# Magnetic field between two concentric inductors

Suppose we have two inductors with $n_1$ and $n_2$ turns, of radii $R_1$ and $R_2$ $(R_1 > R_2)$ respectively and length $l$. They are aligned concentrically and the smaller inductor is wholly inside the bigger one (i.e. the smaller one doesn't stick out). The currents flowing are $I_1, I_2$ respectively (they might be of different signs to indicate different direction of the current).

The magnetic field inside both indutors (inside the smaller one) is of course homogeneous and equals $B = B_{1_{in}} + B_{2_{in}} = \frac{\mu\mu_0}l(I_1n_1 + I_2n_2)$. But why is the field between these inductors $(r \in (R_2, R_1))$ homogeneous as well and what's its induction value equal?

• Because the magnetic field due to a solenoid, outside is zero so in that region there is only magnetic field of the bigger solenoid and which is assumed to be homogenius. – Paul Jan 9 '15 at 6:21

Given the simplification that the magnetic field produced from a solenoid is zero when outside the coils of that solenoid, you can say that there is no magnetic field from the interior inductor in that space. Therefore it is as if that interior solenoid is not there. Assigning the variables to the larger, exterior solenoid with a subscript 1, you get $B = B_{1_{in}} = \frac{\mu\mu_0}l{I_1n_1}$ when $R_2<r<R_1$.