Magnetic Flux in a Solenoid

[DaveTan]

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!

 

[Daeho Ro]

Magnetic Field = 4pi *10^-7 * (1280 loops/ 1.00m) * 2.15A = 0.003458T

Why are you put the length as $1.00m$?

 

[DaveTan]

Why are you put the length as $1.00m$?

Because the question says 1280 turns per meter

 

[Daeho Ro]

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?

 

[DaveTan]

You are right! Thanks