Recent content by BunDa4Th

  1. B

    Find Magnetic Field of Parallel Conductors at Different Locations

    Well this is what I did but not sure if it is correct. B1 = UoI1/2pid = 4pi x 10^-7/2pi(.1) = 2 x 10^-5 B2 = 2 x 10^-5 B1 + B2 = Bp = 4 x 10^-5 b) B1 = 1 x 10^-5 and B2 = 1.5 x 10^-6 B1 distance is .10 m and B2 distance is .20 m then i did B1 - B2 = Bp1 = 5 x 10^-6 c) B1 = UoI/2pi.30m =...
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    Find Magnetic Field of Parallel Conductors at Different Locations

    Okay, I figure out how to do this. For A i was suppose to find B1 and B2 then add them and it is going out of the page i believe. For B it is going into the page and C is out of the page. I didn't realize that I had to find the two B field. I was only thinking of one bar at a time when doing...
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    Find Magnetic Field of Parallel Conductors at Different Locations

    Homework Statement The two wires shown in Figure P19.38 carry currents of 5.00 A in opposite directions and are separated by 10.0 cm. Find the direction and magnitude of the net magnetic field at the following locations. http://www.webassign.net/sf5/p19_38.gif (a) at a point midway...
  4. B

    Radius of the path of a charged particle in a mass spectrometer

    Thanks for all the help and explanation and I was finally able to get the answer correctly.
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    Radius of the path of a charged particle in a mass spectrometer

    Okay, i finally got to working on this problem again with another classmate and came to this. I am not sure if this is correct but this is what i came up with Fm = Fe so, qvB = qE V = E/B so I found the constant velocity which is 850/.910 = 934.07 m/s after it travels into the reflector...
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    Radius of the path of a charged particle in a mass spectrometer

    I decided to look at the equation again and look at this post again and now I am seriously confuse. I think I am going to find a tutor tomorrow to help me but I was told the physics tutor are not very helpful but I am going to give it a try anyways. BTW teacher emailed me back that she is not...
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    Radius of the path of a charged particle in a mass spectrometer

    each time i read this i get more confuse. I think I will wait till Tuesday to give this another try since I emailed the teacher and she said she will go over this with the class since I asked people in my class and they are all stuck at it also. Thanks for the help and explanation. I will be...
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    Radius of the path of a charged particle in a mass spectrometer

    I am still a bit confuse and not really sure how i would draw a diagram for this. Today class did not even cover this either. do i use this and solve for v? qvB = q(vB + E)? I know i just removed the x. Thanks for the help so far, but when it comes to physics it takes me forever to understand it.
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    Radius of the path of a charged particle in a mass spectrometer

    The formula given is a bit confusing. If I use F = q(vxB + E) I am still stuck because I do not know what F is. Would I have to set two equation together? like: qvB = q(vxB + E)? I just realize that will not work either since I do not know what x is. Also I don't think we ever cover the...
  10. B

    Radius of the path of a charged particle in a mass spectrometer

    Homework Statement Consider the mass spectrometer shown schematically in Figure P19.30. The electric field between the plates of the velocity selector is 850 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.910 T. Calculate the radius...
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    Calculating Total Energy Stored in Parallel Capacitors

    Two capacitors, C1 = 25 µF and C2 = 6.0 µF, are connected in parallel and charged with a 150 V power supply. (a) Calculate the total energy stored in two capacitors. energy stored = 1/2C(deltaV)^2 or Q^2/2C I been doing this and its marking me wrong. first i found both charges which...
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    Electric Field at Point 1 cm Left of Middle Charge (-3.5 µC Force)

    Its not so confusing but the given formula you wrote is new to me, since i have not seen that formula before. I will give this problem another try hopefully when i get home today or during the weekend. Thanks for the help and explanation.
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    Electric Field at Point 1 cm Left of Middle Charge (-3.5 µC Force)

    do i just find each individual like: 8.99e9 (6e-6)/.01^2 = 5.394e8 K(1.5e-6)/.01^2 = 1.349e8 k(-2e-6)/.01^2 = -1.798e8
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