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Reading: Chapter 27, MAGNETIC FIELD AND MAGNETIC FORCES
Study guide: Chapter 28
Ancient Indians, Greeks and Chinese (at least) were aware of the singular properties of lodestone (magnetized magnetite): of how it attracted other lodestones, as well as iron.
By the 11th century, the Chinese were using "floating fish": magnetized needles or pieces of iron floating on water, to keep track of direction at night or with low visibility. The Ming dynasty diagram shows the names for different compass headings.
Early navigators thought there was some 'magnetic mountain' up north. In the 1600s the modern definition of the North pole: The place where the magnetic field is perpendicular to Earth's surface
I
have cunningly mapped out a "field" with a compass (not a charge)
around a magnet that looks like the electric field around an electric dipole:
We will call the field $\myv B$: It points in the direction that the "N" end of a compass would point.
This is the direction of force that a 'N' magnetic monopole would feel...but wait: there aren't any magnetic monopoles!
From our experiments with the Crooke's tube... what is the orientation of the $\myv B$ field and the force that a charge feels?
Oersted wondered: Does a current give rise to a magnetic field?
What if you have many loops of wire?
Experiment shows that the force on a moving charged particle is...
$$\myv F_m = q \myv v \times \myv B$$
So, magnetic field must have units of...
$\ \ \ \ \frac{F}{qv}=\frac{[N]}{[C][m/s]}\equiv
T$, "Tesla".
The cross product of two vectors is a vector with magnitude $$|\myv A \times \myv B| = AB\sin\theta$$ which is the same as the area of the trapezoid in the figure at right.
[What is the cross product of two vectors which are parallel?]
But the cross product is a vector with a direction $\uv{n}$ which is perpendicular to both $A$ and $B$ according to a right-hand rule: $$\frac{\myv A \times \myv B}{|\myv A \times \myv B|}\equiv \uv n.$$
$$\myv A \times \myv B \equiv AB\sin\theta \,\uv{n}.$$
With the right hand rule, the order in which the two vectors are "crossed" matters:
$$\myv A \times \myv B = - \myv B \times \myv A.$$
The cross product is not generally distributive. Try coming up with a set of three vectors such that... $$(\myv A \times \myv B) \times \myv C \neq \myv A \times (\myv B \times \myv C).$$
In a right-handed coordinate system: $\uv{e}_1 \times\uv{e}_2 =\uv{e}_3$. We will always take a Cartesian system where: $$\uv{x} \times\uv{y} =\uv{z}.$$
By compiling all the other unit vector products individually, e.g. $\uv{z} \times \uv{z} = 0;\ \uv{y} \times\uv{x} = -\uv{z}$; etc. you can eventually find that
$$\begineq\Rightarrow \myv A \times \myv B &=& (A_x\uv{x} +A_y\uv{y}+A_z\uv{z}) \times (B_x\uv{x} +B_y \uv{y}+B_z\uv{z})\\ &=& (A_yB_z-A_zB_y)\uv{x} + (A_zB_x-A_xB_z)\uv{y} +(A_xB_y-A_yB_x)\uv{z}\endeq.$$
I remember this as the determinant of the three-by-three matrix shown below, which you evaluate by
adding the red products, and subtracting the blue products for this determinant:
[Vector product of $(4,0,0) \times (2,1,0)$, both ways]
$$\myv F_m = q \myv v \times \myv B$$
An electron is happily moving along through empty space with
constant $\myv v$ in a straight line....
when suddenly, a constant magnetic field is suddenly switched
on! (pointing into the page as indicated by the Xs)
Quick! What is the direction of the resulting force?!?!
$\myv v \times \myv B$ points left, but $\myv F_m=q
\myv v \times \myv B$ points right. So how will the electron move next??
Perhaps circular motion? Is $\myv F_m$ still perpendicular
to $\myv v$ as the direction of $\myv v$ changes? Is the speed changing?
Centripetal acceleration:$F=mv^2/r$ so, the radius is:$$r=\frac{mv^2}{qvB_\perp}=\frac{mv}{qB_\perp}$$
How much time for one circle (the 'period')? $T = d/v$
$$T = \frac{2\pi r}{v}=\frac{2\pi mv}{vqB_\perp}=\frac{2\pi m}{qB_\perp}.$$
Initial $v$ doesn't matter!
"Cyclotron frequency" (radians / sec):$$\omega
= 2\pi f=2\pi/T = \frac{qB_\perp}{m}$$
[what if we turn on an electric field such that $\myv F_e=q\myv E = -\myv F_m$?]
[relationship to northern lights:
take velocity apart
into a component of the velocity parallel to the magnetic field and a component
perpendicular to the field:$$\myv v= \myv v_\perp + \myv v_\parallel$$ Does
$\myv v_\parallel$ change as a result of the magnetic field?]
$$\myv F = q\myv v \times \myv B$$
if $\myv v$ and $\myv B$ are perpendicular, then the magnitude of the force is $F=qvB$
The total force on a current-carrying wire carrying a total number $N$ of carriers (all with the same charge and average speed) is the sum of all the forces on all the carriers:$$F=\sum F_i = NqvB$$
Relate this to the current $$I=nqvA$$...
We get total force = $F=I \ell B$
More generally $$\myv F = I\myv \ell \times \myv B$$where $\myv \ell$ points along the wire, in the direction of the current.
[Force between two parallel wires with current flowing in same direction? See this video...]
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