A95.0 a current circulates around a 2.10-mm-diameter superconducting ring. part a) what is the ring's magnetic dipole moment?
part b) what is the on-axis magnetic field strength 4.10 cm from the ring?

Explanation:

It is given that,

Current flowing in the ring, I = 95 A

Diameter of the ring, d = 10 mm

Radius of the ring, r = 5 mm = 0.005 m

(A) We need to find the ring's magnetic dipole moment. The dipole moment is given by :

or

(B) We need to find the magnetic field strength 4.10 cm from the ring, x = 4.10 cm = 0.041 m

The magnetic field at some distance is given by :

B = 0.0000211 T

or

Hence, this is the required solution.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this

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