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Square Coil, Round Hole

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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)
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 im doing wrong and any help with this would be greatly appreciated.
 
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Answers and Replies

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Redbelly98
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Welcome to Physics Forums!

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 im 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.
 
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