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Trip1
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Homework Statement
A solenoid has a radius of 2mm and a length of 1.2cm. If the # of turns per unit length is 200 and the current is 12A, calculate the magnetic flux density at a) the center and b) the ends of the solenoid
Homework Equations
The biot-savart law:
[itex]\vec{B} = \frac{\mu_0}{4 \pi} \ \int \frac{ I \ \vec{dl}\times \hat{r}}{r^2} \ \text{ or } \ \frac{\mu_0}{4 \pi} \ \int \frac{ I \ \vec{dl}\times \vec{r}}{r^3}[/itex]
More importantly, my textbook has taken me through a few examples to wind up with the following general results for a circular loop of current with radius b.
In general, on the axis of a current-carrying loop:
[itex]\vec{B} = \frac{\mu_0 \ I \ {b^2}}{2 \ ({{b^2} + {z^2}})^{3/2}} \hat{z}[/itex]
By setting z=0, we can obtain the magnetic flux density at the center of the loop as
[itex]\vec{B} = \frac{\mu_0 \ I}{2 \ b} \hat{z}[/itex]
I believe i am expected to use only these equations to solve the problem.
The Attempt at a Solution
We have 200 turns per unit length, thus we have a total of N = (200)(1.cm) = 2.4 turns
I'm guessing what I have to do is sum the B from each loop / turn. I know that for the loop that is in the plane of the center, the magnetic flux density is given by
[itex]\vec{B} = \frac{\mu_0 \ I}{2 \ b} \hat{z}[/itex]
and that for all the others, the magnetic flux density is given by
[itex]\vec{B} = \frac{\mu_0 \ I \ {b^2}}{2 \ ({{b^2} + {z^2}})^{3/2}} \hat{z}[/itex]
where b is the radius and is given.
I don't know what to put in for z in the above equation. Furthermore I'm not sure what to do for part b).
I think there must be a simplifying assumption somewhere. Either that or potentially somehow use the general equation for the Biot-Savart law and integrate along the path of the helix
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