A single-turn square loop of side L centered on the axis of a long solenoid. In

Question

A single-turn square loop of side L 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 1440 turns per meter and a diameter of 5.12 cm, and carries a current of 3.28 A.

A) Calculate magnetic flux through loop when L=1.65cm..... I found this part and it is correct,Answer=1.62x10^(-6)Wb B=u(N/L)I and then putting that into Flux=BA---> (.00594)(0.0165)^2 B)

B) Caculate the magnetic flux through the loop when L=5.12cm. I tried using my same method i got for the answer in part A but its not getting me the right answer. Im not sure why i was given the diameter so maybe im missing something there. Answer for part B is not 20.45Wb, 1.56x10^(-5)Wb. I cannot figure it out

Explanation / Answer

B = mu*NI/L = 4*pi*k*N*I/L = 4*pi*10^-7*1440*3.28 = 5.9323392 mT
a) Flux = BA = B*1.65^2*10^-4 = 1.615*10^-6 Wb
b) Flux = BA = B*pi*(5.12/2)^2*10^-4 = 12.207*10^-6 Wb

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.