We will review the main topics up to Biot-Savart's Law.

Magnitude of the force between two charges:

$F=k\frac{Q_{1}Q_{2}}{r^{2}}$

where $k=8.99\times 10^{9}\mathrm{Nm^{2}/C^{2}}$

The electric field $\vec{E}$ is defined as the force $\vec{F}$ a small positive test charge with charge $q$ would experience, divided by that charge, i.e.

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

The SI units of electric field are $\mathrm{N/C}$ or $\mathrm{V/m}$.

The magnitude of the electric field due to a single point charge $Q$ at a distance $r$ from the charge is

$E=k\frac{Qq}{r^{2}}\frac{1}{q}=k\frac{Q}{r^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$

Given the field $\vec{E}$ at a point force $\vec{F}$ on an arbitrary charge $q$ using

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

The field from a point charge is a vector $\vec{E}$ which has magnitude

$E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$

and when $Q$ is positive points radially outwards from the charge.

The field from an element of charge, which also points radially outwards from the charge element has magnitude

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^{2}}$

We define an infinitesimal charge element $dQ$ which is a distance r from the point at which we want to know the field. The field due to this charge element $dE$ is

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{r^{2}}$

The electric field is

$\vec{E}=\int d\vec{E}$

Suppose we want to integrate over a distribution of charge. What does $\int dq$ really mean?

For a line of charge $\int dq = \int \lambda\,dx$

For a sheet of charge $\int dq = \int \sigma\,dA$

For volume of charge $\int dq =\int \rho\,dV$

Depending on the symmetry of the problem we might choose to integrate over Cartesian coordinates or polar coordinates, so for example,

for a rectangular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma \,dx\,dy$

for a circular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma r\,dr\,d\theta$

Often we come across integrals of spherical or cylindrical volumes (as these have easier symmetry than other volumes…)

In cases where $\rho$ is dependent either constant or only depends on r

For spherical volume the integral $\int\int\int\rho\,dV=\int\rho A dr=\int\rho 4\pi r^{2}dr$

For cylindrical volumes the integral $\int\int\int\rho\,dV=\int\rho A dr=\int\rho 2\pi r l dr$

When we have a volume with planar symmetry we instead can divide our volume in to slices of area A, so for a generic prism with area $A$ in which the charge density is uniform in the directions along $A$,

$\int\int\int\rho\,dV=\int\rho A dl$

If we consider a uniform electric field $\vec{E}$ passing through a flat area $A$ we can define the electric flux as $\Phi_{E}=EA\cos\theta$

We represent the area by a vector with magnitude equal to the area and direction normal to the surface.

Now we can go from

$\Phi_{E}=EA\cos\theta$

to

$\Phi_{E}=\vec{E}.\vec{A}$

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Gauss's law will often give us a better way to evaluate the field for a distribution of charge.

The most important step is to choose an appropriate Gaussian surface which reflects the symmetry of the charge.

We should remember that the Gaussian surface must be a closed surface.

If possible we want to choose the surfaces so that the field on each surface is uniform.

It's important to remember that flux has a sign. It is positive when it is leaving the volume, and negative when entering (as the $\vec{A}$ always points out).

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

If the field points out radially from the center of the sphere

$\oint\vec{E}.d\vec{A}=\oint E\,dA$

and if it is uniform over the surface

$\oint E\,dA=E \oint dA=E4\pi r^{2}$

As for the enclosed charge

$Q=\int\int\int\rho\,dV=\int\rho A dr=\int\rho 4\pi r^{2}dr$

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

If we consider the line of charge to be infinitely long then the field points straight out from the line, so the flux through the end circular surfaces is zero and for the curved surface

$\oint\vec{E}.d\vec{A}=\oint E\,dA$

as the field is uniform over the surface

$\oint E\,dA=E \oint dA=E2\pi rl$

To find the enclosed charge

$Q=\int\int\int\rho\,dV=\int\rho A dr=\int\rho 2\pi r l dr$

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

If the plane is infinite then the flux through the curved surfaces is zero and we only need to be concerned about the flux through the flat prism surfaces of area $A$. The flux through those two faces will not always be the same, so in general if we define a field through the left face $E_{L}$ and the right face $E_{R}$

$\Phi_{E}=\oint\vec{E}.d\vec{A}=E_{L}A+E_{R}A$

$Q=\int\int\int\rho\,dV=\int\rho A dl$

or when there is sheet of charge $\sigma$

$Q=\int \sigma\,dA=\sigma A$

The electric potential and field are easily related.

$U_{b}-U_{a} = -\int_{a}^{b}\vec{F}\cdot\,d\vec{l} $

$V_{b}-V_{a}=\frac{U_{b}-U_{a}}{q}=-\int_{a}^{b}\vec{E}\cdot\,d\vec{l} $

When the field is not uniform we can obtain the potential from the field by integration. Note again that the path of the line integral has not been specified: we are free to choose the most convenient path to facilitate the calculation.

We also need to be able to calculate the potential due to a charge distribution

$V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}$

In general, use the first formula when Gauss's law easily gives the field, and the second formula where the distance $r$ is easily written in terms of the spatial variables that define $dq$ and the resulting integral is not too hard to do.

We can get the field from the potential from

$\vec{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}$

If we want to find the change in potential in going from a distance $r_{a}$ to a distance $r_{b}$ from a single point charge we need to evaluate the integral

$V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$

We can recall that the field is

$E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}$

If we integrate along a path which is directly radially outward $\vec{E}$ and $d\vec{l}$ are in the same direction and

$-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=-\frac{Q}{4\pi\varepsilon_{0}}\int_{r_{a}}^{r_{b}}\frac{1}{r^{2}}dr=\frac{Q}{4\pi\varepsilon_{0}}[\frac{1}{r}]_{r_{a}}^{r_{b}}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$

It is common to define the potential as being zero at $\infty$, and if we do this here (setting $r_{a}=\infty$) the potential as a function of the distance $r$ from a point charge is

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$

It can be easier to find the potential due to multiple charges or a distribution of charges than it is to find the field.

We denote the negative charge as $A$ and the positive charge as $B$ and we can then say that the potential is given by

$V=\frac{1}{4\pi\varepsilon_{0}}(-\frac{Q}{R_{A}}+\frac{Q}{R_{B}})$

From the previous example we can conclude that for $n$ point charges each which produces a potential $V_{i}$ at point A the potential at point $A$ will be

$V_{A}=\sum\limits_{i=1}^{n}V_{i}=\frac{1}{4\pi\varepsilon_{0}}\sum\limits_{i=1}^{n}\frac{Q_{i}}{r_{ia}}$

Now if we want to instead consider a continuous charge distribution

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}$

$\Delta U = -\int_c\vec{F}\cdot\vec{dl}\Longrightarrow \vec{F} = -\vec{\nabla}U$.

When we want to find the gradient vector field we use an operator $\vec{\nabla}$ which is read as del

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

When applied to a scalar field we can use the word grad, so we would say “$E$ is equal to minus *grad* $V$”

$\vec{E}=-\nabla V $

$\vec{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}$.

If one moves along an equipotential line, **$dV$ is zero**. This means that the $\vec{\nabla}V$ is perpendicular to that line, which shows why **electric field lines are perpendicular to equipotentials**.

This consists of two metal plates with area $A$ carrying equal and opposite charges $Q$. The field due to one of these sheets of charges can be found from Gauss's law

$E=\frac{Q}{\varepsilon_{0}A}$

The potential due to this sheet of charge is

$V=-\frac{Qd}{\varepsilon_{0}A}$

Here we have defined the zero of potential as the position of the positive sheet of charge.

For capacitors it makes most sense to define the voltage as the potential difference across the capacitor and give it a positive sign, ie.

$V=\frac{Qd}{\varepsilon_{0}A}$

Charges will not spontaneously separate across a capacitor, we need to apply a potential to the capacitor to move charges from one plate to another. The amount of charge $Q$ separated when a given voltage $V$ is applied is obtained from the equation above

$Q=\frac{\epsilon_{0}A}{d}V$

The capacitance $C$ is the amount of charge separated per volt applied

$C=\frac{Q}{V}$

and we can see for a parallel plate capacitor

$C=\frac{\epsilon_{0}A}{d}$

The unit for capacitance is the farad, $\mathrm{F}$ (named after Michael Faraday) which is equivalent to $\mathrm{\frac{C}{V}}$

$E_{dielectric}=\frac{Q}{\varepsilon_{0} A}-\frac{P}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}-\chi E_{dielectric}$

$KE_{dielectric}=\frac{Q}{\varepsilon_{0} A}$

$E_{dielectric}=\frac{V}{d}$

$K\frac{V}{d}=\frac{Q}{\varepsilon_{0} A}$

$C=\frac{Q}{V}=K\frac{\varepsilon_{0}A}{d}$

In general, the capacitance of a configuration with a dielectric is related to the same geometry with a vacuum gap:

$C_{dielectric} = KC_{vacuum}$.

Circuit diagrams are a useful way to represent an electrical circuit. We introduce two symbols for capacitors and batteries so that we can draw circuits with capacitors in series and in parallel.

When capacitors are connected in parallel the potential across each capacitor is the same. Each acquires a charge determined by it's capacitance and the total charge is the sum of these charges

$Q=Q_{1}+Q_{2}+Q_{3}=C_{1}V+C_{2}V+C_{3}V=(C_{1}+C_{2}+C_{3})V$

$Q = C_{eq}V$

We can then consider an equivalent capacitance which is the sum of the capacitors

$C_{eq}=C_{1}+C_{2}+C_{3}$

When capacitors are connected in series it is the charge on each capacitor which is the same, as the region between the capacitors, which is considered to be an ideal conductor must remain overall electrically neutral and cannot have a field.

$Q=C_{eq}V$

The total voltage must equal the voltage across the battery

$V=V_{1}+V_{2}+V_{3}$

For each capacitor $Q=C_{i}V_{i}$

$\frac{Q}{C_{eq}}=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}}$

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$

If an amount of charge $\Delta Q$ flows in a time interval $\Delta t$ the average current in this time is defined as

$\bar{I}=\frac{\Delta Q}{\Delta t}$

The instantaneous current is the limit as $\Delta t \to 0$

$I=\frac{dQ}{dt}$

The unit for current is the Ampere $\mathrm{A}$ which is equivalent to $\mathrm{\frac{C}{s}}$, and is named after Andre Ampere. This is usually abbreviated as an amp.

Even though in metallic wires it is negative electrons that flow the convention for current is for a flow of positive charges, so current flows from high to low potentials, even though when electrons carry the current they are actually moving from low to high potential.

Ohm's law state that

$V=IR$

It should be noted that Ohm's law is not valid for all materials, and typically the resistance $R$ depends on factors such as temperature. For example, as a filament light bulb heats up it's resistance will increase.

When an infinitesimal amount of charge $dq$ moves through a potential $V$ the change in the potential energy of the charge is

$dU=Vdq$

The work done on the charge is $dW=-dU=-Vdq$, the work done **by** the charge is equal to $dU$.

We can recall that power is the rate at which work is done, and so

$P=\frac{dU}{dt}=\frac{dq}{dt}V=IV$

The above is a general relationship, for an ohmic conductor **only**, where $V=IR$

$P=VI=I^{2}R=\frac{V^{2}}{R}$

We should be careful here, the power dissipated in a given component needs to be calculated using only the potential change from one end to other!

We have now discussed the properties of batteries, capacitors and resistors.

We will now look at how the behave when connected together in circuits.

For ease of communication, there are a standard set of symbols used for circuit diagrams.

If we recall our formula for resistance

$R=\rho\frac{l}{A}$

we can expect that if we connect two resistors in series the total resistance should be the sum of the resistances.

The current $I$ which flows through the circuit is the same everywhere.

The potential in each resistor is given by Ohm's law $V=IR$.

The total potential is the sum of the potentials across the resistors.

$V=V_{1}+V_{2}+V_{3}=IR_{1}+IR_{2}+IR_{3}$

We can define an equivalent resistance for the 3 resistors

$R_{eq}=R_{1}+R_{2}+R_{3}$

When resistors are connected in parallel the current splits itself up between the various branches

$I=I_{1}+I_{2}+I_{3}$

As the potential across each resistor is the same $V$

$I_{1}=\frac{V}{R_{1}}$ $I_{2}=\frac{V}{R_{2}}$ $I_{3}=\frac{V}{R_{3}}$

The total current is

$I=\frac{V}{R_{eq}}$

where $R_{eq}$ is the equivalent resistance of the 3 resistors.

As $\frac{V}{R_{eq}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}$

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Kirchoff's **junction rule** comes from conservation of charge and says that for any junction in a circuit the sum of all currents entering the junction must equal the sum of all currents leaving the junction.

Kirchoff's **loop rule** comes from conservation of energy and states that the changes in potential around any closed loop of a circuit must be zero.

The way to use these rules is to use them set up a sufficient number of equations based on the junctions and loops in a circuit to solve for the required unknowns.

$I_{1}+I_{2}=I_{3}$

$\mathcal{E}_{1}-I_{3}R_{3}-I_{1}R_{1}=0$

$\mathcal{E}_{2}-I_{3}R_{3}-I_{2}R_{2}=0$

From Kirchoff's loop rule

$\mathcal{E}-I(t)R-\frac{Q(t)}{C}=0$

Writing this in terms of charge

$\mathcal{E}-\frac{dQ}{dt}R-\frac{1}{C}Q=0$

and then

$\mathcal{E}C-Q=\frac{dQ}{dt}RC$

$\frac{dQ}{\mathcal{E}C-Q}=\frac{dt}{RC}$

At time $t=0$, $Q=0$ and at time $t=t$, $Q=Q$

$\int_{0}^{Q}\frac{dQ}{\mathcal{E}C-Q}=\int_{0}^{t}\frac{dt}{RC}$

$[-\ln(\mathcal{E}C-Q)]_{0}^{Q}=[\frac{t}{RC}]_{0}^{t}$ ⇒ $\ln(\frac{\mathcal{E}C}{\mathcal{E}C-Q})=\frac{t}{RC}$ ⇒ $\ln(\frac{\mathcal{E}C-Q}{\mathcal{E}C})=-\frac{t}{RC}$ ⇒ $\frac{\mathcal{E}C-Q}{\mathcal{E}C}=e^{-\frac{t}{RC}}$

$Q=\mathcal{E}C(1-e^{-\frac{t}{RC}})$

The voltage on the capacitor is given by $V_{C}=\frac{Q}{C}$

$V_{C}=\mathcal{E}(1-e^{-\frac{t}{RC}})$

$I=\frac{dQ}{dt}=\frac{\mathcal{E}}{R}e^{-\frac{t}{RC}}$

If we consider wire of which length $l$ lies within a magnetic field we find that the force depends on $l$ as well as the current $I$.

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

or in the diagram below

$F=IlB\sin\theta$

We can also chop the length up in to infinitesimal pieces which produce infinitesimal forces to accommodate a wire that changes it's direction with respect to a magnetic field, or a non-uniform magnetic field.

$d\vec{F}=I\,d\vec{l}\times\vec{B}$

For a single charge $q$

$\vec{F}=q\vec{v}\times\vec{B}$

As before we can use a right hand rule to determine the direction of the force, we simply replace the current with the direction of the velocity of the charge.

The Lorentz equation combines the electric force and the magnetic force on an charged particle

$F=q(\vec{E}+\vec{v}\times\vec{B})$

The centripetal force is provided by the magnetic field

$\frac{mv^{2}}{r}=qvB$

The radius of the circle is then

$r=\frac{mv}{qB}$

The time it takes for an electron to go round the circle is

$T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}$

and the frequency, which we call the cyclotron frequency, is

$f=\frac{1}{T}=\frac{qB}{2\pi m}$

When the loop is aligned with the field the net torque experienced will be

$\tau=IAB$

Where $A$ is the area of the loop. For $N$ loops the formula just becomes

$\tau=NIAB$

However we can see that when the loop makes an angle $\theta$ with the field

$\tau=NIAB\sin\theta$

A good way to represent the orientation dependence of the torque is to define a new vector quantity, the magnetic dipole moment

$\vec{\mu}=NI\vec{A}$

The direction of the vector can be determined by the right hand rule and we may now write the torque as

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

Ampere's law relates the magnetic field that runs around the edge of a closed surface to the current that passes through that surface and is

$\oint\vec{B}\cdot\,d\vec{l}=\mu_{0}I_{encl}$

$\mu_{0}=4\pi\times10^{-7}\mathrm{Tm/A}$ is the permeability of free space.

$d\vec{l}$ is an infinitesimally small length element which points in the direction of a the path around the edge.While any path/surface can be chosen, smart use of Ampere's law relies on using symmetry to make the integral easy.

For a current carrying wire the right hand rule tells us to choose circular loop centered on the wire, in which case

$\oint\vec{B}\cdot d\vec{l}=\oint B\,dl$

because $\vec{B}$ is always parallel to $d\vec{l}$, and as the path is circular

$\oint B\,dl=B\oint \,dl=B(2\pi r)$

So we find that

$B(2\pi r)=\mu_{0}I\to B=\frac{\mu_{0}I}{2\pi r}$

Within a uniform wire carrying a current $I$ the magnetic field as a function of $r$ can be found from

$\oint\vec{B}\cdot d\vec{l}=B(2\pi r)=\mu_{0}I_{encl}$

Inside the wire $I_{encl}=I\frac{\pi r^{2}}{\pi R^{2}}$ so

$B(2\pi r)=I\frac{\pi r^{2}}{\pi R^{2}}\to B=\frac{\mu_{0}Ir}{2\pi R^{2}}$

Outside the wire $I_{encl}=I$ so

$B=\frac{\mu_{0}I}{2\pi r}$

In tightly wound solenoid we can deduce that the field strength inside is much greater than outside, and that also components of the magnetic field perpendicular to the solenoid axis cancel out. .

We can see that the integral $\oint\vec{B}\cdot d\vec{l}$ reduces to $\int_{c}^{d} B dl$ which is simply $Bl$. If the rectangle includes $N$ wires then

$Bl=\mu_{0}NI$

This is more usefully expressed in terms of the number density of wires in a unit length, $n$

$B=\mu_{0}\frac{N}{l}I=\mu_{0}nI$

The Biot-Savart law provides the basis for obtaining the field (in principle) due to an arbitrary current distribution.

The field at some point in space due to an infinitesimal length $d\vec{l}$ through which a current $I$ flows is

$$d\vec{B}=\frac{\mu_{0}I}{4\pi}\frac{d\vec{l}\times\hat{r}}{r^{2}}$$

$\vec{r}$ is the displacement vector from the element $d\vec{l}$ to to the point and $\hat{r}$ is the unit vector in the direction of $\vec{r}$.

In the above diagram the magnitude of $d\vec{B}$ is

$$dB=\frac{\mu_{0}Idl\sin\theta}{4\pi r^{2}}$$

The total magnetic field is found by integrating over all the elements

$$\vec{B}=\int d\vec{B}=\frac{\mu_{0}I}{4\pi}\int\frac{d\vec{l}\times\hat{r}}{r^{2}}$$

We have to pay attention the directions of the $d\vec{B}$ vectors when we evaluate this integral!

$$\vec{B}=\int d\vec{B}=\frac{\mu_{0}I}{4\pi}\int\frac{d\vec{l}\times\hat{r}}{r^{2}}=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{dy\sin\theta}{r^{2}}$$

since $dy=dl$.

We need to write $r$ in terms of $y$, $r^{2}=R^{2}+y^{2}$ but we also need to realize that $y$ and $\theta$ are related to each other and $\sin\theta=\frac{R}{r}$.

We need to write $y$ in terms of $\theta$ i.e. $y = -R/\tan\theta$ and therefore:

$$dy = +\frac{Rd\theta}{\sin^2\theta} = \frac{Rd\theta}{(R/r)^2} = \frac{r^2d\theta}{R}$$

From the figure, we see that $y=-\infty$ corresponds to $\theta=0$ and $y=+\infty$ corresponds to $\theta=\pi$. Therefore:

$$|\vec{B}| = \frac{\mu_0I}{4\pi R}\int_{\theta=0}^{\pi}\sin\theta\cdot d\theta = \frac{\mu_0I}{2\pi R} $$

Taking in to account the Biot-Savart Law

$$d\vec{B}=\frac{\mu_{0}I}{4\pi}\frac{d\vec{l}\times\hat{r}}{r^{2}}$$

and/or our previous result for an infinite wire

$$B=\frac{\mu_{0}I}{4\pi R}\left[-\cos\theta\right]_{0}^{\pi} =\frac{\mu_{0}I}{4\pi}\left[\frac{1}{R}\frac{y}{(R^{2}+y^{2})^{1/2}}\right]_{-\infty}^{+\infty}=\frac{\mu_{0}I}{4\pi}\frac{2}{R}=\frac{\mu_{0}I}{2\pi R}$$

we can deduce that for any finite straight length of wire starting at $y_{1}$ and ending at $y_{2}$, where the point at which the field is measured is considered to be the origin,

$$B=\frac{\mu_{0}I}{4\pi R}\left[-\cos\theta\right]_{\theta_1}^{\theta_2}=\frac{\mu_{0}I}{4\pi R}\left[\frac{y}{(R^{2}+y^{2})^{1/2}}\right]_{y_{1}}^{y_{2}}$$

The Biot-Savart law $d\vec{B}=\frac{\mu_{0}I}{4\pi}\frac{d\vec{l}\times\hat{r}}{r^{2}}$ tells us that current parallel to $\hat{r}$, or in other words current flowing directly towards or away from a point does not produce a magnetic field.

So in the following example we only need to consider the current due the curved part of the wire.

Everywhere along the wire $d\vec{l}$ is perpendicular to $\hat{r}$ and the contribution from each length element points in the same direction, so $dB=\frac{\mu_{0}I\,dl}{4\pi R^{2}}$.

$B=\int\,dB=\frac{\mu_{0}I}{4\pi R^{2}}\int\,dl=\frac{\mu_{0}I}{4\pi R^{2}}(\frac{1}{4}2\pi R)=\frac{\mu_{0}I}{8R}$

In the case of a current loop the Biot-Savart Law, $d\vec{B}=\frac{\mu_{0}I}{4\pi}\frac{d\vec{l}\times\hat{r}}{r^{2}}$, tells us that magnetic field $d\vec{B}$ for a point on the $x$ axis due to a length element $d\vec{l}$ has magnitude

$dB=\frac{\mu_{0}I\,dl}{4\pi r^{2}}$.

Here we should note that $d\vec{l}$ is always perpendicular to $r$ and the direction of the field that is produced may be determined by the right hand rule.

We can note that the symmetry of the situation will lead to the magnetic field perpendicular to the axis cancelling out when we integrate over all the contributions from the loop, so we can write that

$B=B_{||}=\int\,dB_{||}=\int\,dB\cos\phi=\int\,dB\frac{R}{r}=\int\,dB\frac{R}{(R^{2}+x^{2})^{1/2}}$

$=\int\,\frac{\mu_{0}I\,dl}{4\pi r^{2}}\frac{R}{(R^{2}+x^{2})^{1/2}}=\frac{\mu_{0}I\,dl}{4\pi}\frac{R}{(R^{2}+x^{2})^{3/2}}\int\,dl$

$=\frac{\mu_{0}I\,dl}{4\pi}\frac{R}{(R^{2}+x^{2})^{3/2}}2\pi=\frac{\mu_{0}IR^{2}}{2(R^{2}+x^{2})^{3/2}}$

$\Phi_{B}=\int\vec{B}\cdot d\vec{A}$

or when the magnetic field is uniform

$\Phi_{B}=\vec{B}\cdot\vec{A}$

In the example above the magnetic flux is

$\Phi_{B}=\vec{B}\cdot\vec{A}=BA\cos\theta$

where $A=l^{2}$.

The unit of magnetic flux, is called a weber $\mathrm{Wb}$, where $1\mathrm{Wb}=1\mathrm{Tm^{2}}$

Faraday's law of induction states that the induced emf in a circuit is equal to rate of change of magnetic flux through the circuit.

$\mathcal{E}=-\frac{d\Phi_{B}}{dt}$

If the circuit is made of a number of loops $N$

$\mathcal{E}=-N\frac{d\Phi_{B}}{dt}$

The negative sign in Faraday's law represents Lenz's law which states that

“An induced emf is always in such a direction as to oppose the change in flux causing it”

A conductor moving in a magnetic field will experience a induced emf.

The emf produced is given by the change of flux

$\mathcal{E}=-\frac{d\Phi_{B}}{dt}=B\frac{dA}{dt}=-\frac{Blv\,dt}{dt}=-Blv$

But what about the case where the rails are not there?

In this case the electrons still feel the force and will collect at one end of rod, so there will be a potential difference across it.

It is important to remember that an emf is not a force, but rather a measure of the work done in a circuit, so we can write the emf in terms of an integral over a closed path of the electric field

$\mathcal{E}=\oint\vec{E}\cdot d\vec{l}$

and then

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$

Here we are taking an integral around the path that encloses the area in which magnetic flux is changing.

We should not that the implication of this is that in the presence of a time varying magnetic field the electric force is no longer a conservative force.

Consider the change in flux on a loop as it is rotated by some external torque:

$\mathcal{E}=\frac{d\Phi_{B}}{dt}=-\frac{d}{dt}\int\vec{B}\cdot d\vec{A}=-\frac{d}{dt}BA\cos\theta$

If the loop rotates with a constant angular velocity $\omega=\frac{d\theta}{dt}$ then $\theta=\theta_{0}=\omega t$ and we can say that

$\mathcal{E}=-BA\frac{d}{dt}(\cos\omega t)=BA\omega\sin\omega t$

of course if there are $N$ loops

$\mathcal{E}=-NBA\frac{d}{dt}(\cos\omega t)=NBA\omega\sin\omega t=\mathcal{E}_{0}\sin\omega t$

In a transformer two coils are coupled by an iron core so that the flux through the two coils is the same.

When an AC voltage is applied to the primary coil the magnetic flux passing through it is related to the applied field by

$V_{P}=N_{P}\frac{d\Phi_{B}}{dt}$

if we assume the coil has no resistance. The voltage induced in the secondary coil will have magnitude

$V_{S}=N_{S}\frac{d\Phi_{B}}{dt}$

We can thus see that

$\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}}$

If we assume there is no power loss (which is fairly accurate) then $I_{P}V_{P}=I_{S}V_{S}$ and

$\frac{I_{S}}{I_{P}}=\frac{N_{P}}{N_{S}}$

In general for two coils the relationship between the flux in one coil due to the current in another is described by a parameter called the mutual inductance.

$\Phi_{21}$ is the magnetic flux in each loop of coil 2 created by the current in coil 1. The total flux in the second coil is then $N_{2}\Phi_{21}$ and is related to the current in coil 1, $I_{1}$ by

$N_{2}\Phi_{21}=M_{21}I_{1}$

As, from Faraday's Law, the emf induced in coil 2 is $\mathcal{E}_{2}=-N_{2}\frac{d\Phi_{21}}{dt}$ so

$\mathcal{E}_{2}=-M_{21}\frac{dI_{1}}{dt}$

The mutual inductance of coil 2 with respect to coil 1, $M_{21}$ does not depend on $I_{1}$, but it does depend on factors such as the size, shape and number of turns in each coil, their position relative to each other and whether there is some ferromagnetic material in the vicinity.

In the reverse situation where a current flows in coil 2

$\mathcal{E}_{1}=-M_{12}\frac{dI_{2}}{dt}$

but in fact $M_{12}=M_{21}=M$

The mutual inductance is measured in Henrys ($\mathrm{H}$), $1\mathrm{H}=1\mathrm{\frac{Vs}{A}}=1\mathrm{\Omega s}$

The magnetic flux $\Phi_{B}$ passing through the coil is proportional to the current, and as we did for mutual inductance we can define a constant of proportionality between the current and the flux, the self-inductance $L$

$N\Phi_{B}=LI$

The emf $\mathcal{E}=-N\frac{d\Phi_{B}}{dt}=-L\frac{dI}{dt}$

The self-inductance is also measured in henrys.

A component in a circuit that has significant inductance is shown by the symbol.

We can can calculate the self-inductance of a solenoid from it's field

$B=\mu_{0}\frac{NI}{l}$

The flux in the solenoid is

$\Phi_{B}=BA=\mu_{0}\frac{N_{1}IA}{l}$

so

$L=\frac{N\Phi_{B}}{I}=\frac{\mu_{0}N^{2}A}{l}$

When we take a resistor an inductor in series and connect it to a battery then Kirchoff's loop rule tells us that

$V_{0}-IR-L\frac{dI}{dt}=0$

which we can rearrange and integrate

$\int_{I=0}^{I}\frac{dI}{V_{0}-IR}=\int_{0}^{t}\frac{dt}{L}$

$-\frac{1}{R}\ln(\frac{V_{0}-IR}{V_{0}})=\frac{t}{L}$

$I=\frac{V_{0}}{R}(1-e^{-t/\tau})=I_{0}(1-e^{-t/\tau})$ where $\tau=\frac{L}{R}$

If we then switch back to the closed loop that does not include the battery then Kirchoff's loop rule gives us

$L\frac{dI}{dt}+RI=0$

$\int_{I_{0}}^{I}\frac{dI}{I}=-\int_{0}^{t}\frac{R}{L}dt$

$\ln\frac{I}{I_{0}}=-\frac{R}{L}t$

$I=I_{0}e^{-t/\tau}$

The current changes exponentially according to $e^{-t/\tau}$. For an RC circuit the time constant is $\tau=RC$, for an LR circuit the time constant is $\tau=\frac{L}{R}$.

For a charged capacitor in series with an inductor, according to Kirchoff's loop rule:

$-L\frac{dI}{dt}+\frac{Q}{C}=0$

The current comes from the capacitor so $I=-\frac{dQ}{dt}$ so

$\frac{d^{2}Q}{dt^{2}}+\frac{Q}{LC}=0$

This is like a simple harmonic oscillator, but with charge in place of displacement! A solution to this differential equation is

$Q=Q_{0}\cos(\omega t+\phi)$

To find $\omega$ we substitute the solution in to the equation

$-\omega^{2}Q_{0}\cos(\omega t + \phi)+\frac{Q_{0}}{LC}\cos(\omega t+\phi)=0$

$(-\omega^{2}+\frac{1}{LC})\cos(\omega t+ \phi)=0$

Which requires $\omega=\sqrt{\frac{1}{LC}}$

As we saw previously an actual inductor has some resistance, and thus a real LC circuit is better represented by an LRC circuit in which we represent the resistance of the inductor, or other resistances in the circuit by an in series resistor $R$

Now Kirchoff's loop rule when the circuit is closed is

$-L\frac{dI}{dt}-IR+\frac{Q}{C}=0$

Which can be written in terms of the charge as before

$L\frac{d^{2}Q}{dt^{2}}+R\frac{dQ}{dt}+\frac{Q}{C}=0$

and we can compare this to the differential equation for a damped harmonic oscillator.

When $R^{2}<\frac{4L}{C}$ the system is underdamped and the solution to this equation is

$Q=Q_{0}e^{-\frac{R}{2L}t}\cos(\omega' t+\phi)$

where $\omega'=\sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}$

When $R^{2}>\frac{4L}{C}$ the system is overdamped and the charge will decay slowly.

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We can begin our analysis of this circuit by applying Kirchoff's loop rule to the find potential at any given time.

$V=V_{R}+V_{L}+V_{C}$

However an important consequence of the voltages not being at the same phase in the different components is that the peak voltage is not experienced at the same time in each component, and so the peak source voltage $V_{0}$

$V_{0}\neq V_{R0}+V_{L0}+V_{C0}$

The condition of continuity of current demands that the current throughout the circuit should always be in phase, so at any point in this circuit the current will

$I=I_{0}\cos\omega t$

To understand the Phasor approach to AC circuits we can follow a very nice set of animations from Physclips.

A phasor is a way of representing the voltage across a component taking into account the phase difference between the voltage and the current.

As we saw in our last lecture a resistor has a voltage in phase with the current flowing through it. So if we now represent the current as vector moving in a plane we can also represent the voltage across the resistor as a vector of magnitude $V_{R}=I{R}$ which points in the same direction as the current flowing through it.

We saw in our last lecture that in a capacitor the current leads the voltage by 90^{o}.

Also we saw that the reactance $X_{C}=\frac{1}{\omega C}=\frac{1}{2\pi f C}$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{C}$ also should.

We saw in our last lecture that in an inductor the current lags the voltage by 90^{o}.

Also we saw that the reactance $X_{L}=\omega L=2\pi f L$ depends on frequency and so the size of the voltage phasor $V_{0}=I_{0}X_{L}$ also should.

To find the total voltage in the circuit at time $t$ we add the phasors for the different components together as we would vectors. We can see that for a current $I=I_{0}\cos\omega t$ flowing through through the circuit the voltage will be offset by a phase $\phi$

$V=V_{0}\cos(\omega t + \phi)$

The peak voltage $V_{0}$ is linked to the peak current $I_{0}$ through the impedance $Z$

$V_{0}=I_{0}Z$ and we can also say $V_{rms}=I_{rms}Z$

The value of $Z$ is found by considering the vector sum of the voltages

$V_{0}=\sqrt{V_{R0}^{2}+(V_{L0}-V_{C0})^{2}}=I_{0}\sqrt{R^{2}+(X_{L}-X_{C})^2}$

so

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$

The phase difference between the current and voltage $\phi$ is obtained from

$\tan \phi=\frac{V_{L0}-V_{C0}}{V_{R0}}=\frac{I_{0}(X_{L}-X_{C})}{I_{0}R}=\frac{X_{L}-X_{C}}{R}$

To find the total voltage in the circuit at time $t$ we add the phasors for the different components together as we would vectors. We can see that for a current $I=I_{0}\cos\omega t$ flowing through through the circuit the voltage will be offset by a phase $\phi$

$V=V_{0}\cos(\omega t + \phi)$

The peak voltage $V_{0}$ is linked to the peak current $I_{0}$ through the impedance $Z$

$V_{0}=I_{0}Z$ and we can also say $V_{rms}=I_{rms}Z$

The value of $Z$ is found by considering the vector sum of the voltages

$V_{0}=\sqrt{V_{R0}^{2}+(V_{L0}-V_{C0})^{2}}=I_{0}\sqrt{R^{2}+(X_{L}-X_{C})^2}$

so

$Z=\sqrt{R^{2}+(X_{L}-X_{C})^{2}}=\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}$

The phase difference between the current and voltage $\phi$ is obtained from

$\tan \phi=\frac{V_{L0}-V_{C0}}{V_{R0}}=\frac{I_{0}(X_{L}-X_{C})}{I_{0}R}=\frac{X_{L}-X_{C}}{R}$

Power in an LRC circuit is only dissipated in the resistor, and so the average power dissipated is given by

$\bar{P}=I_{RMS}^{2}R$

but we may want to express this in terms of the impedance of the circuit or the $V_{RMS}$ which is applied.

To do this we write

$\cos \phi =\frac{V_{R0}}{V_{0}}=\frac{I_{0}R}{I_{0}Z}=\frac{R}{Z}$

which means that $R=Z\cos\phi$ and

$\bar{P}=I_{RMS}^{2}Z\cos\phi$

or, as $V_{RMS}=I_{RMS}Z$

$\bar{P}=I_{RMS}V_{RMS}\cos\phi$

The RMS current in the circuit we are considering is given by

$I_{RMS}=\frac{V_{RMS}}{Z}=\frac{V_{RMS}}{\sqrt{R^{2}+(\omega L-\frac{1}{\omega C})^{2}}}$

We can see that the current should be frequency dependent and have a maximum when

$(\omega L - \frac{1}{\omega C})=0$

which gives the resonant frequency

$\omega_{0}=\sqrt{\frac{1}{LC}}$

One can get a feel for the circuit response here and here.

Here is a demo that shows how the resonant frequency is varied by the inductor and capacitor values.

Maxwell's equations, named after James Clerk Maxwell who first expressed them together are a set of four equations from which all electromagnetic theory can be derived.

The integral form of Maxwell's equations in free space (ie., in the absence of dielectric or magnetic materials) are

$\oint\vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_{0}}$ (Gauss's Law)

$\oint \vec{B}\cdot d\vec{A}=0$ (Magnetic equivalent of Gauss's Law)

$\oint\vec{E}\cdot d\vec{l}=-\frac{d\Phi_{B}}{dt}$ (Faraday's Law)

$\oint\vec{B}\cdot d\vec{l}=\mu_{0}I+\mu_{0}\varepsilon_{0}\frac{d\Phi_{E}}{dt}$ (Modified form of Ampere's Law).

From Maxwell's equations

$\frac{\partial^{2} E_{y}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} E_{y}}{\partial x^{2}}$ and $\frac{\partial^{2} B_{z}}{\partial t^{2}}=\frac{1}{\mu_{0}\varepsilon_{0}}\frac{\partial^{2} B_{z}}{\partial x^{2}}$

tell that electromagnetic waves can be described by an electric and magnetic field in phase with one another, but at right angles to one another

$E_{y}(x,t)=E_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

$B_{z}(x,t)=B_{0}\sin\frac{2\pi}{\lambda}(x-\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}t)$

with velocity

$v=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}=3.00\times10^{8}\,\mathrm{m/s}=c$

The energy stored in an electric field is

$u_{E}=\frac{1}{2}\varepsilon_{0}E^{2}$

and in a magnetic field

$u_{B}=\frac{1}{2}\frac{B^{2}}{\mu_{0}}$

The total energy of an EM wave is

$u=u_{E}+u_{B}=\frac{1}{2}\varepsilon_{0}E^{2}+\frac{1}{2}\frac{B^{2}}{\mu_{0}}$

and using

$\frac{E}{B}=c=\sqrt{\frac{1}{\mu_{0}\varepsilon_{0}}}$

$u=\varepsilon_{0}E^{2}=\frac{B^{2}}{\mu_{0}}=\sqrt{\frac{\varepsilon_{0}}{\mu_{0}}}EB$

In an electromagnetic wave the fields are moving with velocity $c$ the amount of energy passing through a unit area at any given time is

$S=\varepsilon_{0}cE^{2}=\frac{cB^{2}}{\mu_{0}}=\sqrt{\frac{EB}{\mu_{0}}}$

More generally, the Poynting vector is a vector $\vec{S}$ which represents the flux of energy in an electromagnetic field

$\vec{S}=\frac{1}{\mu_{0}}\vec{E}\times\vec{B}$

We define an angle of incidence $\theta_{i}$ relative to the surface normal and find that the angle of reflection $\theta_{r}$, also defined relative to the surface normal is given by

$\theta_{r}=\theta_{i}$

$\frac{1}{d_{o}}+\frac{1}{d_{i}}=\frac{1}{f}$

The magnification of a mirror $m$

$m=\frac{h_{i}}{h_{o}}=-\frac{d_{i}}{d_{o}}$

**Sign Conventions:**

- The image height $h_i$ is positive if the image is upright, and negative if inverted.
- $d_i$ or $d_o$ is positive if image/object is in front of the mirror (real image).
- $d_i$ is negative if image is behind mirror (virtual image)
- $r$ and $f$ are also negative when behind the mirror
- Magnification is positive for an upright image and negative for an inverted image

A useful way to describe a material in terms of it's refractive index. The velocity of light in a medium is related to it's velocity on free space by the refractive index $n$ through the equation

$v=\frac{c}{n}$

We can recall that the velocity of a wave is given $v=f\lambda$, when light is traveling in a medium, the frequency $f$ does not change, the wavelength $\lambda$ changes according to

$\lambda=\frac{\lambda_{0}}{n}$

$n_{1}\sin\theta_{1}=n_{2}\sin\theta_{2}$

The refractive index of the medium the light is traveling out of is $n_{1}$, the refractive index of the material the light is traveling in to is $n_{2}$ and both the angle of incidence $\theta_{1}$ and refraction $\theta_{2}$ are defined relative to the normal to the surface.

For light leaving a more optically dense medium and entering a less optically dense one there is a maximum incident angle, the critical angle $\theta_{C}$ above which light is completely reflected, which we refer to as total internal reflection.

The critical angle can be found from the condition that the refracted angle is $90^{o}$

$\sin\theta_{C}=\frac{n_{2}}{n_{1}}\sin 90^{o}=\frac{n_{2}}{n_{1}}$

A lens which is thicker in the center than at the edges is a converging lens, an incoming parallel beam of light will be focused to a point $F$ at $x=f$ from the center of the lens.

To obtain a parallel beam of light light should be radially propagating outwards from the point $F'$ a distance $x=-f$ from the center of the lens.

A lens which is thinner in the center than at the edges is a diverging lens, an incoming parallel beam of light will be focused to a point $F$ at $x=-f$ from the center of the lens, which means that the rays appear to diverge outwards from that point.

To obtain a parallel beam of light light the incoming rays should have a path such that they would go to the point $F'$ a distance $x=f$ from the center of the lens if the lens were not there..

To find the image position for a lens we can use a technique called raytracing. We only need to use 3 rays to find an image for a given object (provided we know the focal length of the lens).

- A ray that leaves the object parallel to the axis and then goes through $F$
- A ray that passes through the center of the lens and is not bent
- A ray that passes through $F'$ and exits parallel to the axis

To find the image position for a lens we can use a technique called raytracing. We only need to use 3 rays to find an image for a given object (provided we know the focal length of the lens).

- A ray that leaves the object parallel to the axis and then goes through $F$
- A ray that passes through the center of the lens and is not bent
- A ray that passes through $F'$ and exits parallel to the axis

$\frac{1}{d_{o}}+\frac{1}{d_{i}}=\frac{1}{f}$

This equation also works for diverging lens (with $d_{i}$ and $f$ negative).

As with mirrors, the magnification $m$ is

$m=\frac{h_{i}}{h_{0}}=-\frac{d_{i}}{d_{o}}$

**Sign Conventions:**

- The focal length is positive for converging lenses and negative for diverging lenses.
- The object distance is positive if the object is on the side of the lens from which the light is coming (this should mostly be the case, but there are exceptions when there are combinations of lenses).
- The image distance is positive if the image is on the opposite side of the lens from which the light is coming (real image) and negative otherwise (virtual image).
- The height of the image is positive if upright and negative if inverted with respect to the object.

Combinations of lens can be treated sequentially, by first finding the image produced by the first lens and then then using it as the object for the next lens.

Applying the lens equation to the first lens

$\frac{1}{d_{iA}}=\frac{1}{f_{A}}-\frac{1}{d_{oA}}$

and to the second lens

$\frac{1}{d_{iB}}=\frac{1}{f_{B}}-\frac{1}{d_{oB}}$

and then using $d_{oB}=l-d_{iA}$ allows the determination of the final image position.

The first lens produces an image that has height $-\frac{d_{iA}}{d_{oA}}h_{o}$, which will then be used as the object height in the next lens, so the final image has height

$\frac{d_{iA}}{d_{oA}}\frac{d_{iB}}{d_{oB}}h_{o}$

When considering the multiplying power of lens combinations we can simply multiply the effects of the individual lens.