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Problems about magnetic/electric fields

  1. Jul 22, 2004 #1
    Hi, I'm self study some physics this summer, and now I'm working on some problems relate to magnetic fields and electric fields.
    And here are 5 problems that I am not sure how to do. I put some of my solution on it, hope you can check the answer if they are right or not.
    Thank you very much.
    ----------------------------------------------------------------------
    Q1:
    A hollow metal sphere of radius R carries a net electric charge Q. It is immersed in a uniform magnetic field B = B hat(y). At time t = 0 a test charge q is placed a distance r > R from the center of the sphere, on the y-axis. Qualitatively describe the motion of q for t > 0.

    A1:
    If both charges have the same sign, they will repel one another, plus the force of the magnetic field. So q will go away from Q on the y-axis direction.
    If both charges have different sign, then:
    If [itex] F_B[/itex] > [itex]F_E[/itex], q will go away from Q on the y-axis direction.
    If [itex] F_B[/itex] < [itex]F_E[/itex], q will go towards Q on the y-axis direction.

    ----------------------------------------------------------------------
    Q2:
    A rectangular loop of wire of width a and mass m hangs vertically in a uniform magnetic field B, which points out of the page. What current I is needed to produce an upward magnetic force that exactly balances the weight of the wire?

    A2:
    The current need to be in the counterclockwise direction for the magnetic force I x B on the horizontal segment in the field to point upward. Then, [itex]F_B = IBa[/itex]
    For [itex]F_B[/itex] to balance the gravitational force, we need to have [itex]I = \frac {mg}{Ba}[/itex]

    ----------------------------------------------------------------------
    Q3:(see attachments)
    A proton is fired along the axis (center) of two circular loops of wire with a speed 2e4 m/s.
    The loops carry identical 10A currents, flowing in the same direction;radius R=0.25.
    a. What magnetic force (magnitude and direction) does the proton experience at point P ? (set up will be like: loop1-distant R- point P –distant R- loop2)

    b. Suppose that proton is fired between the coils into the page. Qualifiedly describe its trajectory as it moves between the coils.
    (set up will be like: loop1-distant R- point P (into the page) –distant R- loop2)

    A3:
    a. From The Biot-Savart law, I get [itex]B(P) =\frac { \mu }{ (4 \Pi)} I \int \frac {dl x \hat r}{r^2} [/itex]; B=8.88e(-6)
    But because there are 2 same loops, and same distant away from point P, so 2 of them cancel each other out, so there is no magnetic field at point P, so there is no magnetic force.

    b. The trajectory of the proton should not change, since both loops’ magnetic field cancel out each other, so there is no magnetic force acting on the proton.

    ----------------------------------------------------------------------
    Q4:
    A 100-turn circular coil has a diameter of 2.0cm and resistance 50ohm. The axis of the coil is parallel to an external magnetic field of magnitude 1.0T. The field decreases to zero, then increases back to 1.0T, but now is pointing in the opposite direction. This process takes a total of 0.1s. Find (a) the induced emf in the wire, and (b) the induced current in the wire during the first 0.05s. (c) Sketch a plot of current vs. time. Be sure to show important values for t on your graph, and indicate the value of I at those times.

    A4
    I don’t know how to do this one. I know this is a solenoid problem, and emf = IR, but that doesn’t help me to solve the problem.


    ----------------------------------------------------------------------
    Q5:
    A circular loop of wire, radius 0.25m surrounds a solenoid, also of radius 0.25m, length 10m and 1000 loops/meter. A current I=1500A is run through the circular loop. Calculate the total magnetic flux in the solenoid.

    A5:
    I know it’s something to do with inductance.
    So, the field inside the loop of wire is constant; B=[itex]\mu N_1 I[/itex] (N = total # of loop of the wire)
    Total flux through the inner solenoid: [itex]\Phi = (\mu \Pi r^2 N_1 N_2 L) I [/itex] (L=length)
    So I get; [itex] \Phi = 1.178 [/itex]
    ----------------------------------------------------------------------
     

    Attached Files:

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    Last edited: Jul 22, 2004
  2. jcsd
  3. Jul 22, 2004 #2

    Doc Al

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    problem #1

    If Q and q are of the same sign, the test charge will move away from Q in the y-direction; if opposite sign, the test charge will move towards Q in the - y-direction. So far, so good. Now how will the magnetic field affect that motion? What's the magnetic force on that moving test charge?
     
  4. Jul 22, 2004 #3

    Doc Al

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    problem #2

    I don't quite understand the set up. Why would a vertical loop have a net force? The force on the top segment would cancel that on the bottom segment. (I'm sure I'm just missing something. How is the loop oriented with respect to the page?) Is only one of the horizontal segments in the field?

    That would be correct if only one horizontal segment was in the field.
     
  5. Jul 22, 2004 #4

    Doc Al

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    problem #3

    I assume point P is on the axis right in the middle between the two loops.

    What makes you think that the magnetic fields from each loop cancel each other?
     
  6. Jul 23, 2004 #5
    Thanks Doc Al,
    Here’re some picture for #1, 2, and 3
    For some reasons the forum disconnect me, so I have to retype the whole thing….. :frown:
    ------------------------------------------------------
    #1
    http://home.comcast.net/~chou55/pic/phys/1.gif
    The magnetic field should not affect that motion, since magnetic field never does work.
    So it’s all depending on the charge Q…..?
    ------------------------------------------------------
    #2
    http://home.comcast.net/~chou55/pic/phys/2.gif
    So the whole rectangular loop is in the magnetic field..

    ------------------------------------------------------
    #3
    a.
    http://home.comcast.net/~chou55/pic/phys/3-1.gif
    b.
    http://home.comcast.net/~chou55/pic/phys/3-1.gif

    a.
    after thinking through this problem, the magnetic fields should not cancel each other because the current on both loops is going the same direction.
    So [tex] B = \frac {\mu I}{2} ( \frac {R^2}{(R^2 + R^2)^(3/2)})[/tex]
    And [tex]F_B = \int I (dl x B) = IB [/tex]
    So the magnetic force at point P should be [itex]2IB[/itex]. And the direction should be along the center line (bule line in the picture a.) Is that correct?

    b.
    The trajectory should move toward the right side, since 2 loops have the same magnetic fields pulling the proton.
    ----------------------------------------------------------
    hope you can give me some pointers for #4, and 5 too.
    Thank you very much!
     
  7. Jul 23, 2004 #6

    Doc Al

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    more on #1, 2, & 3

    The magnetic field will not affect the motion (which is along the y-axis) because the magnetic field exerts no force. Since [itex]F = q \vec{v} \times \vec{B}[/itex], and v and B both are along the y-axis, the magnetic force is zero.

    Note: Just because a magnetic field does no work on a moving charge does not mean that it cannot effect the motion of that charge--it just means that it can't change the KE.

    Then I see no reason why the net magnetic force on the loop isn't zero.

    Much better. That will be the contribution to the field at P due to one of the loops; the net B will be twice that and directed along the axis.
    There is no current, just a moving charge. The force on that charge is [itex]F = q \vec{v} \times \vec{B}[/itex]. Since the proton velocity and the magnetic field are parallel, what will be the magnetic force on the proton?

    Figure out the direction of the magnetic force on the proton. In this case the proton velocity is perpendicular to the B field.

    I'll take a look at them when I get the chance. And welcome to PF, by the way. :smile:
     
  8. Jul 23, 2004 #7

    Doc Al

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    problem #4: Faraday's law

    For this one you need to understand Faraday's law, which describes the induced emf in a coil due to the changing magnetic flux through the coil:
    [tex]{emf} = - N \frac{\Delta \Phi}{\Delta t}[/tex]
     
  9. Jul 24, 2004 #8
    Thanks again, Doc
    ------------------------
    #2
    Now I'm a little confused..
    If the net magnetic force on the loop is zero, does that mean no matter what current we used, we cannot balance the weight of the wire??

    ------------------------
    #4
    [tex]{emf} = - N \frac{\Delta \Phi}{\Delta t} = -NA \frac {\Delta B}{\Delta t}[/tex]
    From the problem: [itex]B = 1T[/itex] and decrease to [itex]0[/itex], then increase back to [itex]1T[/itex], but pointing in the opposite direction.
    So I put [itex] \Delta B = 2T [/itex]
    Therefore, [tex]emf = -NA \frac {2T}{0.1s} [/tex]

    For #4b, We need to get the induced current when t=0.05s;
    [itex] \frac {\Delta B}{\Delta t} = \frac {2T}{0.1s}[/itex], so [itex]\frac {1T}{0.05s} [/itex]
    I use [tex]I = \frac {emf}{R} = \frac {-NA \frac {1T}{0.05s}}{50 \Omega} [/tex]

    hope I'm going the right way this time..
    still working on others...
    Thanks!
     
    Last edited: Jul 24, 2004
  10. Jul 25, 2004 #9

    Doc Al

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    Based on how you described the problem, that's what I would say. I suspect that something's missing in the problem set up.

    Looks good. Note that the problem is worded a little vaguely. The best we can do is assume that the magnetic field changes at a uniform rate, since no details were given.

    Note that the problem statement (as you gave it) talks about "during the first 0.05s" not exactly at t = 0.05s. In any case, if we assume that the field is changing uniformly throughout the 0.1 sec period the induced emf and current will be constant during that time.
     
  11. Jul 26, 2004 #10
    Thank you very much, Doc Al!
    I think I got most of them.

    So for #3a, [itex]F_B = 0[/itex]... Didn't see that coming..

    Thanks again for you help!
    Without your help, I think I won't be able to solve them for another week or two..
    Thanks!!
     
  12. Jul 26, 2004 #11

    Doc Al

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    My pleasure. Keep at it!
     
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