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

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

$k$ can also be written $k=\frac{1}{4\pi \epsilon_{0}}$ where $\epsilon_{0}$ is the permittivity of free space, $\epsilon_{0}=8.85\times10^{-12}\mathrm{C^{2}/Nm^{2}}$.

Like all forces, electric forces are vectors. Note that in any given case a charge exerts an equal and opposite force on another charge to the force that the other charge exerts on it, and thus satisfies Newton's third law.

The field from a point charges is

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

and points radially outwards from the charge.

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

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

If we want to find the field directly from a distribution of charge then we should first find out if there is any cancellation of components of the field from different parts of the charge distribution.

We can consider a situation where the symmetry is quite helpful, a thin ring of charge, total charge $Q$ with radius $a$. We can ask what the field is at a point on the axis going through the center of the ring a distance $x$ from the center of the ring.

Our first step is to define a charge element $dQ$ in terms of a length element $dl$.

The magnitude of the field due to this element is then $dE=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{1}{r^{2}}dl=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{1}{x^{2}+a^{2}}dl$

We cannot however simply add the magnitude of the field from the whole ring together to get the field, because the field is a vector. However, symmetry makes the problem fairly easily because all the $dE_{\perp}$ components cancel.

Therefore,

$E=\int dE_{x}=\int dE{\cos\theta}=\int_{0}^{2\pi a}\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{1}{x^{2}+a^{2}}\cos\theta\,dl$

$=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{x}{(x^{2}+a^{2})^{3/2}}\int_{0}^{2\pi a}dl=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi a}\frac{x}{(x^{2}+a^{2})^{3/2}}2\pi a=\frac{Q}{4\pi\epsilon_{0}}\frac{x}{(x^{2}+a^{2})^{3/2}}$

For an infinitely long line of charge with a linear charge density $\lambda$

$dE=\frac{1}{4\pi\epsilon_{0}r^{2}}\,dQ=\frac{1}{4\pi\epsilon_{0}}\frac{1}{x^{2}+y^{2}}\lambda\,dy$

As before we can invoke symmetry to cancel the y components of the field

$E=E_{x}=\int dE\cos\theta=\frac{\lambda}{4\pi\epsilon_{0}}\int_{-\infty}^{\infty}\frac{\cos\theta}{x^{2}+y^{2}}\,dy$

We want to find the field at a fixed distance $x$, so this may be treated as a constant. However, $\theta$ and $y$ are not independent, $y=x\tan\theta$ so we need to convert this into either an integral in terms of either $\theta$ or $y$ only.To do this we need to find an expression for $dy$ in terms of $x$ and $\theta$

$\frac{dy}{d\theta}=\frac{x}{\cos^{2}\theta}$ ⇒ $dy=\frac{x}{\cos^{2}\theta}d\theta$

$E=\frac{\lambda}{4\pi\epsilon_{0}}\int_{-\infty}^{\infty}\frac{\cos\theta}{x^{2}+y^{2}}dy=\frac{\lambda}{4\pi\epsilon_{0}}\int_{-\pi/2}^{\pi/2}\frac{\cos\theta}{(x^{2}+x^{2}\tan^{2}\theta)}\frac{x}{\cos^{2}\theta}d\theta$

$=\frac{\lambda}{4\pi\epsilon_{0}x}\int_{-\pi/2}^{\pi/2}\cos\theta\frac{\cos^{2}\theta}{\cos^{2}\theta+\sin^{2}\theta}\frac{1}{\cos^{2}\theta}\,d\theta$

$=\frac{\lambda}{4\pi\epsilon_{0}x}\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=\frac{\lambda}{4\pi\epsilon_{0}x}[\sin\theta]_{-\pi/2}^{\pi/2}=\frac{\lambda}{2\pi\epsilon_{0}x}$

To find the field due to a uniform disk of charge which has charge density $\sigma$ we can use our result for a ring of charge and consider the field as an integral over a set of these rings. If we take a ring of charge $dQ$

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{z}{(z^{2}+r^{2})^{3/2}}\,dQ$

We need to relate the charge element to the width of the ring $dr$

$dQ=\sigma 2\pi r\,dr$

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{z\sigma 2\pi r}{(z^{2}+r^{2})^{3/2}}\,dr=\frac{z\sigma r}{2\epsilon_{0}(z^{2}+r^{2})^{3/2}}\,dr$

$E=\frac{z\sigma}{2\epsilon_{0}}\int_{0}^{R}\frac{r}{(z^{2}+r^{2})^{3/2}}\,dr=\frac{z\sigma}{2\epsilon_{0}}[-\frac{1}{(z^{2}+r^{2})^{1/2}}]_{0}^{R}=\frac{\sigma}{2\epsilon_{0}}[1-\frac{z}{(z^{2}+R^{2})^{1/2}}]$

The limit as $R\to\infty$ or when $z<<R$ that $E=\frac{\sigma}{2\epsilon_{0}}$ is valid for any infinite plane of charge.

We are now going to define a new quantity, electric flux, which will allow us to introduce Gauss's Law. We will come to see that Gauss's law is often a much more effective way of calculating the electric field than Coulomb's law.

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

or

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

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

Gauss's law will often give us a better way to evaluate the field for a distribution of charge, because on the right hand side we only need to evaluate the total amount of charge, we don't need to consider the direction or the distance of the field from the charge to the point at which we need to know the field. The trickier part of Gauss's law is writing the flux in terms of the field we want to find. 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, and so it may be composed of more than one surface. If possible we want to choose the surfaces so that the field on each surface is uniform and we don't have to actually any integration over the surface.

It's important to remember that flux has a sign. It is positive when it is leaving the volume, and negative when entering.

The charge per unit length is $\lambda$ and we can reason that the field should always be perpendicular to the line of charge. To reflect the symmetry of the problem we take a cylindrical Gaussian surface of length $l$ and radius $R$ where $R$ is the distance at which we wish to know the field.

To evaluate the total flux passing through the surface we should take into account the ends of the cylinder as well as it's curved surface, but as the field is parallel to the ends of the cylinder the flux through the ends is zero. Gauss's law is then

$\oint\vec{E}.d\vec{A}=E(2\pi R l)=\frac{Q}{\varepsilon_{0}}=\frac{\lambda l}{\varepsilon_{0}}$

which gives

$E=\frac{1}{2\pi\varepsilon_{0}}\frac{\lambda}{R}$

We draw a cylindrical Gaussian surface, but this time it is the flat surfaces that have flux passing through them. The flat surfaces have area $A$, and we can see that in fact we could have taken any kind of prism to be our surface, so long as the sides of the prism are parallel to the field.

$\oint\vec{E}.d\vec{A}=E(2A)=\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}$

$E=\frac{\sigma}{2\varepsilon_{0}}$

Suppose we now consider a thin metal sheet, which has a charge density $\sigma$ on both the front and back sides. If we draw a Gaussian surface that includes both sheets of charge we can now find that

$\oint\vec{E}.d\vec{A}=E(2A)=\frac{Q}{\varepsilon_{0}}=\frac{\sigma 2A}{\varepsilon_{0}}$

$E=\frac{\sigma}{\varepsilon_{0}}$

The same result can also be obtained by drawing a surface that end within the metal, in this case the flux is only through the top surface of the cylinder, as no field can exist in the metal

$\oint\vec{E}.d\vec{A}=E(A)=\frac{Q}{\varepsilon_{0}}=\frac{\sigma A}{\varepsilon_{0}}$

$E=\frac{\sigma}{\varepsilon_{0}}$

In order to exclude electric field from a conductor any Gaussian surface that lies entirely within the conductor must contain no net electric charge. This fact can be used to easily explain the charge distribution we discussed previously for a hollow metal sphere with a charge inside.

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

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

For an infinitely long line of charge with a linear charge density $\lambda$

$V=\frac{1}{4\pi\epsilon_{0}}\int \frac{dQ}{r}=\frac{1}{4\pi\epsilon_{0}}\int_{-\infty}^{+\infty} \frac{\lambda\,dy}{(x^{2}+y^{2})^{1/2}}$

This is a doable, but not particularly easy integral. In this case we are better off using Gauss's law to find the field, which is

$E=\frac{1}{2\pi\varepsilon_{0}}\frac{\lambda}{r}$ pointing radially outward

and using

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

For a point on the x axis

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}=\frac{1}{4\pi\varepsilon_{0}}\frac{1}{(x^{2}+R^{2})^{1/2}}\int dq=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{(x^{2}+R^{2})^{1/2}}$

$E_{x}=-\frac{\partial V}{dx}=\frac{Q}{4\pi\varepsilon_{0}}\frac{x}{(x^{2}+R^{2})^{3/2}}$

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

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

and for a parallel plate capacitor

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

The amount of energy stored in a capacitor is $U=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

The above is true for any capacitor. For a parallel plate capacitor

$U=\frac{1}{2}\frac{Q^{2}d}{\varepsilon_{0}A}$

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

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

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

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 $Q=C_{i}V_{i}$

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

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

The resistance of a wire is proportional to it's length $l$ , inversely proportional to it's cross sectional area $A$ and also depends on the material property, resistivity, $\rho$, which is measured in $\mathrm{\Omega m}$

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

The reciprocal of the resistivity of a material is the conductivity

$\sigma=\frac{1}{\rho}$

measured in $\mathrm{(\Omega m)^{-1}}$

The values of resistivity exhibited by different materials have the widest range of any physical quantity, from about $1\times10^{-8}\,\mathrm{\Omega m}$ for the best conductors to $1\times10^{24}\,\mathrm{\Omega m}$ for the best insulators.

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 $V=IR$

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

The voltage which produces an AC current may be described by the equation

$V=V_{0}\sin2\pi f t=V_{0}\sin\omega t$

$V_{0}$ is the peak voltage, this is not the value of the voltage we typically refer to when talking about mains power. The 120 V we normally refer to is the rms (root-mean square) voltage. If we are considering the potential across an ohmic conductor, then $V=IR$ and

$I=\frac{V}{R}=\frac{V_{0}}{R}\sin\omega t=I_{0}\sin\omega t$

The power is then

$P=VI=I^{2}R=I_{0}^{2}R\sin^{2}\omega t$

We can see that this is always positive, this is because the current changes sign in phase with the voltage. The average value of the power is found by taking the average value of $\sin^{2}\omega t$ which is $\frac{1}{2}$

$\bar{P}=\frac{1}{2}I_{0}^{2}R=\frac{1}{2}\frac{V_{0}^{2}}{R}$

As the root mean square voltage and current are $V_{rms}=\frac{V_{0}}{\sqrt{2}}$ and $I_{rms}=\frac{I_{0}}{\sqrt{2}}$

$\bar{P}=I_{rms}V_{rms}=I_{rms}^{2}R=\frac{V_{rms}^{2}}{R}$

Suppose we have a wire carrying a current $I$. A useful quantity for us to define, especially if we want to consider the microscopic details of conduction, is the current density $\vec{j}$ which describes the magnitude and direction of the amount of current flowing per unit area at a given point inside a material. In general

$I=\int\vec{j}\cdot d\vec{A}$

and for a uniform wire with cross-sectional area $A$

$I=jA$ or $j=\frac{I}{A}$

When dealing with conductors in electrostatics we said there could not be a field inside a conductor. This is only true if there is no current flowing. If we consider a wire of length $l$ with a potential difference $V$ from one end to the other there will be a field inside the wire with magnitude $E=\frac{V}{l}$.

If we take Ohm's law, $V=IR$, we can see it can be rewritten as

$El=(jA)(\rho\frac{l}{A})=j\rho l \to j=\frac{1}{\rho}E=\sigma{E}$

We can note from this equation that in either the case of zero current or zero resistance the field in the conductor should be zero.

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. We can look at this also in terms of a circuit

The current $I$ which flows through the circuit is the same everywhere. The potential in each resistor is given by Ohm's law $V=IR$, and 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.

When we consider a circuit with a resistor and capacitor in series with a battery and a switch and consider the case where the capacitor is initially uncharged we can reason that when the switch is closed a current needs to flow from the battery to the capacitor. The current can be analyzed based on the DC circuit concepts we have just been discussing.

From Kirchoff's loop rule

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

Writing this in terms of charge

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

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

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

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

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

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

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

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 $I$. We can write an equation that contains this information as well as the right hand rule for the direction that we identified earlier. We can give the length of the wire a direction and make it a vector $\vec{l}$. The current is then defined to be positive when it flows in the direction of the length vector. The force is then

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

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.

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

Electric charges in a wire feel a force but are not free to leave the wire, so the effect is force on the wire. Free electrons in a magnetic field also feel a force and are free to respond to it. In the same way as wire only feels a magnetic force when a current flows, charges only feel a magnetic force when they are moving and the force depends on the velocity. 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

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

We can show that the path of an electron moving in a uniform electric field is a circle. As we will recall from when we studied uniform circular motion, a force perpendicular to the velocity changes the direction of the velocity only, not it's magnitude. 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 is a useful way to determine magnetic field in cases where symmetry makes it practical. It is valid in cases where the magnetic field is static, or in other words when steady currents flow. It also does not apply when magnetic materials are present. (It can however be modified to deal with both changing currents and magnetic materials).

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, $n$

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

Within the interior radius of the torodial solenoid the enclosed current is zero, and thus also the magnetic field is zero, as

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

Within the solenoid, which has $N$ turns

$\oint\vec{B}\cdot d\vec{l}=B(2\pi r)=\mu_{0}NI\to B=\frac{\mu_{0}NI}{2\pi r}$

And outside of the solenoid we again find

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

The Biot-Savart law gives a different approach for obtaining the field due to a 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!

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

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 $\sin\theta=\frac{R}{r}$.

$\vec{B}=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{dy\sin\theta}{r^{2}}=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{R\,dy}{r^{3}}=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{R\,dy}{(R^{2}+y^{2})^{3/2}}$

$=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{1}{R}(\frac{R^2\,dy}{(R^{2}+y^{2})^{3/2}})=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{1}{R}(\frac{R^{2}+y^{2}}{(R^{2}+y^{2})^{3/2}}-\frac{y^{2}}{(R^{2}+y^{2})^{3/2}})\,dy$

$=\frac{\mu_{0}I}{4\pi}\int_{-\infty}^{\infty}\frac{1}{R}(\frac{1}{(R^{2}+y^{2})^{1/2}}-\frac{y^{2}}{(R^{2}+y^{2})^{3/2}})\,dy=\frac{\mu_{0}I}{4\pi}[\frac{1}{R}\frac{y}{(R^{2}+y^{2})^{1/2}}]_{-\infty}^{\infty}$

$=\frac{\mu_{0}I}{4\pi}\frac{2}{R}=\frac{\mu_{0}I}{2\pi R}$

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

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

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

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

Faraday based his law of induction around the concept of magnetic lines of force. He found that the induced emf depended on the rate of change of the total amount of magnetic field passing through an area, which we call the magnetic flux.

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

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, $N$ that are wrapped so that the same flux passes through them all the net emf in the circuit is the sum of the emf from each loop

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

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. We can consider the situation for a conducting rod moved backwards and forwards on a pair of rails which are connected to galvanometer, as in this applet.

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. Independent of whether a current flows or not there will be an induced emf in the conducting rod and $\mathcal{E}=-Blv$

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.

We saw in previous lectures that a torque could be produced on a current carrying loop placed in a magnetic field and that when appropriate commutation was used an electric motor could be made.

If we now think about it the other way round and consider the change in flux on a loop as it is rotated by some external torque we can see that it should produce some emf which can be used as the basis for an electric generator.

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

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

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

As an electric motor accelerates under the action of magnetic force the rate of change of flux $\frac{d\Phi_{B}}{dt}$ will get larger, producing an emf which opposes the motion. The total potential driving the current in the circuit is the applied potential and the back emf generated by the changing magnetic flux. When these two are equal the current flowing through the circuit is zero. If there is no other load on the motor the motor would then turn at that speed constantly. Of course any load will reduce the speed, and hence the back emf, so at the constant top speed of a real motor there will still be some current flowing.

In a generator the opposite occurs, if the emf generated is used to produce a current then this produces a counter torque that opposes the motion.

In a transformer two coils are coupled by an iron core so that the flux through the two coils is the same. The iron core is usually layered with insulating materials to prevent eddy currents. 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 emf produced by the changing flux is $\mathcal{E}=-N_{P}\frac{d\Phi_{B}}{dt}$ and this means that the net potential drop in the coil is actually zero (as Kirchoff's rules say it must be if the resistance of the coil is zero). 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 the example of a transformer we assumed that the flux induced in the secondary coil was equal to the flux in the primary coil. 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 ($\mathrm{H}$), $1\mathrm{H}=1\mathrm{\frac{Vs}{A}}=1\mathrm{\Omega s}$

If we now consider a coil to which a time varying current is applied a changing magnetic field is produced, which induces an emf in the coil, which opposes the change in flux. 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 $\tau=\frac{L}{R}$.

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

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

The first rule we have is for reflection. 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 $m$

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

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

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

Refractive indices of materials are typically somewhere between 1 (vacuum) and ~2.5 (diamond, strontium titanate). Glass will typically have a refractive index of about 1.5 though the exact value depends on the type of glass.

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

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

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.

The biggest possible size of an object on the eye without the aid of an optical instrument is obtained by placing it at the near point of the eye $N$. If we bring an object closer than the near point then it will occupy a larger angle, but we won't be able to focus on it. This problem can be addressed by a magnifying glass. If we place an object closer than the focal length of the lens, it will produce a virtual image at a distance $d_{i}$. The maximum magnification is achieved by bringing the lens right up to your eye and then arranging the lens, object and your head so that the image is at the near point. To find the magnification we need to know $d_{o}$ which can be found from the lens equation, taking $d_{i}=-N$

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

In the small angle approximation

$\theta=\frac{h}{N}$ and $\theta'=\frac{h}{d_{o}}$

The angular magnification of the lens, also called the magnifying power is defined as $M=\frac{\theta'}{\theta}$ which we can see here is

$M=\frac{N}{d_{o}}=N(\frac{1}{f}+\frac{1}{N})=\frac{N}{f}+1$

It is not very convenient to use a magnifying glass with the eye focused at the near point, firstly as we are required to constantly maintain the correct positioning of lens, object and head, but also because our eye muscles are at maximum exertion, which is not very comfortable over long periods of time. An alternative way to use a magnifying glass is to place the object at the focal point of the lens producing an image at $\infty$.

In this case

$\theta'=\frac{h}{f}$

and the magnifying power is

$M=\frac{\theta'}{\theta}=\frac{N}{f}$

A magnifying glass is only useful for looking at objects nearby (the maximum object distance from the lens is the focal length). To view distant objects we need to use a telescope. A refracting telescope uses two lens, and objective and an eyepiece. The objective produces an image which is then magnified by the eyepiece.

The original apparent object size is

$\theta\approx\frac{h}{f_{o}}$

If we consider an object at infinity and an eyepiece which is adjusted so that the focus of the eyepiece is at the focus of the objective (this produces a final image at infinity, which is why I chose not to draw it this way on the diagram), then the apparent size of the final image is

$\theta'\approx\frac{h}{f_{e}}$

giving the magnification power of the telescope as

$M=\frac{\theta'}{\theta}=-\frac{f_{o}}{f_{e}}$

with the minus sign signifying that the image is inverted.

If we treat each of the slits as a point source of circular wavefronts. The condition for constructive interference (bright fringes) is

$d\sin\theta=m\lambda$ (m=0,1,2,..)

and for destructive interference (dark fringes)

$d\sin\theta=(m+\frac{1}{2})\lambda$ (m=0,1,2,..)

As in the case of a wave on a rope that is incident on a heavier rope and is reflected with a 180^{o} phase change when a light wave is reflected from a more optically dense media a 180^{o} phase change occurs. This effect is important when we want to consider interference effects in thin films.

As the two reflections both occur from more optically dense media they both experience a $\frac{\lambda}{2}$ phase change on reflection. To have the light be out of phase we need to light that goes through the coating to have advanced by $\frac{\lambda}{2}$ for destructive interference to occur. Critically, when destructive interference occurs the light is not lost, but is instead transmitted. As the wavelength of light in a medium is given by $\lambda=\frac{\lambda_{0}}{n}$ where $n$ is the refractive index of the medium and $\lambda_{0}$ is the wavelength of the light in free space, the thickness of the coating should be $\frac{\lambda}{4n_{2}}$.

In practice the light incident will not all be the same wavelength, so the thickness of the coating is typically chosen to work optimally in the center of the visible band (~550nm).

For a single wavelength dark rings will occur whenever

$2t=m\lambda$ (m=0,1,2,..)

and bright rings will occur whenever

$2t=(m+\frac{1}{2})\lambda$ (m=0,1,2,..)

For white light different colors will experience constructive interference at different thicknesses, leading to the colorful lines we see when an air gap is under normal light.

Rather than writing our equation in terms of k

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{Dk}{2}\sin\theta)$

it is convenient to write it in terms of the wavelength, using $k=\frac{2\pi}{\lambda}$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

This function has a central maxima and then minima at

$D\sin\theta=m\lambda$ $m=\pm 1,\pm 2,\pm 3,..$

Now we have an expression for the intensity from a slit of width $D$

$I_{\theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)$

we can use consider slits as sources instead of the point sources we considered earlier which gave

$I_{\theta}=I_{0}\cos^{2}(\frac{d \pi}{\lambda}\sin \theta)$

The combination of these gives

$I_{theta}=I_{0}\mathrm{sinc^{2}}(\frac{D\pi}{\lambda}\sin\theta)\cos^{2}(\frac{d\pi}{\lambda}\sin \theta)$

Irrespective of the number of slits $n$ the condition for maxima is the same as for the double slit

$d\sin\theta=m\lambda$ (m=0,1,2,..)

but that the larger the number of slits from which diffraction occurs the sharper the maxima will be.

The polarization of the light after it has passed through the polarizer is the direction defined by the polarizer, so the way to approach the calculation of the magnitude of the intensity that passes through the polarizer is by finding the component of the electric field which is in that direction

$E=E_{0}\cos\theta$

the intensity is then

$I=E^{2}=E_{0}^{2}\cos^{2}\theta=I_{0}\cos^{2}\theta$

This formula is known as Malus' Law.