Question

A 95.0 A current circulates around a 2.10-mm-diameter superconducting ring. PART A) What is the ring's magnetic dipole moment?

Answer 1

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 :

$\mu =I\times A$

$\mu =I\pi r^2$

$\mu =95\pi (0.005)^2$

$\mu=0.00746\ A-m^2$

or

$\mu=7.46\times 10^{-3}\ A-m^2$

(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 :

$B=\dfrac{\mu_o Ir^2}{2(r^2+x^2)^{3/2}}$

$B=\dfrac{4\pi\times 10^{-7}\times 95\times (0.005)^2}{2((0.005 )^2+(0.041)^2){3/2}}$

B = 0.0000211 T

or

$B=2.11\times 10^{-5}\ T$

Hence, this is the required solution.