The Magnetic Flux through a Solenoid

[marrone]

Homework Statement

Find the magnetic flux through a solenoid of length 31.2 cm, radius 1.5 cm, and 610 turns that carries a current of 3.8 A.
(in Wb)

Homework Equations

B_solenoid=u₀*N*I/L

Flux= B *A

The Attempt at a Solution

So to find the magnetic field, I plugged the values into the equation like this: u₀*610*3.8/.312=9.34*10^-3.

And I've tried doing pi*(.015²) for the area, and multiplying them together, and I didn't get the right answer.

I tried taking the total surface area of the solenoid, such as 2pi(.015)²* 2pi(.015)*(.312) and I still can't get the answer. What am i missing? I feel like this problem should be really easy...

 

[Gear300]

what was your answer for the flux?

 

[thomate1]

To find the magnetic flux through a solenoid, we can use the equation B_solenoid = u0 * N * I / L, where u0 is the permeability of free space, N is the number of turns, I is the current, and L is the length of the solenoid. Plugging in the given values, we get B_solenoid = (4π * 10^-7 T*m/A) * 610 * 3.8 A / 0.312 m = 9.34 * 10^-3 T.

To find the flux, we need to multiply the magnetic field by the cross-sectional area of the solenoid. The cross-sectional area of a solenoid is given by A = π * r^2, where r is the radius of the solenoid. In this case, r = 0.015 m. Therefore, the flux through the solenoid is given by Flux = B_solenoid * A = (9.34 * 10^-3 T) * (π * (0.015 m)^2) = 2.6 * 10^-6 Wb.

It is important to note that the magnetic flux through a solenoid is directly proportional to the number of turns, current, and length, and inversely proportional to the cross-sectional area. Therefore, by increasing any of these parameters, we can increase the magnetic flux through the solenoid. Additionally, the magnetic flux through a solenoid can also be influenced by the material inside the solenoid and the external magnetic field.