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Charge through a coil in a magnetic field

  1. Jun 24, 2012 #1
    1. The problem statement, all variables and given/known data
    At a certain place, the Earth's magnetic field has a magnitude B = 59 μT and is inclined downward at an angle of 70º to the horizontal. A flat horizontal circular coil of wire with a radius of 13 cm has 950 turns and a total resistance of 85 Ω. It is connected to a galvanometer with 140 Ω resistance. The coil is flipped through a half revolution about a diameter, so it is again horizontal. How much charge flows through the galvanometer during the flip? (Hint: See Problem 11.)

    2. Relevant equations
    The "Problem 11" referred to in the question concludes that, if a circular loop is connected to a resistance R, this configuration is perpendicular to a magnetic field B, Φ(0) is the flux in the loop at time t = 0, and the magnetic field B varies in a continuous but not specified way, so that at time t the flux is represented by ΦB(t), then the net charge q(t) that has passed through resistor R in time t is:
    [tex]q(t)=\frac{1}{R}[\Phi_B(0)-\Phi_B(t)][/tex]

    3. The attempt at a solution
    If the circular coil is horizontal, and Earth's magnetic field at that point is inclined downward at an angle of 70º to the horizontal, then angle between the plane of the coil and Earth's magnetic field is 70º. So, if we adopt the normal to the plane of the coil to be downward, then the angle θ between the normal to the plane and the magnetic field is 160º. So, in its initial position, the magnetic flux through the surface of the N loops should be:
    [tex]\Phi(0)=NAB\cos{160º}[/tex],
    where [tex]A=\pi r^2[/tex] is the area of a loop.
    Thus, [itex]\Phi(t)[/itex] would be the negative of the above, since the loop is 180º from its original inclination.
    Also, I think that the resistance R is 140 Ω + 85 Ω = 225 Ω.
    So, the result would be:
    [tex]q(t)=\frac{1}{R}[NAB\cos{160º}-(-NAB\cos{160º})] = \frac{1}{R}[2NAB\cos{160º})][/tex]
    where R = 85+140 = 225 Ω, A = 13 cm, N = 950, A = π*13² cm², B = 0.000059 T.
    This gives approximately 24.86 μC.
    Does this reasoning seem correct?

    Thank you in advance.
     
  2. jcsd
  3. Jun 24, 2012 #2

    BruceW

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    The question says that the magnetic field is inclined downward at an angle of 70º to the horizontal. So surely if we adopt the normal to the plane of the coil to be downward, then the angle between the normal and the magnetic field would be 30º. In other words, the normal to the coil plane is straight down, and the magnetic field is downwards, but not straight down, which is why the angle between them is less than 90º
     
  4. Jun 24, 2012 #3

    BruceW

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    That stuff about the angle should only cause the sign of the final answer to be different, which doesn't really matter anyway because we don't care which way the charge is going. The rest of your working looks good, I'm going to check through to see if I get the same (numerical) answer.

    EDIT: Actually no, the angle does matter. If the angle is what I think it is (30º) then that will give a different answer to your 160º because cosine of these two angles will not give the same value. (The reason that I thought that only a sign change would result is because somehow I thought that 180-30=60, bad arithmetic on my part).
     
    Last edited: Jun 24, 2012
  5. Jun 24, 2012 #4

    BruceW

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    I've checked it through, and I'd say your working is good apart from I think the angle is not correct, which then affects the final answer.
     
  6. Jun 24, 2012 #5
    Thank you for checking it.
    About the angle, I think that it shouldn't matter, because, if I choose the normal to the coil to be downward instead of upward (which, I think, shouldn't matter), then the angle between the normal and B would be (90º - 70º) = 20º, and cos(20º) = -cos(160º) ≈ 0,94. So, I guess the way the charge goes through the galvanometer doesn't matter here.
     
    Last edited: Jun 24, 2012
  7. Jun 25, 2012 #6

    BruceW

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    Yes, you're totally right. I made another arithmetic error in thinking that 90-30=70 which is why I thought it should be 30 degrees. Sorry about that. I must have been more tired than I realised to have made two arithmetic errors that evening.
     
  8. Jun 25, 2012 #7
    No problem, thank you for confirming it.
     
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