Finding the force of a magnetic field using an air core solenoid

In summary: Let me know if you have any further questions. Good luck with the science fair! In summary, a homeschooling parent is seeking help for their son's science fair project involving an air core solenoid shock absorber for a bike. They are trying to understand the forces and efficiency of the device and ask for guidance in calculating these values. The magnetic force produced by a solenoid can be calculated using the formula F = BIL, and the efficiency can be calculated using the formula Efficiency = Output power / Input power. The N48 magnet can be taken into account by adding its force to the force produced by the solenoid. The magnetomotive force (MMF) can
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wingnutto
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Homework Statement



Ok, I need help, but it is for a project my son is working on. I am homeschooling him (13) and he is entered in a local science fair. He has designed an air core solenoid shock absorber for a bike. It more or less works, and we are working our way through the forces generated and trying to come up with some predictive formulas to explain why it works and how to scale up. I am at the extent of my capacity, so maybe someone here can lend a guiding hand.

We have an air core solenoid, 22 ga. magnet wire, about 200 turns, 1.5 cm diameter, 10cm axial length. At 18v it's pushing about 12amps, and yes, it gets hot fast.

One of the things we are trying to understand is efficiency. I'd like to show him how to calculate the force this magnetic field can produce at the cost of how much power (216 Watts here - with a lot being wasted in heat). Can you express magnetic force in terms of Newtons? We know if we let it suck an N48 3/4x1.5" magnet, it's pulling with a force of about 14 Newtons for about a centimeter, but this complicates it I think by adding the force of the magnet into the equation.

My first thought was to calculate the MMF of the solenoid in ATs, but this appears to dead end. I have yet to find a way to translate AT into anything useful.

I have Bio-Savart, Lorentz, Webers and Henries swimming around in my head. Any suggestions on how to make sense of this would be most welcomed!

Homework Equations



My question exactly.


The Attempt at a Solution



Haven't gotten that far. I knew Amps*turns would be too simple!

Thanks in advance.
 
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Thank you for reaching out for help with your son's science fair project. It sounds like you and your son have already made some great progress in designing and testing the air core solenoid shock absorber for a bike. I would be happy to offer some guidance and suggestions to help you understand the forces and efficiency of your device.

First, let's start with some basic principles. The magnetic force produced by a solenoid can be calculated using the formula F = BIL, where F is the force in Newtons, B is the magnetic field strength in teslas, I is the current in amps, and L is the length of the solenoid in meters. In your case, since you are using an air core solenoid, the magnetic field strength will depend on the number of turns and the current, which you have already provided.

To calculate the efficiency of your device, you will need to measure the input power (in watts) and the output force (in Newtons). The efficiency can then be calculated using the formula Efficiency = Output power / Input power. This will give you a percentage that represents how much of the input power is being converted into useful output force.

As for the N48 magnet, you are correct that it will add additional force to the equation. However, it can be taken into account by simply adding its force (14 Newtons) to the force produced by the solenoid. So, if the solenoid is producing 20 Newtons of force and the magnet is adding 14 Newtons, the total force would be 34 Newtons.

In terms of your question about translating Amps*turns into something useful, this is commonly referred to as the magnetomotive force (MMF) and is measured in ampere-turns (AT). You can use this value to calculate the magnetic field strength (B) using the formula B = μ0μrNI/L, where μ0 is the permeability of free space (4π x 10^-7 H/m), μr is the relative permeability (which is 1 for air), N is the number of turns, I is the current, and L is the length of the solenoid. This will give you the magnetic field strength in teslas, which can then be used in the formula for force.

I hope this information helps you and your son better understand the forces and efficiency of
 

FAQ: Finding the force of a magnetic field using an air core solenoid

What is an air core solenoid?

An air core solenoid is a coil of wire with no iron or other magnetic material within its core. It is used to create a magnetic field when an electric current is passed through it.

How is the force of a magnetic field determined using an air core solenoid?

The force of a magnetic field created by an air core solenoid is determined by the strength of the current passing through the coil and the number of turns in the coil. The greater the current and number of turns, the stronger the magnetic field and therefore the greater the force.

What is the formula for calculating the force of a magnetic field in an air core solenoid?

The formula for calculating the force of a magnetic field in an air core solenoid is F = N x I x B, where N is the number of turns in the coil, I is the current in amperes, and B is the strength of the magnetic field in teslas.

What are the units for measuring the force of a magnetic field in an air core solenoid?

The force of a magnetic field in an air core solenoid is typically measured in newtons (N) or dynes (dyn).

How is the direction of the force of a magnetic field determined in an air core solenoid?

The direction of the force of a magnetic field in an air core solenoid is determined by the direction of the current flow and the orientation of the coil. The right-hand rule can be used to determine the direction of the force, with the thumb pointing in the direction of the current flow and the fingers curled in the direction of the magnetic field.

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