Question

Two coaxial solenoids of different radius carry current I in the same direction. ${\vec F_1}$ 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
(A) ${\vec F_1} = {\vec F_2} = 0$
(B) ${\vec F_1}$ is radially inwards and ${\vec F_2}$ is radially outwards
(C) ${\vec F_1}$ is radially inwards and ${\vec F_2} = 0$
(D) ${\vec F_1}$ is radially outwards and ${\vec F_2} = 0$.

Answer 1

Hint: The magnetic field is constant along the axis of a solenoid. Force on an element of a coil of the inner solenoid will be radially inward, resulting in a zero force on the coil.
Formula Used: The formulae used in the solution are given here.
$dF = i\vec l \times \vec B$ where $i$ is the current passing through the solenoid, $l$ is length, $\vec B$ is the magnetic field and $F$ is the magnetic force.

Complete Step by Step Solution: Solenoid is the generic term for a coil of wire used as an electromagnet. It also refers to any device that converts electrical energy to mechanical energy using a solenoid. The device creates a magnetic field from electric current and uses the magnetic field to create linear motion.
It has been given that two coaxial solenoids of different radius carry current $i$ in the same direction. ${\vec F_1}$ is the magnetic force on the inner solenoid due to the outer one and ${\vec F_2}$ is the magnetic force on the outer solenoid due to the inner one.
Let us assume that the axis of the solenoid is along the x-axis. Let z-axis be in the upward direction.
Consider the force on the inner solenoid, the magnetic field is constant along the axis of solenoid. This is because the force of a magnetic field is maximized when the current moves perpendicularly to the field lines, but is zero when it moves parallel to it. Therefore, we can assume that in a solenoid the magnetic field is essentially uniform or constant because the wires are nearly parallel.
The force on element 1 of solenoid coil is given as $dF = i\vec l \times \vec B$ where $i$ is the current passing through the solenoid, $l$ is length, $\vec B$ is the magnetic field and $F$ is the magnetic force.
When the element is taken on the coil, $dF = ilB \cdot \hat k$, the force will be in positive z-direction.
For diagonally opposite element 2, the magnitude of force will be the same but the direction will be negative z direction. Force on the element of any coil of inner solenoid will be radially inward which will cancel out and net result will be a zero force on the coil.
${\vec F_1} = 0$.
The magnetic field of inner solenoid which is symmetric about its axis, but not exactly along the axis of outer solenoid.
Thus, $dF = - ilB \cdot \hat j \times \left( {\hat k + \hat i} \right)$.
The radial component of force will again be zero.
Thus, ${\vec F_2} = 0$
There will be a net pull or push on the coil depending on position of coil of outer solenoid w.r.t. inner solenoid.

Hence, the correct answer is Option A.

Note: Consider the two coaxial solenoids, Due to one of the solenoid's magnetic fields at the centre of the other can be assumed to be constant.
Due to symmetry, forces on the upper and lower part of a solenoid will be equal and opposite and hence resultant is zero.
Therefore, ${\vec F_1} = {\vec F_2} = 0$.