Magnetism

Magnetic Field

A bar magnet was placed on a horizontal whiteboard and a compass was placed near its north pole. We noticed that the needle pointed directly at the north pole of the magnet. We repeated this for many positions around the magnet (along the blue line) and drew a red arrow on the whiteboard representing the orientation of the compass needle.

Here we see the resulting arrows at various positions around the magnet.

We figured the magnetic field of the bar magnet must look similar to the diagram above.

Prof. Mason sprinkled some iron filings evenly around a magnet and used a projector to show the results. We can see that the magnetic field looks very similar to the one we had hypothesized.

Magnetic Flux

Magnetic flux follows some of the same principles as electric flux. We can determine if there is a net flux by counting the magnetic field lines going in and out. Here we show a magnetic field in red and three areas of interest circled in black and labeled 1,2,3. In area 1 there are two lines coming in and two coming out, therefore the magnetic flux is zero at that point. Area 2 encases the south pole and contains 0 lines coming in and 7 going out, therefore there is a net magnetic flux. In area 3, which encases both north and south poles, there are 7 lines going in and 7 coming out, therefore the net flux is zero.

Magnetic Field, Force, and Velocity

In this given problem, we were given that the magnetic field was 2.6E-3 T and the velocity of an electron was at 30 degrees from it with a magnitude of 3E6 m/s. We found the magnitude of the force using the formula shown above where q is the charge of an electron. We also used the fact that the cross product with the magnetic field can also be taken by taking the magnetic field and multiplying it by the sin of the angle between it and the charge velocity.

The diagram above shows that force is magnetic field is always perpendicular to the force and charge velocity.
Below we begin deriving equations that have to do with centripetal force.

We start by stating that the centripetal force is equal to the mass multiplied by the velocity squared all divided by the radius of the circle. We also know that the force is equal to qv X B so setting them equal to each other cancels out a factor of v from both sides. We got rid of the cross product since it is equal to sin(90)=1 in this case (not cos(90) as in picture). We solved for R and changed translation velocity into angular velocity (v = Rw). The R's cancel out and we can solve for w.

We have another formula for angular velocity which states that it is directly proportional to the frequency and a factor of 2pi. Setting these two formulas gives us a direct relationship between frequency (given), mass of an electron, charge of an electron, and magnetic field (which we need to find). Solving for magnetic field and plugging in the known values we calculate the magnetic field to be 0.0876 T.