
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where:
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- $dl$ is the line along the loop/boundary in an arbitrary direction
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- $I_{enc}$ is the sum of all enclosed currents
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$dl$ (along the loop) and $dS$ are related in direction with each other per the **right hand rule**.
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!!! warning
|
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Ampere's law is only true in when dealing with DC.
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For each enclosed $I$, if its direction is:
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- the same as $\vec dS$, it is positive in the sum term
|
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- opposite $\vec dS$, it is negative in the sum term
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|
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1. Use $dl$ to find $dS$ or vice versa
|
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2. Determine $I_{enc}$
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3. Solve
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The angle of a cut to a surface does not affect any equations and can be treated identically. Any imaginary closed loop such that $\vec B$ **is constant over the loop and parallel to the loop** is usable with Ampere's law as $B$ can be reduced to a constant scalar.
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The geometries that work include:
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- Infinite cylinders with $J$ that may vary with $r$ but not $\phi$
|
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- Infinite sheets/slabs where $J$ may vary with $z$ but not $x,y$
|
|
- Infinite selenoids
|
|
- Toroids (a selenoid bent into a donut shape)
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|
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1. Create a cross-section perpendicular to the current and determine if symmetry of the loop can meet conditions for geometry
|
|
2. Choose $dl$ in the direction of $B$ (counterclockwise)
|
|
3. Determine $dS$ (out of the page) and apply Ampere's law
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|
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$$\hat\phi=\hat z\times\hat r_1$$
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|
|
!!! warning
|
|
A spinning cylinder rotates faster along its outer ring, forcing an integral setup.
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|
|
### Faraday's law
|
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|
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Faraday's law states relates magnetic flux similarly to electric flux. Where $s$ is the open surface bounded by the conductor:
|
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$$\phi_m=\int_s\vec B\bullet\vec{dS}$$
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A flux that changes with time results in an **induced voltage** across the terminals of the conductor. Per Faraday's law of electromagnetic induction, magnetic energy is convertible to electric energy.
|
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|
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$$V_{ind}=-\frac{d}{dt}\phi_m$$
|
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|
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As the electric field is always perpendicular to a magnetic field, this indicates that it will curl around a straight magnetic field.
|
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|
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Relating $dl$ and $dS$ with the right-hand rule accounts for **Lenz's law**, which creates a $\vec E$ to create a $\vec B$ to oppose the change in $\phi_m$ that created the current.
|
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|
|
$$\boxed{\oint\vec E\bullet\vec{d\ell}=\frac{d}{dt}\int\vec B\bullet\vec{dS}}$$
|
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|
|
If there is a conducting loop in a time-varying magnetic field, a $V_{ind}$ is formed such that the current is in the direction of the induced field:
|
|
|
|
$$V_{ind}=\oint\vec E\bullet\vec{d\ell}=-\frac{d}{dt}\int\vec B\bullet\vec{dS}$$
|
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|
|
Time-varying magnetic fields are formed if the field or charge is moving or if bounds change.
|
|
|
|
## Inductance
|
|
|
|
Kirchoff's voltage law is a simplification of Faraday's law, valid when there is no fluctuating magnetic field within the closed loop, so it's used with low frequency waves with less time variation.
|
|
|
|
The **inductance** is the flux travelling through a medium over its current.
|
|
|
|
$$L=\frac{\phi_m}{i}$$
|
|
|
|
If there are $N$ loops in a selenoid, where $\Lambda=N\phi_m$ is the total flux/**flux linkage**, $i$ is the current in one loop, and $I$ is the current of all loops:
|
|
|
|
$$L=\frac{\phi_m}{i}=\frac{\Lambda}{I_{eff}}$$
|
|
|
|
The **energy density** per unit volume is $u_m$.
|
|
|
|
$$u_m=\frac 1 2 \frac {B^2}{\mu_0}$$
|
|
|
|
The **total work** $U_m$ done to charge current from $0$ to $I$ is related to energy density.
|
|
|
|
$$U_m=\sqrt u_m=\frac 1 2 LI^2$$
|
|
|
|
$$\boxed{\frac 1 2 LI^2=\frac 1 2\int_{volume}U_mdV}$$
|
|
|
|
### Self-inductance
|
|
|
|
A magnetic flux that passes through the current that created it will induce voltage if $I$ changes.
|
|
|
|
**Mutual inductance** is wireless charging as changing current in one coil produces a changing magnetic flux in another, creating a voltage $\epsilon_{1\to 2}$.
|
|
|
|
$$V_{ind}=\epsilon_{1\to 2}=N_2\frac{d\phi_{1\to 2}}{dt}=-\frac{d}{dt}\int \vec B_1\bullet\vec{dS}_2$$
|
|
|
|
The mutual inductance is the rate of change of magnetic flux proportional to the rate of change of current. It is equal regardless of direction.
|
|
|
|
$$\boxed{M_{1\to 2}=\frac{N_2\phi_{1\to 2}}{I_1}}$$
|