# Coaxial solenoids (Griffiths)

Given the two coaxial solenoids in the image with the inside solenoid (of radius $a$ and $n1$ turns)having a current $I$ and the outside solenoid (of radius $b$ and turns $n2$) a current $2I$, in opposite directions. Find the magnetic field a) inside the small solenoid, b) between the two solenoids, c) outside

The field of a coil is only axial (from the symmetry of the coil). (In example 5.9 this is discussed in detail)

Consider one has the 3 Ampere loops from the figure above. On loop 1:

$∮ B*d L=0$ since there are no currents enclosed

$B_1*L-B_2*L=0$ so that $B_1=B_2$ in all points of space outside the coil

But $B_{outside}→0$ at infinite distance so that $B_{outside}=0$ in all points

On loop 2:

$∮ B d L=μ_0$ or $(I_{red}-I_{black} )=μ_0((n_1*L-n_2*L)*I$

$B_3 L+0+0+0=μ_0 (n_2-n_1)L*I$ so that $(overrightarrow{B}_3 =μ_0 (n_2-n_1)I*hat z$

On loop 3:

$∮ B d L=μ_0*I_{black}$ or $B_4 L+0+0+0=μ_0*n_2 L*I$ or $(overrightarrow{B_4} =μ_0*n_2 I*hat z$

Observation: the field follows the rule of right screw; the screw is rotating as the current and advances in the direction of filed.