Two coaxial solenoids of different radii carry current $$I$$ in the same direction. Let $$\vec{F_1}$$ be the magnetic force on the inner solenoid due to the outer one and $$\vec{F_2}$$ be the magnetic force on the outer solenoid due to the inner one. Then:

We have two long, concentric (co-axial) solenoids. Let the inner solenoid have radius $$r_1$$ and the outer one have radius $$r_2$$ with $$r_1 < r_2$$. Both carry the same steady current $$I$$ in the same sense along their windings.

For an ideal, infinitely long solenoid, the magnetic field produced by that solenoid is well known. First we state the standard formula:

For a solenoid having $$n$$ turns per unit length and carrying current $$I$$, the magnetic field at any point on its axis and anywhere inside its own cross-section is

$$B = \mu_0\,n\,I,$$

directed along the common axis. Outside the solenoid (that is, at any point whose radial distance from the axis is greater than the solenoid’s radius) the magnetic field is practically zero for an infinitely long solenoid.

Now we analyse the forces one by one.

Force on the inner solenoid due to the outer solenoid

The outer solenoid produces a uniform axial magnetic field of magnitude $$B_2 = \mu_0 n_2 I$$ everywhere inside its own radius $$r_2$$. The entire volume of the inner solenoid (radius $$r_1$$) lies inside this uniform field. A current-carrying loop (or a collection of such loops, i.e., a solenoid) placed in a perfectly uniform magnetic field experiences no net translational force because the elemental magnetic forces on opposite segments cancel pairwise.

Mathematically, for a small element $$d\vec{l}$$ of a loop carrying current $$I$$ in a uniform field $$\vec{B}$$, the differential force is

$$d\vec{F} = I\,d\vec{l}\times\vec{B}.$$

Since $$\vec{B}$$ has the same magnitude and direction at every point, integrating $$d\vec{F}$$ over the complete closed path gives

$$\vec{F} = I\oint d\vec{l}\times\vec{B} = \vec{0},$$

because the vector sum of all $$d\vec{l}$$ over a closed loop is zero. Hence, the inner solenoid experiences no net magnetic force due to the outer solenoid:

$$\vec{F_1} = 0.$$

Force on the outer solenoid due to the inner solenoid

The inner solenoid itself generates a magnetic field $$B_1 = \mu_0 n_1 I$$ only within its own radius $$r_1$$. For an ideal infinitely long solenoid, the field outside (i.e., for radial distance > $$r_1$$) vanishes:

$$\vec{B}_{\text{outside inner}} = 0.$$

Every turn of the outer solenoid lies at a radial distance $$r_2 > r_1,$$ squarely in the zone where the inner solenoid’s field is zero. So each elemental segment of the outer solenoid sits in zero magnetic field produced by the inner solenoid, and therefore

$$\vec{F_2} = I\oint d\vec{l}\times\vec{B}_{\text{outside inner}} = \vec{0}.$$

Thus, the outer solenoid also experiences no net magnetic force due to the inner solenoid.

Putting both results together, we conclude

$$\vec{F_1} = \vec{F_2} = 0.$$

Hence, the correct answer is Option B.