Coaxial Solenoids: Finding Magnetic Field

[Reshma]

This one is from Griffiths.

Two coaxial long solenoids each carry current I, but in opposite directions.
The inner solenoid has radius 'a' and has 'n₁' turns per unit length.
The outer solenoid has radius 'b' and has 'n₂' turns per unit length.
Find the magnetic field $\vec B$ in three regions:
1] inside the inner solenoid
2] between them
3] outside both

My work:
I worked out the solution for these. Someone verify if my answers are correct.
General formula for magnetic field for a solenoid of 'n' turns is:
$$\vec B = \mu_0 nI \hat k$$

1] For inner solenoid:
$$\vec B = \mu_0 I n_1 \hat k$$

2]Between the solenoids:
$$\vec B = \mu_0 I n_1\hat k - \mu_0 I n_2\hat k$$

$$\vec B = \mu_0 I \left(n_1 - n_2\right)\hat k$$

3]Outside both:
$$\vec B = 0$$

 

[qbert]

not quite right.
why would the fields add between them but cancel outside both?
[especially what is the field of the "inner" solenoid?
use superposition.]

 

[Reshma]

You mean the fields superimpose at the inner solenoid and not between them?

 

[qbert]

the fields superimpose everywhere.
but the fields are only nonzero inside
the respective solenoids.

 

[Reshma]

the fields superimpose everywhere.
but the fields are only nonzero inside
the respective solenoids.

So the field inside the inner solenoid would be:
$$\vec B = \mu_0 I \left(n_1 - n_2\right)\hat k$$
& the field between them would be:
$$\vec B = -\mu_0 I n_2\hat k$$

Hope, I got it right & thanks for the help!

 

[Reshma]

No clarifications so far...then I suppose my answer is correct .