Sources of the magnetic field

Reading: Chapter 28, SOURCES OF MAGNETIC FIELD

Study guide: Chapter 29

  1. Right hand rule for fields around wires
  2. Magnetic field of a moving charge
  3. What happened to action and re-action?
  4. Magnetic field from a current

Right hand rule for fields around wires


  1. Point thumb in direction of current,,

  2. Curl fingers around wire,

  3. Magnetic field direction of $\myv B$ is in direction of fingers.

Results of worksheet: Currents parallel...





so force is...

[MIT demo video]

Magnetic field of a moving charge

Electric Field:$$\myv E = \frac{1}{4\pi\epsilon_0}\frac{q\uv r}{r^2}$$

Magnetic field might look similar...?$$\myv B = [k]\frac{\text{source strength}*[\text{not }\,\uv r]}{r^2} $$

Turns out to be...$$\myv B = \frac{\mu_0}{4\pi}\frac{q\myv v\times \uv r}{r^2}.$$ $\mu_0=4\pi\times10^{-7}T\cdot m/A$.

[Magnetic field from an alpha particle at origin, moving at 600 km/s $\uv x$, at a location (in meters) $(x, y, z)=(1,2,0)$]:

$$\myv B = 10^{-7} *\frac{3.2 \times 10^{-19}*5.36\times10^{5}}{5}(-\uv z) = -3.44\times10^{-21}\,T\,\uv z$$

Using 1 gauss = $10^{-4}$T, this is a field strength of $3.44\times10^{-17} g=34.4\times10^{-18} g=17.2 ag$.

 

Force between two moving charges...

$$\begineq \myv F_{12}&=&q_2\myv v_2\times \myv B\\ &=&q_2\myv v_2\times \frac{\mu_0}{4\pi}\frac{q_1\myv v_1\times \uv r_{12}}{r^2}\\ &=&\frac{\mu_0}{4\pi}\frac{q_2 q_1}{r^2}\myv v_2\times (\myv v_1\times\uv r_{12})\endeq$$

Very peculiar...

  1. Forces do not obey Newton's law of action and equal and opposite re-action
  2. Forces depend on what reference frame we're in (relative velocities)...

Biot-Savart

I'd like to be able to calculate the magnetic field due to the current in a wire.

If we have lots of moving charge, then we need to integrate. If we separate current from a wire into many little 'chunks' of wire containing little chunks of moving charge, what becomes small and what does not become small when we start from$$\myv B = \frac{\mu_0}{4\pi}\frac{q\,\myv v}{r^2}\times \uv r?$$ to $d\myv B$?

$$\Rightarrow d\myv B = \frac{\mu_0}{4\pi}\frac{dq\,\myv v \times \uv r}{r^2}$$

From moving charge to current...

$$dq = \lambda\,d\ell \Rightarrow \lambda = \frac{dq}{d\ell}.$$

If line charge $\lambda$ moves at speed $v$, then, in $\Delta t$ a length of line charge $\Delta \ell = v\Delta t$ passes by a point, containing a total charge $\Delta q = \Delta \ell *\lambda$, so overall we have a current...

$$I=\frac{\Delta q}{\Delta t}=\frac{v\Delta t \,\lambda}{\Delta t} =\frac{v\,dq}{d\ell}$$

Rearranging a bit... $I\,d\ell=v\,dq$, or $I\,d\myv\ell=\myv v\,dq$, so our expression for $d\myv B$ becomes

$$d\myv B=\frac{\mu_0}{4\pi}\frac{I\,d\myv \ell \times \uv r}{r^2}.$$

Field of a straight wire segment

$$d\myv B=\frac{\mu_0}{4\pi}\frac{I\,d\myv \ell \times \uv r}{r^2}.$$

We'll do the derivation in class, but the important results are...

Field at a distance $x$ from the center of a current-segment of length $2a$: $$B=\frac{\mu_0 I}{4\pi}\frac{2a}{x\sqrt{x^2+a^2}}=\frac{\mu_0 I}{2\pi x}\sin \theta_\text{max}$$where you have to remember how to come up with the direction...

Field at a distance $x$ from an infinitely long straight wire: Let $\theta_\text{max}\to 90^o$ in the expression above, then $$B=\frac{\mu_0 I}{2\pi x}$$

[Example: two long wires 10 cm apart with opposite currents flowing. What is field 2 cm from one of them??]

Field at center of loop

Make-up question for integration question on Exam #1:

Find the field at the center of a wire loop of radius $a$ with a current I in it.

The answer is $$B=\frac{\mu_0 I}{2a}.$$

And, if you have many--$N$--loops of wire, all at approximately the same position (not a solenoid...) the fields add to give..$$B=\frac{\mu_0 NI}{2a}.$$

The general result for the field along the line piercing the center of a ring of current, but a distance $x$ away from the center is: $$B=\frac{\mu_0 I a^2}{2(x^2+a^2)^{3/2}}.$$

WA #4

[Another right-hand rule for field inside loop]

Gauss's law for $\myv B$?

Outline of Gauss's law for $\myv E$

Magnetic flux: $\Phi_m = \int_{\cal S}\myv B\cdot d\myv a=\int_{\cal S}B_\perp\,da.$

Magnetic field lines: never start or end (no monopoles) so we suspect....

$$\oint B_\perp\,da = 0$$

The "sources" of magnetic fields are not monopoles, but rather moving charge (either $dq\,v$ or $I\,d\ell$).

Is there some other sort of relationship...:

Notice that the field around a long, straight current is:

So it looks like for any of these circles (which are each a closed path) the following relation holds for a closed path around the current...$$\oint \myv B\cdot d\myv \ell =\oint B_\parallel\,d\ell= \mu_0 I_\text{enc}!$$ Where $d \myv \ell$ is a little "step" along the path chosen (circle in this case).

It can be shown that this relationship holds for any closed path. Though it's most useful in cases where we can make a symmetry argument about the field...., and where for each segment of the path either:

This is called Ampere's Law.

[Use it to derive field inside of a toroidal solenoid]

toroid

Example: Co-axial cable

Using Ampere's law...$$\oint B_\parallel\,d\ell= \mu_0 I_\text{enc}$$

What is the magnetic field outside the coax cable?

What is the magnetic field in the insulator? (between cylinders?

First step: argue that, from symmetry, the magnetic field $\myv B$ must circulate around the cable in circles, like the field around a long wire.

Second: Draw a path where the field is always parallel to the path and always has the same magnitude.

(should be a circular path...)

Example: Field in the center of a solenoid

A solenoid consists of many loops of wire, all carrying current in the same direction.

Inside the wire loops:

Using the path shown below and Ampere's law...$$\oint B_\parallel\,d\ell= \mu_0 I_\text{enc}$$

Show that the field in the middle (along the $x-$axis) has strength: $$B=\mu_0nI$$

Current inside a wire

The magnetic field outside a long straight wire. is $\frac{\mu_0 I}{2\pi s}$.

Using Ampere's Law to find the current inside the wire: Let's say that we know the magnetic field inside a wire of radius $S$ with a circular cross section--$$\myv B(s)=B_0 \uv \phi$$. What's the amount of current flowing through a circle of radius $s$ inside the wire? And what's the total current flowing through the wire?

Using Ampere's law...$$\oint B_\parallel\,d\ell= \mu_0 I_\text{enc}$$ ...Find $I_\text{enc}(s)$, Find total $I$ flowing through wire.


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