F = qvB sin theta, equation for a particle in a field.
For a straight wire, F = IlB sin theta.
B = permeability of free space/2pi * I/r

B is for magnetic field, it is measured in Teslas (T).
v is velocity (m/s) of particle in the field, q is its charge in Coulombs (C), F is in Newtons (N) of course. I is current measured in A, l is the length of the wire in meters. The permeability of free space is also designated by "mu sub 0," it is 4pi x 10^-7 Tm/A
An important concept is that when theta is 90 (perpendicular), maximum force is achieved.

Electroducks in a field move clockwise, protoducks move counterclockwise.

Here are some notes.. hehe
1. The characteristics of a magnetic field of a permanent magnet are that the field goes from South to North internally. When considering the external magnetic field, it is North to South. At the ends of the magnet, it appears that a magnetic force line extends to infinity from the North pole and out and a line from infinity enters from infinity into the South pole.

2. What is believed to be the cause of all magnetism is the concept of a moving charge.

3. Magnetic domains are tiny regions on an object which when are randomly arranged, causes the object to not exhibit a magnetic field. In order to cause it to have a magnetic field, the object must have all its domains arranged in an organized fashion in order to exhibit a magnetic field.

4. In order to use your right hand to determine the direction of the magnetic field about a straight wire, point your thumb along the wire in the direction of the conventional current, curl your fingers around the wire, this will be the direction of the magnetic field. In other words, it can only be CCW or CW about a straight wire, thus it will have no poles. For a solenoid, you must wrap your finger in the orientation of the wire wrapping. Your thumb will then point in the direction of the North pole.

5. The factors that affect the strength of an electromagnet are the battery’s provided output, the field magnet’s strength, the proximity of the field magnets with the electromagnet, the coil looping space.

6. In order to find the direction of the force of a moving POSITIVE particle in a magnetic field, point your fingers in the direction of the particle’s motion, use the flat of your palm to point in the direction of the magnetic field, use your thumb to point in the direction of the force. You can solve for the magnitude of the force by F = qvB sin theta. If the charge is negative, use your left hand instead.

7. The major components of a motor are the armature and the motorbase. It is common to also have a commutator segment, but it is not present in an AC motor. The armature is the part of the motor that rotates. The motorbase is what holds the battery as well as the armature. A simple DC Motor works on the principle of electromagnetism. When a current is induced in a wire, an magnetic field is produced. The major parts of a DC motor are the armature and the motor base. The motor base holds the armature as well as the battery; the armature is the part of the motor that rotates, comprised of an “iron core” and a winding of wire about
it. The armature is mounted between two field magnets of opposite magnetic poles facing the armature. There is a commutator segment that is essential to making the armature rotate. What happens is that a current passes through which causes deflection, but will not keep deflecting since the current is at the moment DC. The current does not automatically alternate. It is the commutator segment that makes the current alternate every half revolution which causes the rotation since the flow of electrons is then circular. To reverse the direction of rotation, you merely have to switch the battery’s connection to the battery terminals on the motor base.

8. In order to solve problems for charged particles under the influence of magnetic fields, convert to the appropriate units first, and use the equation F = qvB sin theta. In order to determine the direction, you must determine if the particle has a positive or negative charge. Use your right hand rule if it’s positive, and use left hand if it’s negative.

9. When two current carrying wires with current running in the same direction are placed parallel, they will exhibit an attractive force between each other. If the currents are running in opposite directions, then the force will be repulsive.

10. The definition of an ampere is the current flowing in each of two long parallel conductors 1m apart which results in a force of exactly 2 x 10^-7 N/m of length of each conductor.

11. Electric motors, meters, and speakers work all on the principle of electromagnetism. Electric motors convert electrical energy into mechanical energy by using an alternating current to reverse the polarities every half revolution of the armature. Electric meters work because when the current is applied through, the needle will deflect since it is sensitive to the magnetic field. Speakers work by responding to the alternating current that runs though, which converts electric energy into sound energy.

Oh this is a fun one. For a straight wire, point your thumb in the direction of the conventional current, curl your fingers around the wire(not literally if its plugged in!) Your fingers will curl in the direction of the magnetic field. Same goes for a curled wire. For a particle in a field, use your fingers to point in the direction of the velocity vector of the particle, use your thumb for the Force, use the flat of your palm for the direction of the magnetic field.
That was for a positive particle, use your left hand for a negatives!

Problem 1
What is the magnitude of the force on a 140m length of wire that's stretched between two towers if it carries a 200A current? The earth's magnetic field is 5.00 x 10^-5T, it makes an angle of 60 degrees with the wire.

Use F = IlB sin theta..
F = 200*140*5E-5 sin 60.
F = 1.21 N

Problem 2
Find the max force on an electroduck moving 3.24 x 10^5m/s when a magnetic field is introduced with magnitude 2.0T.

F = qvB sin 90
F = qvB.
F = 1.602x10^-19 * 3.24E5 * 2.0T
F = 1.0E-13 N

Problem 3
A protoduck moves in a circular path perpendicular to a 0.5T field. The radius is 0.0052m. Calculate the energy in electronVolts.

F = qvB sin theta = mv^2/r
qvB = mv^2/r, qBr/m = v
(1.602x10^-19 * 0.5 * 0.0052) / 1.67 x 10^-27 = 249413 m/s
KE = 1/2mv^2.
KE = 1/2 * 1.67 x 10^-27 * 249413^2 = 5.19E-17 J
1 eV = 1.602x10^-19 J.
5.19E-17 * 1eV/1.602x10^-19J => 324.237 eV

Problem 4 (You thought this was tough? quaCK on you! (^)<)
A 5.0 kg dareduckil moves with a speed of 100m/s perpendicular to the Earth's magnetic field of 5.00 x 10^-5 T. If the dareduckil possesses a net charge of 10 x 10^-9 C, by what distance will it be deflected from its path due to the magnetic field after it has traveled 1000m?

Okay, First you should determine which direction the force will act. Using the right hand rule, you will find that the force acts "up."
So, we have the equation F = ma. Since the force acts up, let a be a_y to designate vertical acceleration.
W e also know that F = qvB sin theta.
So, ma_y = qvB sin theta.
a_y = (qvB sin theta)/m
Cool! NOW, back to kinematics!
Considering projectile motion, it should make sense to use time as the bargaining factor.
So, lets use y-y0 = v0t + 1/2at^2.
Of course, time = d/v(using _x to designate HORIZONTAL component now), so we arrive at y-y0 = v0t + 1/2a(d_X/v_x)^2.
Substitute in a_y, and v0. y = (qBx^2)/mv_x
Cool, plug it all in. y = 1/2 * (10E-9 * 5E-5 * 1000^2) / (5 * 100)
y = 0.0000000005m

Problem 5
A long pair of wires serves to conduct 25.0 A of DC current to and from an instrument. If the wires are of negligible diameter, but are 2.0 mm apart, what is the magnetic field 10.0 cm from their midpoint in their plane?

B1 = mu_0/2pi * I/r. Plug it in, r is 2pi*.1, should get 0.00005T
B2 = Same formula, but use r+d for "r," .1 + .002, should get 4.9 x 10^-5.
Subtract B2 from B1, arriving at 9.8 x 10^-7N.

Problem 6
A protoduck and an alphaduck have identical circular orbits in a magnetic field. The protoduck has a speed of 4.4x10^5 m/s. Find the speed of the alphaduck.

F_centripetal = F_mag
mv^2/r = qvB sin theta
Let theta = pi/2, thus sin theta = 1.
We want to find the magnetic field vector B since it is the same for both ducks.
(1.67x10^-27kg)(4.4x10^5m/s)/(1.6x10 ^-19C)(1m)=4.6x10^-3T
One meter is used because it doesn't really matter the radius since it is also the same for both ducks.
Solve for v from F_centripetal = F_mag.
qBr/m = v
We have to DOUBLE q because we are using an Alpha particle(like two protons).
(3.2x10^-19C)(4.6x10^-3T)(1m)/(6.65x10^-27kg) = 2.2x10^5 m/s

Problem 7
A protoduck moving in a circular path perpendicular to a constant, uniform magnetic field takes 4.0msec to complete one revolution. Determine the magnitude of the magnetic field.

this was asked in #physics (yay #physics on EFnet) by tornado_, he was worried about getting rid of unknowns.

F_centripetal = F due to particle in mag field.
mv^2/r = qvB sin theta.
Pay careful attention to the givens and what we need to eliminate in order to get the answer!
We're given T, we apply it to this problem by the circumference of the circular path divided by its linear velocity.
T = 2*pi*r/v. cool, this will be useful.
2*pi*r/T = qvB sin 90.
qB/m = 2*pi/T
B = 2*pi*m/qT.
Substitute and evaluate!

Problem 8
A metal rod having mass per unit length 60g/m carries a current of 9.0A. The rod hangs from two (massless) wires in a uniform magnetic field B, as shown above.