Magnetism

5.3.1 Draw the pattern of magnetic field lines of an isolated bar magnet.

A magnetic field represents the direction that a compass needle would line up if it were placed at that point. The arrows point in the direction that the north pole of the compass needle would point.

MagField.jpg

In the picture to the right the iron filings line up with the magnetic field. Each iron atom can be though of a tiny bar magnetic, so when iron is exposed to a magnetic field all the tiny bar magnetic try to line up with the magnetic field lines.

5.3.2 Draw the magnetic field for the Earth

MagFieldEarth.jpg

Notice that the north pole of the earth is actually a magnetic south pole.

5.3.3 Draw and annotate magnetic fields due to currents

Here’s a great site with pictures of magnetic fields due to currents in a straight wire, a flat circular coil and a solenoid: [http://www.ux1.eiu.edu/fc~cfadd/1160/Ch20Mag/FldCur.html]

The direction of the magnetic field due to a current can be found using the right hand rule. Using your right hand (go figure) point your thumb in the direct of the current (not electron flow) then wrap your fingers around the wire. Your fingers point or represent the direction of the magnetic field.

The strength of the magnetic field in Teslas (T) produced by a current carrying wire is given by the following equation:

(1)
\begin{align} B = \frac{\mu_0 I}{2 \pi r} \end{align}

Where I is the current, r is the distance from the wire and μ0 is called the permeability of free space (not to be confused with the permittivity of free space for electric fields).

It is interesting to note that the speed of light, which is an electromagnetic wave, in a vaccum is equal to:

(2)
\begin{align} c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \end{align}

5.3.4 Determine the direction of the force on a current carrying conductor in a magnetic field.

5.3.5 Determine the direction of the force on a charge moving in a magnetic field

5.3.6 Define the magnitude of the magnetic field strength B

When a current carrying wire is placed in a between the poles of the U-shaped magnet the wire feels a force. This is caused by the interaction of the magnetic field and is called the motor force.

The direction of force can be found using several different “hand rules,” maybe the simplest being the right-hand-palm-rule. Your thumb points in the direction of the current, your fingers in the direction of the magnetic field (external not that caused by the current) and your palm faces in the direction of the force.

The force on a current carrying wire in a magnetic field B is given by:

(3)
\begin{align} F = I l B \sin \theta \end{align}

Where l is the length of the conductor in the magnetic field, I is the current in the wire and θ is the angle between the wire and the magnetic field.

The force on the wire from the magnet causes the wire to accelerate and thus "jump."

If a charged particle is moving through a magnetic field the force on that particle can also be found using the right-hand-palm-rule, but remember that when we talk about current we are talking about the direction of a positive charge, so if an electron is moving through an electric field we can find the force by pretending it’s a proton and when we find the direction of the force we simply reverse it.

The force on a moving charged particle can be found by noting that:

(4)
\begin{align} v=\frac{l}{t} \rightarrow l=vt \end{align}

Where l is the length of the path. And

(5)
\begin{align} I = \frac{q}{t} \end{align}

Substituting into the equation for force on a current carrying wire:

(6)
\begin{align} F = qvB \sin \theta \end{align}

This gives an expression for the force on a charge q moving with velocity v. It is important to realize that the force and the velocity are perpendicular. Therefore the motion of the particle is uniform circular motion. So we know that the force on the particle can also be written as:

(7)
\begin{align} F=m \frac{v^2}{r} \end{align}

Equating the two expression and solving for r:

(8)
\begin{align} qvB \sin \theta =m \frac{v^2}{r} \end{align}
(9)
\begin{align} r = \frac{m v}{qB \sin \theta} \end{align}

If the velocity is perpendicular to the magnetic field so that $\theta = 90^{\circ}$, then the expression simplifies to:

5.3.7 Solve problem involving the magnetic forces on currents and moving charges

5.3.8 Draw the magnetic forces between two parallel current carrying wires

Why draw them when you can see an animation? Courtesy of MIT (video can be slow to load):

Current in the same direction attracting

Current in the opposite direction repelling.

5.3.9 Solve problems involving the magnetic forces between two parallel current carrying wires

5.3.10 State and explain the definition of the ampere

When two long parallel current-carrying wires are placed near each other they feel a force. As seen above the force is attractive or repulsive depending on the direction of the current. If the currents flow in the same direction the force is attractive if the current is in opposite direction the force is repulsive.

The force on a wire of length l in a magnetic field B is:

(10)
\begin{align} F = I l B \sin \theta \end{align}

The force is generated by the magnetic field of the other wire which is given by:

(11)
\begin{align} B = \frac{\mu_0 I}{2 \pi r} \end{align}

Combining the two equations:

(12)
\begin{align} F = I_1 l \left (\frac{\mu_0 I_2}{2 \pi r}\right ) \sin \theta \end{align}

Where the subscripts 1 and 2 are there to allow different currents in each wire. Since the wires are parallel:

(13)
\begin{align} F = \frac{\mu_0 }{2 \pi} \frac{I_2 I_1 l}{r} \end{align}

The force per unit length of wire is given by:

(14)
\begin{align} \frac{F}{l} = \frac{\mu_0 }{2 \pi} \frac{I_2 I_1}{r} \end{align}

Where l is the length of the wires, $I_1$ and $I_2$ are the currents in the wires, r is the distance between the wires. This last equation is given in the IB formula handbook.

5.3.11 Explain the operation of a simple direct current (dc) motor

SimpleDCMotor.jpg

Stolen From: [http://www.pc-control.co.uk/dc-motors.htm]

The magnetic field is from south to north. The current flows in a clockwise direction. This causes a torque on the loop of wire. The torque causes the loop to rotate, when the loop has rotated 90°, the loop will have no net torque. This is where the commutator comes in… When the loop rotates 90° the commutator reverses the direction of the current, which causes the torque to reverse and the loop continues to rotate. As the loop rotates the current in the loop changes every 180°, which keeps the loop rotating.

5.3.12 Solve problems involving the magnetic field strength around a straight wire

5.3.13 Solve problems involving the magnetic field strength within a solenoid

This would be a good thing to add. :)


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