why is the magnetic field inside an ideal solenoid uniform


by jayman16
Tags: field, ideal, inside, magnetic, solenoid, uniform
jayman16
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#1
May3-13, 12:35 AM
P: 9
I understand that the magnetic field at points inside the solenoid is the vector sum of the B field due to each ring. How can the field inside be uniform then since if you consider just one loop in the solenoid the value of the B field is different at different points in that circle or loop.
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ModusPwnd
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#2
May3-13, 01:23 AM
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But a solenoid is not one loop right? Ideally its an infinitely long line of loops. At each point your B field is the sum of the contribution from each of the infinite number of loops.
jayman16
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#3
May3-13, 01:59 AM
P: 9
ya but at different points in each of the loops let it be the center or a point that is off center, the B is different is it not?

vanhees71
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#4
May3-13, 03:52 AM
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Thanks
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why is the magnetic field inside an ideal solenoid uniform


Just use Ampere's Law in integral form. Due to symmetry of a very long coil, ##\vec{H}## must be along the coil's axis, and you can assume it's 0 outside.

For the closed line in the integral take a rectangle with one side (length ##l##) along the axis, somewhere well inside the coil and the parallel side outside. Let there be ##\lambda## windings per unit length. Then you have, according to Ampere's Law (I neglect the signs here; you easily find the direction of the field, using the right-hand rule):
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{H}=\frac{\lambda l}{c} I,$$
where ##I## is the current through the coil. This gives
$$|\vec{H}|=\frac{\lambda}{c},$$
independent of where you locate the rectangle's side within the coil. That's why ##\vec{H}## is uniform.

You can also argue with the differential form of Ampere's Law,
$$\vec{\nabla} \times \vec{H}=\frac{1}{c} \vec{j}.$$
In cylindrical coordinates, with the [itex]z[/itex] axis along the solenoid's axis and with the ansatz due to the symmetry of the problem [itex]\vec{H}=\vec{e}_z H(r)[/itex] you find, using the formulas for the curl in cylindrical coordinates ##\vec{\nabla} \times \vec{H}=-H'(r)##. Since inside the coil there is no current density you get ##H(r)=\text{const}##.


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