Rowan-Classes/5th-Semester-Fall-2023/EEMAGS/Notes/Chapter1.md

Chapter 1

1.1 Vector Analysis

Scalar

A measure described by one real number. Examples include temperature, size, and mass. A scalar is a 1 \times 1 matrix.

Vector

A measure described by more than one real number (direction and magnitude). Examples include force, velocity, and it's derivatives. Vectors are often described by n \times 1 or 1 \times n matricies.

Unit Vector

A unit (direction or normalized) vector is signified with a \hat{ } symbol. Common unit vectors include \hat{x}, \hat{y}, and \hat{z}. The normalized version of any vector is defined as:

\hat{a} = \frac{\vec{A}}{\lVert\vec{A}\rVert}

Dot Product

The dot product is a measure of how parallel two vectors are, scaled by the magnitudes of the two vectors. To compute it, find the sum of the products of the like components of two vectors. It is also defined as the product of the magnitudes of the vectors normalized by the cosine of the angle between them. It is defined as:

W = \vec{F} \cdot \vec{r} =  \lVert\vec{F}\rVert\lVert\vec{r}\rVert \cos{\alpha}

Cross Product

The cross product is a measure of how perpendicular two vectors are. This operation yeilds a vector quantity orthoganal to both original vectors. The direction vector for the cross product is \hat{a}_c, and its magnitude is the product of the magnitudes and the sine of the angle between them. It is defined as:

\vec{C} = \vec{A} \times \vec{B} = \lVert\vec{A}\rVert \lVert\vec{B}\rVert \sin(\alpha) \hat{a}_c = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}

Right Hand Rule

The right hand rule is a quick way to find \hat{a}_c.

\overrightarrow{thumb} = \overrightarrow{pointer} \times \overrightarrow{middle}

Cartesian to Cylindrical

\vec{A} = 4\hat{x} + 4\hat{y} - 2\hat{z}

r = \sqrt{x^2 + y^2}

\varphi = \arctan\left(\frac{y}{x}\right)

z = z

\vec{A} = A_\rho \hat{\rho} + A_\varphi \hat{\varphi} + A_z \hat{z}

A_\rho = A_x \cos(\varphi) + A_y\sin(\varphi) = 4\sqrt{2}

A_\varphi = -A_x\sin(\varphi) + A_y\cos(\varphi) = 0

A_z = A_z = -2

1.2 Vector Calculus

The differential along some path, d\vec{l}, is defined as:

d\vec{l} = dx\hat{x} + dy\hat{y} + dz\hat{z} = d\vec{x} + d\vec{y} + d\vec{z}

The "Del" \left(\nabla\right) operator

The gradient of the scalar field \left(\nabla f\right).

\nabla = \frac{\partial}{\partial x} \hat{x} + \frac{\partial}{\partial y} \hat{y} + \frac{\partial}{\partial z} \hat{z}

df = \nabla f \cdot d\vec{l} = \left(\frac{\partial f}{\partial x}\right)dx + \left(\frac{\partial f}{\partial y}\right)dy + \left(\frac{\partial f}{\partial z}\right)dz

Directional Derivative:

The directional derivative is used to find the change of a function along some infinatesimal direction and is defined as:

\Delta \varphi = {\nabla}_l \varphi \cdot \Delta \vec{l}

\therefore {\nabla}_l \varphi = \frac{\Delta \varphi}{\delta \vec{l}} = \frac{d \varphi}{d \vec{l}} = \nabla \varphi \cdot \frac{\vec{l}}{\lVert\vec{l}\rVert}

Example 1.1

A function, f(x,y,z) = x^2 y^2 + xyz. Find \nabla f.

f(x,y,z) = f(r)

r = \sqrt{x^2 + y^2 + z^2}

\nabla f = (2xy^2)\hat{x} + (2x^2y + xz)\hat{y} + xy\hat{z}

Example 1.2

Consider a function, w = x^2y^2 + xyz. Find \nabla_l w at (2, -1, 0) in the direction, \vec{l} = 3\hat{x} + 4\hat{y} + 12\hat{z}.

\lVert\vec{l}\rVert = \sqrt{3^2 + 4^2 + 12^2} = 13

{\nabla}_l w = \nabla w \cdot \frac{\vec{l}}{\lVert\vec{l}\rVert}

Solving at (2, -1, 0):

\nabla w = 2(2)(-1)^2\hat{x} + 2(2)^2(-1)\hat{y} + 2(-1)\hat{z} = 4\hat{x} - 8\hat{y} - 2\hat{z}

{\nabla}_l w= (4\hat{x} - 8\hat{y} - 2\hat{z}) \cdot \left(\frac{3}{13}\hat{x} + \frac{4}{13}\hat{y} + \frac{12}{13}\hat{z}\right) = \frac{-44}{13}

Divergence and Curl

Divergence of a Vector Field

The divergence of a vector field is a measure of outward flux. It is defined as:

\nabla \cdot f = \left( \frac{\partial}{\partial x} + \frac{\partial}{\partial y} + \frac{\partial}{\partial z} \right) \cdot (f_x + f_y + f_z) = \frac{\partial f_x}{\partial x} + \frac{\partial f_y}{\partial y} + \frac{\partial f_z}{\partial z}

If \nabla \cdot \vec{A} = 0, there is no divergence.

Curl of a Vector Field

The curl of a vector field is a measure of circulation in each infinatesimally small region of the field. It is defined as:

\nabla \times f = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f_x & f_y & f_z \\ \end{vmatrix} = \left< \frac{\partial f_z}{\partial y} - \frac{\partial f_y}{\partial z}, \frac{\partial f_x}{\partial z} - \frac{\partial f_z}{\partial x}, \frac{\partial f_y}{\partial z} - \frac{\partial f_x}{\partial z} \right>

Solenoidal Field

A solenoidal field is a vector field without divergence, defined as:

\nabla \cdot \vec{f} = 0

Conservative / Rotational Field

A conservative field is a vector field without curl, defined as:

\nabla \times \vec{f} = 0

Example 1.3

Consider the vector field, \vec{F} = k \hat{x}, where both the direction and magnitude are uniform in all space.

\nabla \cdot \vec{F} = 0

\nabla \times \vec{F} = 0

Example 1.4

Consider the vector field, \vec{F} = k \hat{r}, where magnitude is constant and direction is away from a central point.

\hat{r} = \sqrt{\hat{x}^2 + \hat{y}^2 + \hat{z}^2}

\nabla \cdot \vec{F} = \sqrt{3} k

\nabla \times \vec{F} = 0

Example 1.5

Consider the vector field, \vec{F} = k \times \hat{r}, where magnitude is uniform and the direction is perpendicular to the distance from a central point for all space.

\nabla \cdot \vec{F} = 0

\nabla \times \vec{F} = 2k

Curl and Divergence Identities

1. The Laplacian: can operate on a scalar or vector field

\nabla \cdot (\nabla f) = \nabla^2 f

\frac{\partial^2 f}{\partial x} + \frac{\partial^2 f}{\partial y} + \frac{\partial^2 f}{\partial z}

2. The curl of a gradient is 0

\nabla \times (\nabla f) = 0

3. The gradient of divergence is a scalar

\nabla (\nabla \cdot \vec{f})

4. The divergence of curl is 0

\nabla \cdot (\nabla \times \vec{v}) = 0

5. Curl of curl

\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla \cdot \vec{A}) - \nabla^2 \vec{A}

The Line Integral

The line integral is the integral of the tangential component of a vector field along a path.

The line integral is defined as:

\int \vec{A} \cdot d\vec{l} = \int \lVert \vec{A} \rVert \cos(\alpha) \lVert d\vec{l} \rVert

Where:

• \vec{A} is some vector field

• d\vec{l} = dx\hat{x} + dy\hat{y} + dz\hat{z}

If the path of the integral is a closed curve, it is said to be the circulation of \vec{A} around \vec{l}, defined as:

\oint_c \vec{A} \cdot d\vec{l}

Example 1.6

Calculate the circulation of \vec{F} around the path.

\vec{F} = x^2 \hat{x} - xy\hat{y} - y^2\hat{z}

Circulation of \vec{F} around the path:

\oint \vec{F} \cdot d \vec{l} = \int_1 + \int_2 + \int_3 + \int_4

Path 1

Straight line from (1,0,0) to (0,0,0)

x varies, z=0, y=0.

Plug into \vec{F}:

\vec{F}_1 = x^2\hat{x} - x(0)\hat{y} - (0)^2\hat{z} = x^2\hat{x}

Since x varies at some rate, dx exists, and since y and z are constant, dy and dz are both 0.

Plug into d\vec{l}:

d\vec{l} = dx\hat{x} + (0)\hat{y} + (0)\hat{z} = dx\hat{x}

\int \vec{F}_1 \cdot d\vec{l}_1 = \int x^2\hat{x} \cdot dx\hat{x} = \int x^2dx

For this specific path:

\int_{1}^{0} x^2dx = \left. \frac{x^3}{3} \right\vert_1^0 = 0 - \frac{1}{3} = -\frac{1}{3}

Path 2

Straight line from (0,0,0) to (0,1,0)

y varies, x=0, z=0

Plug into \vec{F}:

\vec{F}_2 = (0)^2\hat{x} - (0)y\hat{y} - y^2\hat{z} = -y^2\hat{z}

Since y varies at some rate, dy exists, but since x and z are constant, dx and dz are both 0.

Plug into d\vec{l}:

d\vec{l}_2 = (0)\hat{x} + dy\hat{y} + (0)\hat{z} = dy\hat{y}

\int \vec{F}_2 \cdot d\vec{l} = \int -y^2\hat{z} \cdot dy\hat{y} = \int\limits_0^1 0 = 0

Path 3

Straight line from (0,1,0) to (1,1,1)

x and z vary at the same rate and always have the same value, y=1

Surface Integrals

\hat{n} is the unit normal vector of a surface.

The total flux crossing an area, \Delta s, is given by the function:

\Delta s \left[ \lVert \vec{F} \rVert \cos(\alpha)\right] = \vec{F} \cdot \hat{n} \Delta s

Total flux is defined as:

\sum_{i=1}^N \vec{F}_i \cos(\alpha_i) \Delta s_i = \int_s \vec{F} \cdot d\vec{s}

Gradient Theorem

The line integral through a gradient field is the difference of the values a the end points.

\int\limits_{r_1}^{r_2} \nabla \varphi \cdot d\vec{l} = \varphi(r_2) - \varphi(r_1)

\oint \nabla \varphi \cdot d\vec{l} = 0

Divergence Theorem

The divergence in some volume is the same as the flux through its surface.

\iiint_{vol} \nabla \cdot \vec{A} d\tau = \oiint \vec{A} \cdot d\vec{s}

Stokes Theorem

The curl in some region is the same as the circulation of the region's border.

\iint_A (\nabla \times \vec{A}) \cdot d\vec{s} = \oint \vec{A} \cdot d\vec{l}

1.3 Coulomb's Law

Initial observation:

\vec{F} \propto q_1 q_2

Vacuum Permittivity:

\varepsilon_0 = 8.854 \times 10^{-12}

Coulomb's Constant:

k = \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^{-9}

Coulomb's Law:

\vec{F} = k \frac{q_1 q_2}{r^2} \hat{a}_{1 2}

Superposition of Coulomb's Law:

\vec{F}_{Net} = \sum_{i=1}^{N} k_i \frac{Q q_i}{r_i^2} \hat{r}

Note: k_i depends on material properties. When in a vacuum, k_i = k.

Example 1.7

Two point charges, q_1 and q_2 are spaced 2cm apart on the x-axis. A third charge, q_3 is placed between the first two with a distance x_1 between it and q_1 and x_2 between it and q_2 such that q_3 is in static equilibrium.

Known:

\vec{F}_{1 3} + \vec{F}_{2 3} = 0

By Coulomb's Law:

\vec{F}_{1 3} = \frac{k_1 q_3 q_1}{r_{1 3}^2} \hat{z}

\vec{F}_{2 3} = -\frac{k_2 q_3 q_2}{r_{1 3}^2} \hat{z}

\therefore \vec{F}_{1 3} + \vec{F}_{2 3} = k q_3 \left( \frac{q_1}{r_{1 3}^2} - \frac{q_2}{r_{2 3}^2} \right)\hat{z} = 0

\therefore \frac{q_1}{r_{1 3}^2} - \frac{q_2}{r_{2 3}^2} = 0

Solve for r_{1 3}:

r_{1 3} = \pm r_{2 3}\sqrt{\frac{q_1}{q_2}}

Known:

r_{1 3} + r_{2 3} = 2

\therefore r_{2 3} = 2 - r_{1 3}

Plug into first equation:

r_{1 3} = \pm (2 - r_{1 3}) \sqrt{\frac{q_1}{q_2}}

Expand:

r_{1 3} = \pm \left(2 \sqrt{\frac{q_1}{q_2}} - r_{1 3}\sqrt{\frac{q_1}{q_2}} \right)

\therefore r_{1 3} \left(1 \pm \sqrt{\frac{q_1}{q_2}} \right) = \pm 2\sqrt{\frac{q_1}{q_2}}

\therefore r_{1 3} = \pm 2\sqrt{\frac{q_1}{q_2}} \left( 1 \pm \sqrt{\frac{q_1}{q_2}} \right)^{-1}

Electric Field Intensity

Call a test charge at the point of measurement Q_2.

Coulomb's Law:

\vec{F}_{1 2} = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 R^2} \hat{a}_{1 2}

The electric field intensisty:

\vec{E}_{1 2} = \frac{\vec{F}_{1 2}}{Q_1}

\vec{E} = \frac{Q_1}{4 \pi \varepsilon_0 r^2} \hat{a}_{1 2}

\vec{E}_{NET} = \sum_{i=0}^{N} \frac{k_i Q_i}{R_i^2} \hat{a}_{R_i}

Force due to electric field:

\vec{F} = q \vec{E}

Electric field

Always in the same direction as the electric field force.

Electric field of a dipole

At a point equidistant to each pole:

\lVert \vec{r}_1 \rVert = \lVert \vec{r_2} \rVert = r

By Coulomb's Law:

\vec{E}_1 = \frac{kq}{r^2} \hat{r}_1

\vec{E}_2 = \frac{kq}{r^2} \hat{r}_2

\therefore \lVert \vec{E}_1 \rVert = \lVert \vec{E}_2 \rVert

In terms of the component distances:

r^2 = a^2 + y^2

\therefore E = \frac{kq}{a^2 + y^2}

\vec{E}_{NET_y} = \vec{E}_{1_y} + \vec{E}_{2_y} = 2E\cos{\theta}

By definition:

\cos(\theta) = \frac{a}{r}

\therefore \vec{E}_{NET_y} = \frac{2kq}{a^2 + y^2} \frac{a}{\sqrt{a^2 + y^2}}\hat{y}

\vec{E}_{NET_y} = 2 \frac{kqa}{(x^2 + y^2)^{3/2}} \hat{y}

If y \gg a (far field):

\vec{E}_{NET} = 2 \frac{kqa}{y^3}

Takeaway:

\vec{E}_{monopole} \propto \frac{1}{r^2}

\vec{E}_{dipole} \propto \frac{1}{r^3}

Charge Densities

\lambda - Linear charge density

\sigma - Surface charge density

\rho - Volume charge density

Electric flux density:

\vec{D} = \varepsilon_0 \vec{E}

Where:

\vec{E} = \lim_{\Delta q \to 0} \frac{k \sum_i \Delta q_i}{\lVert \vec{r}_i \rVert ^2} \hat{r}_i = \int \frac{k}{\lVert \vec{r}_i \rVert ^2} dq

Example 1.8

A straight line segment of length L with uniform charge density \lambda extends from the origin in the \hat{x} direction. Find the strength of the electric field, \vec{E} at some arbitrary point, p along the ray from the origin in the \hat{z} direction.

The distance along the line segment in the \hat{x} direction is denoted as x, and the distance from the origin to point p is denoted as z. The vector, \vec{r} has length \sqrt{x^2 + z^2} and makes an angle \theta with the ray in the -\hat{z} direction.

The contribution d\vec{E} to the total electric field, \vec{E}, at point p is defined as:

d\vec{E} = \frac{k dq}{\lVert \vec{r} \rVert ^2} \hat{r}

For a linear charge distribution:

dq = \lambda dl

The distance, r, to each x alond the line:

\lVert \vec{r} \rVert ^2 = x^2 + z^2

The components of the $\vec{E}$-field at point p:

d\vec{E}_x = \lVert d\vec{E} \rVert \sin(\theta) \hat{x}

d\vec{E}_z = \lVert d\vec{E} \rVert \cos(\theta) \hat{z}

Gauss's Law

E-Flux Density

EMF

Measured in volts, electromotive force (EMF), is denoted by \mathcal{E}. The value for EMF is defined as:

\mathcal{E} = \oint \vec{E} \cdot d \vec{l} = -\frac{d}{dt} \int_s B_z(t) \cdot ds = -\frac{d}{dt}\psi_m

Where:

\psi_m = \int_s \vec{B} \cdot d\vec{s}

A perfectly conducting ring with radius, \rho_0, centered on the origin in the x-y plane.

The charge distribution:

\rho = \rho_0 + \rho_0 \sin(\omega t)

\vec{B}(t) = B_0 \cos(\omega t)\hat{z}

\mathcal{E} = \oint \vec{E} \cdot d\vec{l} = \iint\limits_{\phi R} B_0 \cos(\omega t) dr d\phi

Where:

\varphi: [0, 2\pi]

R: [0, \rho(t)]

Therefore:

\psi_m = (B_0 \cos(\omega t) \hat{z})(2\pi \rho(t))

\mathcal{E} = -\frac{d}{dt} B_0 2\pi \cos(\omega t)(\rho_0 + \rho_0\sin(\omega t))

Filling in some Gaps

\vec{F} = -\nabla \vec{u}

Where: \vec{u} is potential.

\vec{E} = -\nabla \vec{v}

Where: \vec{v} is electric potential.

W = \vec{F} \cdot \vec{d} = q\vec{E} \cdot \vec{d}

Work done by the $\vec{E}$-field on a charge will reduce the electric potential:

-\Delta u = u_B - u_A = -\Delta W = -qEd

The total change is:

\Delta u = -q \int\limits_A^B \vec{E} \cdot d\vec{l}

\therefore \frac{\Delta u}{q} = \int\limits_A^B \vec{E} \cdot d\vec{l} = \Delta V

W = q \int \vec{E} \cdot d\vec{l}

V(r) = \frac{kq}{r}

For N discrete charges,

V = \sum\limits_{i=1}^{N} \frac{k q_i}{r_i}

In cartesian coordinates:

\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{z}

Where:

F_x = \frac{dw}{dx}

F_y = \frac{dw}{dy}

F_z = \frac{dw}{dz}

Therefore:

\vec{F} = \frac{\partial w}{\partial x}\hat{i} + \frac{\partial w}{\partial y}\hat{j} + \frac{\partial w}{\partial z}\hat{k} = \nabla \vec{w}

q\vec{E} = \nabla (-\vec{u})

\vec{F} = -\nabla \vec{u}

\vec{E} = -\nabla \left( \frac{\vec{u}}{q} \right) = -\nabla \vec{V}

Electric Flux Density

\vec{D} = \varepsilon \vec{E}

Where \varepsilon = \varepsilon_r \varepsilon_0 (permittivity).

Magnetic Flux Density

\vec{B} = \mu\vec{H}

Where \mu = \mu_r \mu_0 (permeability).

Ampere's Law

The total current crossing an area, s, that is enclosed by the contour C:

\oint_C \frac{\vec{B}}{\mu_0} \cdot d\vec{l}

The total current is the sum of the current due to charge flow and the current due to the time rate of change of the electric flux crossing an area, s. Maxwell was able to unify electricity and magnetism by adding the current due to the time rate of change of electric flux.

\oint_C \vec{H} \cdot d\vec{l} = \int_S \vec{J} \cdot d\vec{s} + \frac{d}{dt}\int \varepsilon_0\vec{E} \cdot d\vec{s}

Simplified Ampere's Law

\oint_C \vec{H} \cdot d\vec{l} = I = \int_s \vec{J} \cdot d\vec{s}

The charge density, J = \rho v, has units \left[ \frac{A}{m^2} \right].

Example

A current, I, in an infinitely long cylindrical wire with radius, R.

\int \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

Measuring the magnetic field at some distance, r, from the center of the conductor:

B\int dl = \mu_0 I

Where \int dl is the circumfrece of measurement.

B(2\pi r) = \mu_0 I

B_{out} = \frac{\mu_0 I}{2\pi r} \hat{\varphi}

Inside the wire:

\int \vec{B} \cdot d\vec{l} = \mu_0 I

B\int dl = \frac{\mu_0 I r^2}{R^2} = 2\pi rB

B = \frac{\mu_0 I}{2\pi R^2} r

\vec{B} = \frac{\mu_0 I}{2\pi R^2}r \hat{\varphi}

Coulomb's Law

The total displacement flux of charge:

\int_s \varepsilon_0 \vec{E} \cdot d\vec{s}

The total current (charge with respect to time):

I = \frac{d}{dt} \int_s \varepsilon_0 \vec{E} \cdot d\vec{s}

Faraday's Law

Work done in moving a unit positive test charge around a closed path, C:

\oint_C \vec{E} \cdot d\vec{l}

Magnetic force on a poving charge and is directed perpendicular to both the direction of the motion of the charge and the magnetic field.

\oint_C \vec{B} \cdot d\vec{l}

Solenoid (Ideal)

For an ideal solenoid with constant current, I, assume uniform \vec{B} inside, \vec{B} = 0 outside, and infinite length.

By Ampere's Law:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

For a square loop with one side in the solenoid, and it's parallel side outside the loop:

\oint \vec{B} \cdot d\vec{l} = \int_1 + \int_2 + \int_3 + \int_4

Sides 2 and 4 are parallel, and side 3 is outside the solenoid, so:

\oint \vec{B} \cdot d\vec{l} = \int_1 = Bl

Back to Ampere's Law:

Bl = \mu_0 I N

\therefore B = \frac{\mu_0 I N}{l} = \mu_0 I n

Where:

N is the total number of windings,

l is the sidelength of the Amperian loop,

n is the number of windings per unit length \frac{N}{l}

Toroid (Ideal)

From a symmetric $\vec{B}$-field, lines form concentric circles inside the toroid. For an ideal toroid, assume \vec{B} = 0 outside, and Ampere's law inside.

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

For some circular Amperian loop inside the toroid:

\vec{B} = B\hat{\varphi}

d\vec{l} = 2\pi r \vec{\varphi}

B(2\pi r) = \mu_0 N I

\therefore B = \frac{\mu_0 N I}{2\pi r}