Question

Review What is the ring's magnetic dipole moment? A 101 A current circulates around a 2.20-mm diameter superconducting ring Express your answer with the appropriate units. Value Units Previous Answers Request Answer X Incorrect; Try Again; 2 attempts remaining Enter your answer using units of magnetic dipole. Part B What is the on-axis magnetic field strength 6.00 cm from the ring? Express your answer with the appropriate units.

Answer 1

Solution) current i = 101 A

Diameter d = 2.20 mm

Radius r = d/2 = 2.20/2 = 1.10 mm =1.10×10^(-3) m

Magnetic dipole moment = i.A = i((pi)(r^2))

Magnetic dipole moment = (101)(pi)(1.10×10^(-3))^2

Magnetic dipole moment , M = 383.93×10^(-6) Am^2

(B) z = 6cm = 6×10^(-2) m=0.06m

Magnetic field strength B = (Uo/4(pi))(2M/z^3)

B = (4(pi))×10^(-7)/(4(pi))((2×383.93×10^(-6))/(0.06)^3)

B = 3.55×10^(-7) T

B =