How Is Magnetic Flux Calculated Through a Square Loop Inside a Solenoid?

AI Thread Summary
To calculate the magnetic flux through a square loop inside a solenoid, the magnetic field is determined using the formula B = 4π * 10^-7 * (number of loops per meter) * current. For a solenoid with 1280 turns per meter and a current of 2.15 A, the magnetic field is calculated to be 0.003458 T. The magnetic flux is then computed using the area of the loop, resulting in a value of 1.18 x 10^-5 Wb. Clarifications arise regarding the use of the length in calculations, as it should reflect the number of turns per meter specified in the problem. Accurate calculations are essential for determining the correct magnetic flux.
DaveTan
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Homework Statement


A single-turn square loop of side L is centered on the axis of a long solenoid. In addition, the plane of the square loop is perpendicular to the axis of the solenoid. The solenoid has 1280 turns per meter and a diameter of 5.85 cm , and carries a current of 2.15 A .

Find the magnetic flux through the loop when L= 5.85 cm .

Homework Equations


Magnetic Field = 4pi *10^-7 * (Num loops/ length) * current
Magnetic Flux = Magnetic Field * Area * cos(theta)

The Attempt at a Solution


Magnetic Field = 4pi *10^-7 * (1280 loops/ 1.00m) * 2.15A = 0.003458T
Magnetic Flux = B * area = 0.003458 * 0.0585 * 0.0585 = 1.18x10-5

The answer is wrong, I would appreciate any help! Thanks!
 
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DaveTan said:
Magnetic Field = 4pi *10^-7 * (1280 loops/ 1.00m) * 2.15A = 0.003458T
Why are you put the length as ## 1.00m##?
 
Daeho Ro said:
Why are you put the length as ## 1.00m##?

Because the question says 1280 turns per meter
 
DaveTan said:
Magnetic Flux = B * area = 0.003458 * 0.0585 * 0.0585 = 1.18x10-5
Here the area is maybe the area of solenoid, isn't it?
 
You are right! Thanks
 

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