<p>could someone help me on these sets of problems.</p>
<p>A magnetic field of 0.1 T forces a proton beam of 1.5 mA to move in a circle of radius .1 m. The plane of the circle is perpendicular to the magnetic field. </p>
<li><p>What’s the best estimate of the work done by the magnetic field on the protons during one complete orbit of the circle?
a. 0 J b. 10^-22 J c. 10^-5 J d. 10^2 J e. 10^20 J
Id assume its bc work done in circular motion is 0</p></li>
<li><p>What’s the best estimate of the speed of a proton in the beam as it moves in the circle?
a. 10^-2 m b. 10^3 m c. 10^6 m d. 10^8 m e. 10^15 m </p></li>
</ol>
<p>Also, many ?s ask how many protons were emitted per minute or so with a certain field … How do you answer those?</p>
<p>Thanks.</p>
<p>can someone explain to me why work done in circular motion is 0? PLzzz</p>
<p>i think its no net work bc 1/2 is positive and 1/2 is negative so they cancel out.</p>
<p>"can someone explain to me why work done in circular motion is 0? PLzzz"</p>
<p>B/c your net displacement is 0. </p>
<p>For the second q, you want to equate Fm and Fc (centripetal)
Also, many ?s ask how many protons were emitted per minute or so with a certain field ... How do you answer those?</p>
<p>You find the total energy and divide it by the energy of one proton to get the number of protons.</p>
<p>oh and i asked another person and the ans is a and c respectively for ur question</p>
<p>One complete orbit means if you start at point A, you end up at Point A. And, work = force X displacement. Your displacement is 0 since you are at the same point where you started. you haven't moved anywhere.</p>
<p>another ?
The disk-shaped head of a pin is 1 mm in diameter
how many atoms are on the top layer of hte pinhead?
10^4
10^14
10^24
10^34
10^50</p>
<p>oh i see that now. thz stranger07! but how would u do the second part
i set F= BIL = mv^2/r </p>
<p>so F = .1 * .0015 * .2 = 10^(-27) v^2 / .1</p>
<p>is my conclusion right</p>
<p>I set f=bvq=mv^2/r</p>
<p>and i got v = 958083 m/s rounded to 1.0X10^6</p>
<p>I don't think F=bil is the right eqn to use for this question because when I think of that one, it's of induced current and such right? </p>
<p>Anyone want to look at 3 c ii from the 2004 exam and tell me how to do it. I got the right answer counterclockwise as a guess but how would you do it?</p>
<p><a href="http://www.collegeboard.com/prod_downloads/ap/students/physics/ap04_frq_physics_b.pdf%5B/url%5D">http://www.collegeboard.com/prod_downloads/ap/students/physics/ap04_frq_physics_b.pdf</a></p>
<p>hard to explain without showing u my hand... but field is going into page. so the loop is going to want to oppose that changeby making field go out of page. right hand curl for field going out of page means that thumb points to the right or counter clockwise</p>
<p>how did u get the value of q. i know that A = C/s ; but how would we know the time it takes to complete the circle.</p>
<p>as for ur prob, since the field is getting stronger, there is going to be a opposing force in the opposite direction, mainly out of the page. so by using the curl rule, u know that the current is counter clockwise</p>
<p>q= charge of a proton... 1.6x10^-19... its just a fact</p>
<p>q in the previous question? It's a proton..elementary charge of 1.6X10^-19 C (on your constant sheet btw)</p>
<p>They're asking for speed so the V is the speed when you set the two equations to equal each other. </p>
<p>Ok, for the field one..the third hand rule? Is that what I'm supposed to use? Fingers pointing into the page...but there's no velocity so that can't be the rule..can it? Which hand rule am I supposed to use? Is it the solenoid one?</p>
<p>how bout the pin ?</p>
<p>and the answer would be C to ur second question... not B unless i screwed something up which i dont think i did</p>
<p>umm as for the hand rule... i learned it as curl rule... make like a thumbs up... direction of fingers is field direction... thumb is current... wire makes to oppose that it goes in page so point fingers out, ie wrapping over top of hand and thumb points right... so current goes right</p>
<p>k on the pin one... the surface area of that pin is (3.14)(r^2)/(surface area of an atom)= (1 x 10^-10/2)^2x(3.14)</p>