Question

A 100 A current circulates around a 1.60-mm-diameter superconducting ring. Part A. What is the ring's magnetic dipole moment?

Part B. What is the on-axis magnetic field strength 5.40 cm from the ring?

Answer 1

u = 2.01*10^{-4} Am^2

magnetic field strength = 3.91*10^{-8} m

Step-by-step explanation:

a ) magnetic dipole moment i = u / πr^2

where,

i is current in the ring = 100 A

r is radius of ring = 1.60/2 =0.80 mm

`100 A = (u)/(\pi * ( 0.8 *10^(-3 m) )^2)`

u = 2.01*10^{-4} Am^2

b )
`x = (u_0)/(4\pi * ( (2 u)/(z^3) ))`

`= (4\pi *10^(-7))/(4 \pi (2 * 2.01*10^(-4))/(( 5.40 *10^(-2) )^3))`

magnetic field strength = 3.91*10^{-8} m