# Why is the magnetic field inside an ideal solenoid uniform

by jayman16
Tags: field, ideal, inside, magnetic, solenoid, uniform
 Sci Advisor Thanks P: 2,305 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 $z$ axis along the solenoid's axis and with the ansatz due to the symmetry of the problem $\vec{H}=\vec{e}_z H(r)$ 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}##.