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pbj_sweg
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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?