electromagnetism – Elektromagnetisme http://elektromagnetisme.no The home of FYS1120 Mon, 20 Oct 2014 11:23:54 +0000 en-US hourly 1 https://wordpress.org/?v=4.9.3 28429679 Online lectures from MIT with Walter Lewin http://elektromagnetisme.no/2012/09/18/online-lectures-from-mit-with-walter-lewin/ http://elektromagnetisme.no/2012/09/18/online-lectures-from-mit-with-walter-lewin/#respond Tue, 18 Sep 2012 14:47:05 +0000 http://elektromagnetisme.no/?p=1503 Continue reading →]]>

If you have missed out on some of the lectures or feel you could need a second view on some subjects, we recommend having a look at Walter Lewin’s lectures from MIT:

Our syllabus is not the same as in their course, but some lectures are very relevant. For instance there are lectures on Gauss law, electric flux, magnetism and much more that will follow in our course too.

Spending some time watching a few lectures when you have spare time or are tired of reading can help you grasp concepts that you find hard to understand. It could also be very inspiring!

Should you find some lectures especially useful, don’t forget to tip your fellow students about it by leaving a comment below.

]]> http://elektromagnetisme.no/2012/09/18/online-lectures-from-mit-with-walter-lewin/feed/ 0 1503 Note on the magnetic vector potential http://elektromagnetisme.no/2011/10/27/note-on-the-magnetic-vector-potential/ http://elektromagnetisme.no/2011/10/27/note-on-the-magnetic-vector-potential/#respond Wed, 26 Oct 2011 22:29:56 +0000 http://mindseye.no/?p=992 Continue reading →]]>

Magnetic dipole (Source: Wikipedia*)

In electrostatics we found it very convenient to introduce the concept of the electric potential. It gave us a straight forward way of calculating electric fields without doing any vector calculations or using any symmetry arguments. Can we introduce something similar for magnetic fields?  It turns out that because magnetic fields are divergence less we can find a vector potential who’s curl gives us the magnetic field! Even though this magnetic vector potential is not as useful as the electrostatic potential in elementary applications, it turns out to be of major importance in electrodynamics as well as classical mechanics and quantum mechanics. It might therefore be a good idea to get familiar with the concept and some of it’s properties already, especially if you are taking a degree in physics. In this note I explain how to find the vector potential, the concept of a gauge transformation and it’s fundamental equations relating it to currents in both electrostatics and electrodynamics. Read more here:

* Image found at http://en.wikipedia.org/wiki/File:VFPt_dipole_magnetic3.svg / CC BY-SA 3.0

]]> http://elektromagnetisme.no/2011/10/27/note-on-the-magnetic-vector-potential/feed/ 0 992 The FYS1120 Weekly Assignment Post #3 http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/ http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/#comments Wed, 06 Oct 2010 05:38:57 +0000 http://mindseye.no/?p=284 Continue reading →]]> This week the problems focuses mainly on circuits and Kirchhoff’s laws.

Assignment #7

1)

a) I_1 = \frac{11}{3} A, \ \ I_2 = -\frac{4}{3} A, \ \ I_3 = -\frac{7}{3}A.

Indicating that current nr.2 and 3 is directed in the opposite way of what was assumed.

b) P_{\varepsilon_1} = \varepsilon _1 I_1 = 44 W

P_{dissapated} = R (I_1 ^2 + I_2 ^2 + I_3 ^2) = 41.33 W.

c) Of course the second voltage supply is also doing work on the charges in the current, so this is the reason that P_{dissapated} \neq P_{\varepsilon_1} In this case it is actually doing negative work.

2)

a) At t = 0, the moment when the switch is closed, there is no charge on the capacitor and therefore no potential difference over it. This means that the current at this time will be as if there were no capacitor in the circuit, so I = \frac{\varepsilon}{R}

As time goes to infinity, the capacitor gets charged up. At some point it will reach a potential difference equal to the potential difference of the voltage supply and the current will be zero.

b) At t = 0 it is a ressistance R in one subcircuit, while there is no ressistance in the other, so the current will go trough the circuit with no ressistance. Since there is no resistance I = \frac{\varepsilon}{0} = \infty,  which means that the capacitor gets charged up infinitely fast.

When the capacitor is fully charged, no current will go towards it and therefore all current will go trough the resistance. This means that we now have I = \frac{\varepsilon}{R}

3)

a) \tau = RC = 1 \times 10^{-6} s

b) q(t=30s) = \varepsilon C \left[ 1 - e^{-\frac{t}{\tau}} \right] = 1.2 \times 10^{-11} C

c) I(t=30s) = \frac{dq}{dt}(30s) = 0

4)

a) Here I used Kirchhoff’s laws to get 5 equations which I put in a matrix, did rowreduction and got I_1 = 6A, \ \ I_2 = 5A, \ \ I_3 = 1 A, \ \ I_4 = 7A, \ \ I_5 = 4 A.

b) R = \frac{\varepsilon}{I} = \frac{\varepsilon}{I_1 + I_2} = \frac{\varepsilon}{I_4 + I_5} = 1.812.. (correction: Filip Sund)

]]> http://elektromagnetisme.no/2010/10/06/the-fys1120-weekly-assignment-post-3/feed/ 4 284