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Some more magnetic field/induction questions

  1. Mar 23, 2008 #1
    1. The problem statement, all variables and given/known data

    Problem 33.74
    The switch in the figure has been open for a long time. It is closed at t = 0s.

    PART A - What is the current through the 20[tex]\Omega[/tex] resistor immediately after the switch is closed?

    PART B - What is the current through the 20[tex]\Omega[/tex] resistor immediately after the switch has been closed for a long time?

    PART C - What is the current through the 20[tex]\Omega[/tex] resistor immediately after the switch is reopened?

    Problem 32.62

    An electron travels with speed 0.900 x 10^7 m/s between the two parallel charged plates shown in the figure. The plates are separated by 1.0 cm and are charged by a 200 V battery.


    What magnetic field strength (and direction) will allow the electron to pass between the plates without being deflected

    Problem 32.66

    The uniform 30.0 mT magnetic field in the figure points in the positive z-direction. An electron enters the region of magnetic field with a speed of 5.30 x 10^6 m/s and at an angle of 30 degrees above the xy-plane


    PART A - Find the radius (in mm) of the electron's spiral trajectory.

    PART B - Find the pitch of the electron's spiral trajectory.

    Problem 32.74

    A bar magnet experiences a torque of magnitude 0.075 Nm when it is perpendicular to a 0.50 T external magnetic field. What is the strength of the bar magnet's on-axis magnetic field at a point 20 cm from the center of the magnet?

    2. Relevant equations

    I = V/R

    [tex]\Delta V = w v_d B[/tex]

    [tex]r = \frac{mv}{qB}[/tex]

    [tex]\tau = \vec{\mu} B sin\vartheta[/tex]

    [tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3}[/tex]

    3. The attempt at a solution

    Problem 33.74

    PART A

    I = V/R = 30/20 = 3/2 A

    PART B
    Same has part A

    PART C

    Since there is an inductor in the circuit, the current would still be there if the switch has just been opened - so same as the other two parts.

    Problem 32.62

    [tex]\Delta V = w v_d B \Rightarrow B = \frac{\Delta V}{w v_d} = \frac{200}{(0.01)(0.9*10^7)} = 2.22*10^-4[/tex]

    For the direction, using the right hand rule that related magnetic field and speed; since the index finger points to the right, the middle finger (which is perpindicular to the index finger) is pointing out of the page.

    Problem 32.66

    PART A

    [tex]r = \frac{mv}{qB} = \frac{(9.11*10^-31)(5.3*10^6)}{(1.6*10^-19)(30*10^-3)} = 10^-3[/tex]

    PART B

    I've got no idea. I can't even find pitch in my textbook.

    Problem 32.74

    since its perpendicular, sin90 = 1 so:

    [tex]\tau = \vec{\mu} B \Rightarrow \vec{\mu} = \frac{\tau}{B} = \frac{0.075}{0.5} = 0.15[/tex]

    [tex]B = \frac{\mu_0}{4 \pi} \frac{2 \vec{\mu}}{z^3} = 10^-7 \frac{2(0.15)}{0.2^3} = 37.5 T[/tex]

    Any help would be appreciated. Thank You.
  2. jcsd
  3. Mar 24, 2008 #2


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    I am going to let you know something about how things work in the Homework Forum. It is a better idea to post each problem in its own individual thread than to post a stack of problems all in one long one. Most of the helpers here do this (unpaid...) work in bursts and brief visits, so a thread with four problems in it is going to discourage just about everyone from starting it, if there are other people around asking for less to look at. Just a suggestion...

    So I'm going to look at what I can for the moment and will have to come back to do more later. (Just so you know...)

    Look at this circuit again. When the switch is first closed, that inductor is just going to block current flow down that branch. What is the total resistance that the current will run through at that moment?

    Once the circuit with the switch closed has reached its steady-state (equilibrium) condition, the ideal inductor passes current as if it has no resistance (this is a "physics problem" inductor, not a real one...). So which branch is the current going to run through by that time? How much total resistance in the circuit does the current encounter now?

    The inductor is now happily passing current freely and will not readily change from doing that. So when the switch is first opened again, will anything be different at that moment from Part B? (Your idea is right, but your answer to B is wrong.)

    I would hardly call 9·10^6 m/sec a drift velocity, and I presume you're calling the plate separation w, but this result is correct. (The units are

    V-sec/(m^2) = J-sec/C-(m^2) = N-m-sec/C-(m^2) = N-m/A-(m^2)
    = N/A-m = T .)

    The electric force on the electron is upward, so the magnetic force on the electron must act downward. Since the electron is negative, that means we want v x B to be upward. If the electron is traveling to the right in the diagram, which way must B point? (Be careful about applying the EM force rules. They are always defined for positive charges -- a favorite exam problem trap is to make the particle an electron, so students will forget to reverse the directions of the forces...)
    Last edited: Mar 24, 2008
  4. Mar 24, 2008 #3


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    Another warning here -- avoid posting images or using images on sites that require a user to log in. A typical forum user often has too many $&(%*! log-ins and passwords to keep track of already; they will be very reluctant -- or flat-out refuse -- to join yet one more site, especially one they don't plan to use themselves, just to look at one image. Is there some way you could attach the image for this problem?

    I will take your word for the values in part A, in the meantime, and say that the equation for r and the result look all right.

    As for part B, you were concerned about the definition of pitch angle. You probably already realize that the electron is going to travel on a helix, meaning it will climb at some constant axial speed while also circling the axis of the helix at a constant tangential speed. If you think about the velocity vectors involved in this motion, the axial velocity is always perpendicular to the instantaneous tangential velocity. The pitch angle [tex]\alpha[/tex] will be given by

    [tex]tan \alpha[/tex] = axial speed / tangential speed .
    Last edited: Mar 24, 2008
  5. Mar 24, 2008 #4
    Thanks for the tip. I thought it would be easier if all the problems were together, but your logic makes more sense, and I would do that in the future.

    and the image for 32.66 is:


    posted the url to the actual assigment by mistake

    so if the magnetic field is down, and the velocity is to the right, it would be out of the page, but sence this is an electron, it would be the opposite direction, hence out of the page?
    Last edited: Mar 24, 2008
  6. Mar 25, 2008 #5
    to find the pitch ... took a while to figure this out.

    1st find the frequency (f)
    f = qB / 2(pi)m ,
    all these are known q=charge of electron, m=mass of electron B=electric field in T (dont forget to convert from mT)

    2nd find the T (time)
    T= 1/f

    3rd find the velocity in the vertical direction
    V=vsin() where v=velocity they give you, and angle=30

    4th use d=VT
    where V and T is what you calculated, and d is the pitch (or the height it takes for 1 complete "revoution")
  7. Mar 25, 2008 #6
    problem 33-74 (a) ----

    This is funny, but it may help somebody here. I once had a college instructor who made us repeat this like a chant: The voltage across a capacitor cannot jump instantaneously, but its current can jump. The current through an inductor cannot jump instantaneously, but its voltage can jump. Say it! Say it again!
  8. Mar 25, 2008 #7
    So for part A where I have to find the radius, I inputed 1 mm (since 10^-3 m = 1 mm) and it says I'm wrong.

    this is what I got for part B:

    [tex]f = \frac{qB}{2 \pi m} = \frac{(1.6*10^{-19})(30*10^{-3})}{2 \pi 9.11*10^{-31} } = 838577087.6[/tex]

    [tex]T = f_{-1} = 10^{-9}[/tex]

    V = vsin30 = 5.3*10^6 = 2650000

    d = VT = (2650000)(10^-19) = 0.00265m and since the answer needs to be in mm = 2.65 mm

    did I make any mistakes?

    So since the current cannot instantly jump in an inductor when the switch has been closed, then the current at that point would be 0?
    Last edited: Mar 25, 2008
  9. Mar 26, 2008 #8


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    We're kind of getting all over the place here. This is for Problem 32.62:

    The electron would be deflected upward toward the positive plate by the electric field; if it is not to be deflected, the magnetic force supplied must provide an equal downward force. The magnetic force is given by qvB , since everything is on perpendiculars in the problem, so the sine term in the cross product is just equal to 1. But the electron is negative, so the vector for the magnetic force is given by

    F_mag = -e(v x B).

    This means that the magnetic force would point in the opposite direction of the vector v x B . Since we know the direction of v and the direction we want F_mag to be, we can approach this in a couple of way. One would be to point the right-hand thumb to the right for v and turn the palm so it faces downward; the fingers will point toward you or "out of the page". But this is an electron we're dealing with, so we have to reverse the direction of B; thus, the magnetic field has to be away from you or "into the page".
  10. Mar 26, 2008 #9


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    OK, this is for Problem 33.74(A):

    Yes, since an inductor exploits Faraday's Law and resists changes in its internal magnetic field, if there is no field there to start with, it will fight the creation of one. Thus, initially, it will (ideally) pass no current, so it acts like a break in the circuit. (The "counter-emf" nearly cancels the applied current.)

    What I was saying earlier is that you did not count all the resistances that the current will encounter in returning to the battery...
  11. Mar 26, 2008 #10
    So when the switch is immidatly closed, the total current in the 20 [tex]\Omega[/tex] would be 0?
  12. Mar 26, 2008 #11
    Just the opposite line of reasoning. Immediately after the switch is closed, ALL of the current that goes though the 10 [tex]\Omega[/tex] is forced to go through the 20 [tex]\Omega[/tex] also.
    Last edited: Mar 26, 2008
  13. Mar 26, 2008 #12


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    This is for Problem 32.66:

    A) Ah, be careful about what you use in the equation! You have

    [tex]r = \frac{mv}{qB} = \frac{(9.11*10^-31)(5.3*10^6)}{(1.6*10^-19)(30*10^-3)} = 10^-3[/tex]

    The formula is right, but keep in mind that the magnetic force only involves the component of the electron's velocity that is perpendicular to the magnetic field. The component parallel to the field has no effect. So the speed you should be using in the gyration radius equation is

    v cos 30º = 4.59·10^6 m/sec .

    B) Harvardd's method is correct, but a bit roundabout. The period of gyration of the electron is indeed

    [tex]T = \frac {2\pi m}{eB}[/tex]

    which is equal to

    [tex]\frac {2\pi r}{v_{perp}}[/tex] ,

    which is just the circumference of the "orbit" divided by the velocity component of the electron perpendicular to the magnetic field. So the "pitch" of the helical path of the electron will be

    [tex]d = \frac {2\pi m}{eB} v sin \theta[/tex] ,

    but, since it's also equal to

    [tex]d = \frac {2\pi r}{v_{perp}} v sin \theta[/tex] ,

    we can also write

    [tex]d = \frac {v_{para}}{v_{perp}} 2 \pi r[/tex] .

    In other words, the pitch is just the ratio of the parallel and perpendicular velocity components of the electron times the circumference of the circle. As I mentioned earlier, this velocity component ratio is the tangent of the "pitch angle" of the helix, which is the same as the angle that the electron's entry velocity makes to the perpendicular to the magnetic field direction (or the complement of the angle between the entry velocity and the direction of B). The magnetic force does not change this angle, so the pitch angle is just 30º.

    So if the question is asking for the pitch angle, the statement of the problem already gives that to you. If it is asking for the pitch according to the definition Harvardd gives, it will just be (tan 30º)·(circumference of the gyration).
    Last edited: Mar 26, 2008
  14. Mar 26, 2008 #13
    With switches, it's better to refer to "immediately before" and "immediately after", and not say "when", which is ambiguous.

    Whenever you say "the current" you have to indicate which branch is meant.

    Immediately after the switch is thrown, the current through the inductor will be zero. The current coming out of the voltage source will NOT be zero, because it has another path that it can take without going through the inductor.
  15. Mar 26, 2008 #14


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    Quite so -- the conditions this kind of problem refers to are "instantaneous", so it has to refer to a situation that occurs "a moment" (an infinitesimal amount of time) after the circuit is changed by opening or closing a switch. (These are mathematical idealizations, of course. The physical behavior for real switches and circuit changes is rather more complicated...)
    Last edited: Mar 26, 2008
  16. Mar 26, 2008 #15
    so the radius r =

    [tex]r = \frac{mv}{qB} = \frac{(9.11*10^{-31})[(5.3*10^6)cos30]}{(1.6*10^{-19})(30*10^{-3})} = 8.7 * 10^{-4} m = 0.87 mm[/tex]

    [tex]v_{para} = 5.3*10^6 * sin30 = 2650000[/tex]

    [tex]v_{perp} = 5.3*10^6 * cos30 = 4589934.64 [/tex]

    [tex]d = \frac {v_{para}}{v_{perp}} 2 \pi r = (2 \pi * 8.7 * 10^{-4}) \frac{2650000}{4589934.64} = 0.00315 m[/tex]

    so the total resistence the current passed is 30 [tex]\Omega[/tex] which would make I = V/R = 30/30 = 1 A?
  17. Mar 26, 2008 #16


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    That looks like what I got, using both my method and Harvardd's.

  18. Mar 26, 2008 #17
    PROBLEM 33-74 PART A:
    Your number is correct. That's the starting point of an exponential current function (I as a function of time) that begins right after the switch is closed.

    PROBLEM 33-74 PART B:
    The next thing to consider is: what value does that exponential decay of the current function move toward asymptotically (and actually reach only at time infinity)? The rules for transients circuits that you want to memorize is: After a "very long time" after the throwing of switches, capactors turn into open circuits (zero current through them) and inductors turn into short circuits (zero voltage across them). After a very long time the inductor will be just a piece of conducting wire. So now consider this circuit when time is infinity. The current from the battery reaches a junction where it could conceivable split into two branches, one branch being a 20 ohm resistor, and other branch being an inductor that after a "long time" has turned into a short (perfect conductor). How will that current divide between those two branches?
    Last edited: Mar 26, 2008
  19. Mar 26, 2008 #18
    Once the switch has beel closed for an time t = infinite, the current would pass through the circuit with no resistence, would that make the current I = V voltage?
  20. Mar 26, 2008 #19


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    This looks to be correct, the units being A·(m^2).

    I'd been holding off on this one because I thought the magnetic field strength looked (phenomenally) high and there must be something wrong with the analysis. (But that doesn't seem to be the case.)

    I believe the magnetic dipole equation is correct, but please look at the numbers you've used and what you've written for an answer...
    Last edited: Mar 26, 2008
  21. Mar 26, 2008 #20


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    Is the inductor the only component the current encounters before returning to the battery?

    BTW, if there were no resistance, Ohm's Law would tell you there'd be an infinite current...
    Last edited: Mar 26, 2008
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