91Ó°ÊÓ

Let's first derive the magnetic field B inside a solenoid using Ampere's Law. For an ideal solenoid with n turns per unit length, we have: \[B = \mu_0 n I\]

As both solenoids carry current in the same direction, their magnetic fields will be in the same direction inside the solenoids. We can use the right-hand rule to find the direction of the magnetic field, which will be parallel to the axis of the solenoids.

To calculate the magnetic force on a solenoid due to another solenoid, we will use the expression: \[\vec{F} = I\int\vec{dl}\times\vec{B}\] Where \(\vec{dl}\) is an infinitesimal element of the solenoid and \(\vec{B}\) is the magnetic field. Since the magnetic fields inside both solenoids are parallel to their axis and too their directions, the magnetic force acting on the outer solenoid due to the inner solenoid \(\vec{F}_{2}\) is zero. Now, we need to analyze the force on the inner solenoid due to the outer solenoid \(\vec{F}_{1}\). Considering the magnetic field generated by the outer solenoid outside its region has a radial component, but inside the outer solenoid (and also inside the inner solenoid), it is nearly parallel to the solenoid's axis. Thus, the force \(\vec{F}_{1}\) will also have no radial component, and it will be zero. With this information, we can see that the correct option is: (D) \(\vec{F}_{1}=\vec{F}_{2}=0\).