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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, $n$

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

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

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

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

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

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.

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

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

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

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