Question

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

Answer 1

Answer:

 u = 2.01*10^{-4} Am^2

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

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

 $100 A = \frac{u}{\pi * ( 0.8 *10^{-3 m} )^2}$

            u = 2.01*10^{-4} Am^2

b ) $x = \frac{u_0}{4\pi * ( \frac{2 u}{z^3} )}$

$= \frac{4\pi *10^{-7}}{4 \pi \frac{2 * 2.01*10^{-4}}{( 5.40 *10^{-2} )^3}}$

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