Why Is Magnetic Flux Calculated Using the Solenoid's Area Instead of the Loop's?

[pbj_sweg]

Homework Statement

Question asks for the flux through the loop when the loop is both perpendicular and at an angle to the solenoid.
solenoid diameter = 2.2 cm
loop diameter = 6.8 cm
B inside solenoid = 0.22 T

Homework Equations

∫B⋅dA = Φ_B

The Attempt at a Solution

B is constant so can pull out of the equation. ∫dA is the area of the solenoid, and therefore π(r_sol)^2. I got my flux to be 8.4*10^-5 Wb in both cases, which is correct. My only question is, why do we use the area of the solenoid and not the area of the loop if we're asked to find the flux through the loop? By using the radius of the solenoid, aren't we neglected all the parts of the loop that are outside of the solenoid and have no B field going through it?

 

[rude man]

Homework Statement

Question asks for the flux through the loop when the loop is both perpendicular and at an angle to the solenoid.
B is constant so can pull out of the equation. ∫dA is the area of the solenoid, and therefore π(r_sol)^2. I got my flux to be 8.4*10^-5 Wb in both cases, which is correct.

Flux = ∫B⋅dA. Say the angle were 90 deg, then the B field would be in the plane of the loop and the dot product would be zero.
So I don't agree with the given answer.

My only question is, why do we use the area of the solenoid and not the area of the loop if we're asked to find the flux through the loop? By using the radius of the solenoid, aren't we neglected all the parts of the loop that are outside of the solenoid and have no B field going through it?

Faraday says emf = dΦ/dt and you can't have any dΦ/dt where there is never any Φ.

 

[pbj_sweg]

Flux = ∫B⋅dA. Say the angle were 90 deg, then the B field would be in the plane of the loop and the dot product would be zero.
So I don't agree with the given answer.

I'm sorry. I didn't specify. The angle in question is 60°. I agree that if the B field was 90° to the loop's area vector, then the flux would be zero.

Faraday says emf = dΦ/dt and you can't have any dΦ/dt where there is never any Φ.

I understand your reasoning here, but isn't the definition of flux based on the magnitude of the area which the field is going through? If we wanted to calculate the flux of the solenoid, we'd take the cross-sectional area of the solenoid, but because we're being asked to calculate through the loop, why don't we use the area of the loop? I guess I'm just confused about what flux fundamentally is.

 

[rude man]

I'm sorry. I didn't specify. The angle in question is 60°. I agree that if the B field was 90° to the loop's area vector, then the flux would be zero.
I understand your reasoning here, but isn't the definition of flux based on the magnitude of the area which the field is going through? If we wanted to calculate the flux of the solenoid, we'd take the cross-sectional area of the solenoid, but because we're being asked to calculate through the loop, why don't we use the area of the loop? I guess I'm just confused about what flux fundamentally is.

Flux is the dot-product of B dot dA integrated over your surface, your surface A being the area of the loop. But how can you expect flux where there's no B? So the area of integration is that of the solenoid cross-section, not that of the loop.

And the fact that flux is a dot-product means if the angle of the solenoid axis is 60 degrees to the loop axis then your flux thru the loop will be ∫B⋅dA = BπR²cos(60 deg.) with R = solenoid cross-section radius.

Think: why would there be no reduction in flux until you get to 90 degrees, then all of a sudden the flux would drop to zero? Doesn't make sense, does it?