What Force is Needed to Pull a Solenoid in a Magnetic Field?

[Crotos]

Homework Statement

Two solenoids are placed in a cartesian xy-coordinate manner with their bottoms pointing in the negative y-direction, and their tops pointing in the positive y-direction. The solenoids are placed a distance x from each other (on the x-axis).

The solenoids are identical, each having length L, diameter d, N turns, resistance R and negligible self-inductance L.

The left-most solenoid has current I running through it.

Find the force, as a function of distance x between the solenoids, that is required to pull the right-most solenoid away from the left, with a constant velocity.

Homework Equations

I am not sure.

The Attempt at a Solution

My understanding is that the left-most solenoid having current running through it will produce a magnetic field, drawing the right-most solenoid towards it. I'm looking for a force that exceeds this magnetic force in magnitude, in order to get the solenoid moving in the +x-direction.

It makes sense to me that the force needed should be less for increasing distance from the left-most solenoid, since the magnetic field from it is weaker when moving away from it.

However, I can't seem to find a good expression for the magnetic field, radially outward from the left-most solenoid. All I find on the internet is some crazy formula containing "elliptical integrals of 1st and 2nd order"..

I'm starting to think I'm looking at the problem the wrong way.

Help! =)

 

[Simon Bridge]

Hint: think induction.

 

[Crotos]

Hint: think induction.

Hi Simon and thank you for your reply! I gave your hint a shot, and this is what I came up with:

Since the coil is of N turns, the induced emf in the coil running current through it should be given by

ε = -N(dø/dt).​

ε is also given by Ohm's law, ε = IR, meaning

IR = -N(dø/dt) ⇔ dB/dt = -IR/(NA)​

since the cross-sectional area of the coil is constant I took it out of the derivative. To clarify, A is the cross-sectional area I'm talking about,

A = (d/2)²π → dB/dt = -IR/(N(d/2)²π).​

From this I think it's pretty fair to say that dB/dt is constantly decreasing, meaning B(t) is decreasing linearly.

Is this decrease along the symmetry axis of the solenoid though, or is it actually appliable to the radial axis, the axis I think I'm looking to get the B-field for?

How does this help me in terms of finding the force required to push the right-most solenoid away from the left-most? I don't see the connection force-induction.

 

[Simon Bridge]

Sketch the situation... How would you normally treat that?
I'd try either treating as two magnets or use work-energy.
You are right... You need the radial field or some approx. I'm thinking you have a set of notes to deal with that.

 

[paranormal]

You are correct in your understanding that the left-most solenoid will produce a magnetic field that will attract the right-most solenoid towards it. To find the force between two solenoids, we can use the equation F = μ0N1N2I1I2A/d, where μ0 is the permeability of free space, N is the number of turns, I is the current, A is the cross-sectional area, and d is the distance between the solenoids.

In this case, both solenoids have the same number of turns, current, and cross-sectional area, so we can simplify the equation to F = μ0N^2I^2A/d. The force will vary with distance x, as you mentioned, and can be written as F(x) = μ0N^2I^2A/x.

To find the force required to pull the right-most solenoid away from the left with a constant velocity, we can use the equation F = ma, where m is the mass of the solenoid and a is the acceleration. Since we want the solenoid to move with a constant velocity, the acceleration will be zero and the force required will also be zero. This means that the magnetic force produced by the left-most solenoid will be enough to keep the right-most solenoid in place.

I hope this helps clarify the problem for you. Remember to always start with the basic equations and simplify as much as possible before trying to solve a problem. Good luck!