Honors Homework for Sunday 1/18

Read this and do the problems as your homework for today ( Sunday). I 've given you a link but also printed it here. Note the ws are omegas; blogspot can't write Greek.

http://h1.ripway.com/DrCherdack/Rotational%20Dynamics%20Eqs.doc

Rotational Motion in Magnetic Fields Notes

Magnetic fields are peculiar in that the force they create is felt only by moving charged particles and the forces are perpendicular to the field and the velocity of the charged particles. We will explain all this next quarter. For now, just note that the direction of the forces makes them useful for making charged particles to go around in circles since the force to make something go in a circle is perpendicular to velocity ( the velocity is tangential to the circle and the force must be inward radially).

The magnetic force is given by Fmag = Qv X B. Here B is the magnetic field strength ( I have no clue as to why its B)and it has a direction just as electric and gravity fields do. The X is not simple multiplication. A X B is equal yto AB sin angle between A and B. [The strength of B is measured in Tesla, T. If this makes it clear as to why its called B, let me know.] The X operation ends up with a vector which is perpendicular to both V and B. Let’s just focus on the strength of the force and the fact that it is cnetripetal for now.

We have demonstrated that a centripetal, ac, = wvtan or v2/r or w2r since vtan =wr.


Mac = QvB or Mvsqrd/r =QvB dividing both sides by v we get Mv/r = QB or r =Mv/QB

WE can also solve for w using v = wr. We have Mac = Mvsqrd2r=QwrB divide both sides by w and we have Mw=QB or w = QB/M

Now lets get some numbers. If we know the force and mass we can find r for say a deuteron moving at 2 x106 m/s in a field whose strength is 5 Tesla ( this is about 100,000 time the earth’s magnetic field the one that makes compasses work). Go ahead and do this. Also find w. Try it on your own and then check. If you need to find Q and M from notes go ahead and do it. I’m not telling you any more.

The answers are r = 8.53 x10-3 m or .85 cm (a third of an inch) and w= 2.4 x108 rad/sec.

Now you do it for tritium nucleus, T, traveling at 1x106 m/s and then for a T 2.5 x106 m/s.

Do it again for a field of strength B = 3 T to see how r and w change with B.