A 100 A current circulates around a 2.00-mm-diameter superconducting ring. A. What is the ring's magnetic dipole moment?
B. What is the on-axis magnetic field strength 4.70 cm from the ring?

0.000314 Am²

6.049*10^-7 T

Explanation:

A

From the definitions of magnetic dipole moment, we can establish that

= , where

= the magnetic dipole moment in itself

= Current, 100 A

= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²

On solving, we have

= ,

= 100 * 3.14 * 0.000001

= 0.000314 Am²

B

= (0)/4 * 2 / ³, where

(0) = constant of permeability = 1.256*10^-6

z = 4.7 cm = 0.047 m

B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]

B = 1*10^-7 * 0.000628/1.038*10^-4

B = 1*10^-7 * 6.049

B = 6.049*10^-7 T

the answer is b

explanation:

the answer is 12 as i may remember

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