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Induced Electric Fields

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A long solenoid has a diameter of 12.0 cm. When a current i exists in its windings, a uniform magnetic field = 25.0 mT is produced in its interior. By decreasing i, the field is caused to decrease at the rate of 6.30 mT/s. Calculate the magnitude of the induced electric field at the following distances from the axis of the solenoid's magnetic field.

    a.) 2.20cm
    b.) 8.20cm

    2. Relevant equations

    [tex]\int \vec E \cdot d\vec S = -\frac{d\phi}{dt}[/tex]

    3. The attempt at a solution

    I got the first part of the problem right. I can't stand this crappy WebAssign garbage homework system.

    The equation to use to find the E-field at 8.20cm should be the same one used to find the E-field at 2.20cm because those distances still reside inside the solenoid.

    [tex]\int \vec E \cdot d\vec S = ES = 2\pi rE[/tex]

    [tex]\phi = \int B \cdot d\vec A = BA = B\pi r^2[/tex]

    [tex]-\frac{d\phi}{dt} = -\pi r^2 \frac{dB}{dt}[/tex]

    [tex] 2\pi rE = - \pi r^2 \frac{dB}{dt}[/tex]

    [tex]E= \frac{-r}{2} \frac{dB}{dt} = -\frac{(.082)(-6.3x10^{-3})}{2} = 2.58x10^{-4}[/tex]

    What's wrong with this? It only depends on the radius from the center because I'm still inside the magnetic field produced by the solenoid. If I was outside the solenoid's magnetic field, I'd end up with (-R^2/2r)(dB/dt) but I'm not. so what's wrong with this??
     
  2. jcsd
  3. Nov 10, 2009 #2

    ideasrule

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    Homework Helper

    Looks fine to me. Are you sure it isn't a problem with the sign (try a negative number) o with sig figs?
     
  4. Nov 10, 2009 #3
    It most likely IS something stupid like that. I've encountered other issues with the submissions that aren't my fault too. Like, the book says [itex]\mu_o[/itex] is 4pix10^-7 = 1.26e-6... but if you use 1.26e-6, you end up with a rounding error that the software calls out to be wrong.

    This is a terrible way to administer homework. I come here thinking my answer and solution is wrong, taking over an hour to check and double dcheck it only to find out that it's RIGHT. It's insane.

    Thanks for looking it over though.

    And it can't be the negative sign. There's a negative change of the B-field, in an equation that has a -dB/dt. so it should be a positive answer.

    I tried submitting different answers (negatives, sig figs) nothing worked. I got locked out and got no credit for it. It's retarded.
     
    Last edited: Nov 10, 2009
  5. Mar 30, 2010 #4
    I see the problem.

    You put 2.58x10^-4 above but if you check your math it's actually:
    2.583x10^-4.
     
  6. Apr 21, 2010 #5
    I know its an old problem but hopefully it will help people that have this same problem in the future on webassigns.
    The problem is that for part a, you are calculating the induced electric field inside of a solenoid (r=.06m). Therefore your math is correct if the r of part b were less than the r of the solenoid.
    For part B, we are outside the solenoid. Therefore the magnetic flux is only going through that of the solenoid.
    On the left side of the equation, 2(pi)r*E should use the radius of the point at which you are measuring at. On the right side of the equation -(pi)r2 should use the r of the solenoid because that is the only area through which the B field is going.
     
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