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Homework Help: B field at the center of a large charges sheet

  1. Jul 2, 2011 #1
    1. The problem statement, all variables and given/known data

    In a plastic film factory, a wide belt of
    thin plastic material is traveling between two successive
    rollers with the speed v. In the manufacturing process, the
    film has accumulated a uniform surface electric charge
    density σ. What is B near the surface of the belt in the
    middle of a large flat span?

    2. Relevant equations

    B(x) = [itex] \frac{\mu I}{4\pi}\int \frac{dx' \times (x-x')}{|x-x'|^{3}}[/itex]

    3. The attempt at a solution

    I tried to calculate the integral, starting with assuming that the sheet can be represented via cylindrical coordinates - to better use the r and sin(theta) - a result from the cross product.
    But then I've got in the integrand [itex]\frac{sin(\theta)}{r} and then solved the dr part with log(r)|{inf to 0} and couldn't progress from there, so I suppose I made an error somewhere but can't find it.
  2. jcsd
  3. Jul 2, 2011 #2


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    Science Advisor
    Gold Member

    Since you are in the middle of a large flat span of uniform charge density, it is perhaps simpler to apply Ampere's law than try to brute force it with Biot-Savart.
  4. Jul 3, 2011 #3
    Maybe you are right, but the lecturer pointed out specifically to use Biot-Savart on this one...:yuck:
  5. Jul 3, 2011 #4
    I think it would be right, if you first find the electric field. Then use Maxwell's equation to find the B.
  6. Jul 3, 2011 #5
    I approached him and asked it today, he said that he expects us to solve it only by Bio-Savart...
  7. Jul 4, 2011 #6

    Philip Wood

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    Gold Member

    Regard the moving charged sheet as currents running in parallel wires. A strip of width [itex]\Delta[/itex]w will be equivalent to a wire carrying current [itex]\sigma[/itex] [itex]\Delta[/itex]w v. Agreed?

    Now, using the B-S law to find B due to a long straight wire is a standard derivation. I expect you've learnt it already. [The result can be derived in one line from A's Law.]

    So now you need to integrate the fields at a 'central point', due these strips of the sheet. A diagram is essential here. You'll need to use the right hand grip rule to get the direction of these fields, and then you'll need to add the fields as vectors. This isn't as hard as it sounds because field components perpendicular to the sheet cancel. Again, your answer can be checked in one line using A's Law.
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