\(\)
Magnetic fields
In the 5th century B.C., the Greeks discovered rocks that attract bits of iron. They are named after the area Magnesia where they are plentiful. The rocks contain iron oxide, which we call magnetite.
Magnetites have two places of maximum attraction, the poles. Let’s call them A and B.
- A and A repel each other,
- B and B repel each other, but
- A and B attract each other
There is a huge difference between electricity and magnetism
- Electric monopoles exist. If you have a plus charge or minus, that is an electric monopole. If you have a plus and minus of equal strength, you have an electric dipole.
- Magnetic poles come in pairs. Magnetic monopoles do not exist, as far as we know. You always have a magnetic dipole.
Poles
The earth is a giant magnet. By convention the part of a magnet that points to the North Canada is called the north pole.
Since A and B attract, that means that North Canada is the magnetic South pole of the earth.
Discovery
In 1819, the Danish physicist Hans Christian Ørsted, discovered that a magnetic needle responds to an electric current in a wire. This linked magnetism with electricity. In this important discovery, Ørsted concluded that the current in the wire produces a magnetic field, and that the magnetic needle moves in response to that.
This discovery caused an explosion of activity in the 19th century, by André-Marie Ampère, Michael Faraday and Joseph Henry. It culminated in the brilliant work of Scottish theoretician James Clerk Maxwell. Maxwell composed a unified field theory which connects electricity with magnetism. The heart of this course. You will see all “vier” (4) equation at the end of this course.
Coulomb’s law already described the force between electric charges, at rest or in motion. But, these physicists found that the magnetic force is simply an extra electric force acting between moving electric charges.
Moving charges
A potential difference generates an electric field along and inside the conductor, that makes the free electrons move. We call this move of charge: current.
Notation convention
Let the wire with a electric current be perpendicular to your screen, so the current goes into the screen.
We put ⊗ to indicate the rear of an arrow. If it would come out of the screen we would the arrow point ⊙.
Current exerts a force on magnets
If we put magnetized needles in the vicinity of the wire with an electric current, they align to form a circle.
This is how we define the magnetic field and its direction, \(\vec B\)
When the current goes into the screen ⊗, by convention, the magnetic field is in the clockwise direction. To remember that, you can use the righthand corkscrew rule: if you turn it clockwise it goes into the screen.
Magnet exerts a force on wire
Action equals reaction. If the current exerts a force on magnet, then the magnet must exert a force on the wire.
It is an experimental fact that the force on the wire \(\vec F\) is always the vectorial cross-product $$ \shaded{ \hat F = \hat I \times \hat B } \nonumber $$
Force between two wires
If I run a current through two wires, it will product a magnetic field. Current \(\vec I_1\) will produce \(\vec B_1\).
A current \(\vec I_2\), based on the righthand rule for the cross-product, \(\hat F = \hat I \times \hat B\), the force will be up.
Since action equals minus reaction, the top wire will come down.
Lorenz force
With electricity we defined the strength of an electric field \(\vec E\) as the force \(\vec F\) on an electric charge \(q\)
$$ \vec F_{el} = q\,\vec E \nonumber $$
It would be nice if we could define the strength of the magnetic force \(\vec F_B\) as a magnetic charge \(q_B\) times the magnetic field \(\vec B\) $$ \bcancel{ \vec F_B = q_B\,\vec B } \nonumber $$
But there are no magnetic monopoles \(q_B\)
Force on a charge
Instead, the magnetic field is defined based on an electric charge \(q\) moving with velocity \(\vec v\) through a magnetic field \(\vec B\).
It is an experimental fact, that the force \(\vec F_B\) is always perpendicular to \(\vec v\). The magnitude of \(\vec F_B\) is proportional to the magnitude of \(\vec v\), and proportional to the charge \(q\) $$ \begin{align*} \vec F_B &\perp \vec v \\ F_B &\propto v \\ F_B &\propto q \end{align*} $$
We define the magnetic field strength as $$ \shaded{ \vec F_B = q\,\left(\vec v \times \vec B\right) } \tag{Lorenz force} $$
This is called the Lorenz force, after the Dutch physicist Hendrik Antoon Lorentz.
The unit is \(\frac{N\,s}{C\,m}\) that we call Tesla, \(\rm T\). We occasionally use the unit Gauss \(\rm G=10^{-4}\,\rm T\).
If you have both electric and magnetic fields, the total force on a particle \(\vec F\) $$ \shaded{ \vec F = q\,\left(\vec E + \vec v\times\vec B\right) } \tag{Lorenz force} $$ This is also called the Lorenz force.
An electric field can do work on a charge $$ W = q\,\Delta V \nonumber $$ It can change the kinetic energy of the charge.
Magnetic fields can never do work on a moving charge, because the force \(F_B\) is always perpendicular to the velocity \(\vec v\). You can change the direction, but you can’t change the kinetic energy.
Force on a whole wire
Calculate the force on a wire with current \(I\) through magnetic field \(\vec B\). In the wire is charge \(dq\) with drift velocity \(\vec v_d\)
Thought experiments
- If the current is zero, at room temperature, the electrons have a huge speed 3 million meters per second, but they are in all chaotic directions (random thermal motion). So on each individual charge there will be a force, but they average out to zero.
- It is not until I run a current, that they charges are going to walk through with a very slow drift velocity. Now, the net force will not be zero.
The angle \(\theta\) will come in later, because the force is the cross-product between \(\vec v\) and \(\vec B\).
Side note: in reality negative charges (electrons) move in the opposite direction as the current. However, a negative charge going opposite to the current, is mathematically the same as a positive charge going with the current.
On charge \(dq\) there is a force \(d\vec F_B\) $$ \begin{align*} d\vec F_B &= dq \, \left(\vec v_d \times \vec B\right), & \left(I = \frac{dq}{dt}\right) \\ &= I\,dt \, \left(\vec v_d \times \vec B\right) \end{align*} $$
The drift velocity \(\vec v_d\) is the derivative of the distance \(d\vec l\) in time,
With \(v_d = \dfrac{d\vec l}{dt}\) \(\Longrightarrow\) \(d\vec l = \vec v\, dt\) $$ \shaded{ d\vec F_B = I\, \left( d\vec l \times \vec B \right) } \tag{Force} \label{eq:forcedq} $$
This is the force of a small segment of the wire, with length \(d\vec l\). \(I\) is the current through the wire, and \(\vec B\) is the local magnetic field at location \(d\vec l\).
If you want to know the entire force on the wire, you have to take the integral along the whole wire. At every potion \(d\vec l\), you have to determine what \(\vec B\) is. That will give you a force vector that you have to add those vectors vectorially. Not easy, but that’s the idea. $$ \begin{align*} \int_\rm{wire} d\vec F_B &= \int_\rm{wire} I\, \left( d\vec l \times \vec B \right) \end{align*} $$
Example
Simplifying the geometry so we can solve the integral. Assume the magnetic field was constant over the portion of the wire.
With \(\vec B\perp d\vec l\), so \(\sin\theta=1\). And \(\vec B\) is constant \((B)\). $$ \begin{align*} F_B &= \int_\rm{wire} I\, \left( d\vec l \times \vec B \right) \\ &= I\, \left( \int_\rm{wire} d\vec l \right) \times B \end{align*} $$
The current \(\vec I\) and magnetic field \(\vec B\) are constant, so only \(d\vec l\) stays in the integral $$ \shaded{ F_B = I\, l \, B } \tag{Force on wire} \label{eq:forcewire} $$
Filling in the values $$ \begin{align*} F_B &= I\, \left( \int_\rm{wire} d\vec l \right) \times B \\ &= I\, l \, B \\ &= (300)(0.1)(0.2) = 6\,\rm N \end{align*} $$
DC Motors
Current meter
A constant current loop in a constant magnetic field. The force \(\vec F\) is in the direction \(\vec I\times\vec B\)
If the wire has length \(a\), then \(B\) with \(\vec B\perp \vec I\), the force is what we just derived in equation \(\eqref{eq:forcewire}\) $$ \begin{align*} F &= I\,a\,B \end{align*} $$
On the front and rear sections, the force is zero, because the magnetic field and current are parallel, so their cross-product is zero.
As a result of the forces, there is a torque that wants to rotate it in the counter clockwise direction. The the distance between the forces \(b\), the magnitude of the torque at this moment in time $$ \tau = 2F\,\frac{1}{2}b = I\,a\,b\,B \nonumber $$
As it rotates, the forces come closer, so the torque will be come less. There will come a time, 90° later, that the torque is zero.
At that time, the whole loop is perpendicular to the magnetic field. The forces follow from the cross-product rule
At that point in time, there is again no net force on the system, but now there is no torque either. At it reaches 90°, it has enough inertia, so that it rotates a little further. Now the torque will reverse, causing it to rotate back.
Commutator
How do you get over this torque reversal and the wires to \(A\) and \(D\) intertwining?
The solution is slippery contacts, called brushes.
This not only keeps the wires from tangling up, it also reverses the potential. It it rotates 180° then A will be on the minus side of the potential. Now every half a rotation, the current will reverse direction. We call that a commutator.
Now, the torque reversal will not occur. The torque will always want to rotate the loop in exactly the same direction.