# Square Coil, Round Hole

• nb121307
In summary, a square wire coil is placed inside a solenoid with 113 turns and a current of 3.29 A. The flux through the coil is calculated using the equation Φ = BA, where B is the strength of the magnetic field and A is the area of the coil. The strength of the magnetic field is calculated using the equation B = μ0NI, where N is the number of turns.

## Homework Statement

A square, single-turn wire coil L = 1.75 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 2.82 cm.
http://capa8.phy.ohiou.edu/res/ohiou/serwaylib/Graphics/Graph20/serw2014.gif
The solenoid is 22.0 cm long and wound with 113 turns of wire. If the current in the solenoid is 3.29 A, find the flux through the coil.

Є=(ΔNΔΦ/Δt)
ΔΦ=BAcosΘ
B=μ0NI

## The Attempt at a Solution

I have found the strength of the magnetic field using the constant μ0 = 12.57x10^-7, the number of turns 113, and the current 3.29 A, and because time is not applied Δt can be taken out of the equation to get Є=ΔNΔΦ=ΔN(BAcosΘ). This gives me B = 4.6731489x10^-4 or B = 4.67x10^-4 The next part of the question asks for the flux and the strength of the magnetic field is required to do so. However, when applying the strength of the magnetic field and the area of the square coil using the equation Φ = BA the answer I am given is Φ = 1.43x10^-7 Wb which is apparently wrong. I'm not sure what I am doing wrong and any help with this would be greatly appreciated.

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nb121307 said:

## Homework Statement

A square, single-turn wire coil L = 1.75 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 2.82 cm.
http://capa8.phy.ohiou.edu/res/ohiou/serwaylib/Graphics/Graph20/serw2014.gif
The solenoid is 22.0 cm long and wound with 113 turns of wire. If the current in the solenoid is 3.29 A, find the flux through the coil.

## Homework Equations

Є=(ΔNΔΦ/Δt)
That equation isn't relevant, since they are not asking for Є.

ΔΦ=BAcosΘ
Okay. But strictly speaking, this would be Φ and not ΔΦ. Nothing is changing in this situtation.

B=μ0 N I
If N means the number of turns, then this expression is missing something. Check this out:
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c2

## The Attempt at a Solution

I have found the strength of the magnetic field using the constant μ0 = 12.57x10^-7, the number of turns 113, and the current 3.29 A, and because time is not applied Δt can be taken out of the equation to get Є=ΔNΔΦ=ΔN(BAcosΘ). This gives me B = 4.6731489x10^-4 or B = 4.67x10^-4 The next part of the question asks for the flux and the strength of the magnetic field is required to do so. However, when applying the strength of the magnetic field and the area of the square coil using the equation Φ = BA the answer I am given is Φ = 1.43x10^-7 Wb which is apparently wrong. I'm not sure what I am doing wrong and any help with this would be greatly appreciated.

Try calculating B using the link I give above, then you can use the flux equation you had to find the flux.

Last edited by a moderator:

I would suggest double-checking your calculations and making sure you are using the correct units for all the variables. It is also important to consider the orientation of the square coil with respect to the solenoid and make sure you are using the correct angle in the equation for flux (Φ = BAcosΘ). Additionally, it may be helpful to draw a diagram and label all the given information to better visualize the problem. If you are still having trouble, I would suggest seeking help from a colleague or instructor.

## 1. What is a square coil, round hole?

A square coil, round hole is a type of electrical coil that is designed to fit into a round hole in a circuit board or other electrical device. It is typically used in electronic devices to create a magnetic field.

## 2. How does a square coil, round hole work?

A square coil, round hole works by passing an electric current through the coil, which creates a magnetic field. This magnetic field can then be used for various purposes, such as powering motors or generating electricity.

## 3. What are the advantages of using a square coil, round hole?

There are several advantages to using a square coil, round hole. First, it allows for a more compact design, as the coil can fit snugly into the round hole without any wasted space. Additionally, the square shape of the coil can provide better stability and more efficient magnetic field generation.

## 4. How is a square coil, round hole made?

A square coil, round hole is typically made by winding copper wire around a square-shaped form, such as a bobbin or mandrel. The ends of the wire are then connected to leads that can be attached to a circuit board or other electrical device. The coil may also be coated with insulating material to protect it from damage.

## 5. What are the applications of a square coil, round hole?

A square coil, round hole has a wide range of applications in various industries, including electronics, automotive, and healthcare. Some common uses include generating magnetic fields for motors, transformers, and generators, as well as in sensors, speakers, and medical devices.