Respuesta :
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 :
[tex]\mu =I\times A[/tex]
[tex]\mu =I\pi r^2[/tex]
[tex]\mu =95\pi (0.005)^2[/tex]
[tex]\mu=0.00746\ A-m^2[/tex]
or
[tex]\mu=7.46\times 10^{-3}\ A-m^2[/tex]
(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 :
[tex]B=\dfrac{\mu_o Ir^2}{2(r^2+x^2)^{3/2}}[/tex]
[tex]B=\dfrac{4\pi\times 10^{-7}\times 95\times (0.005)^2}{2((0.005 )^2+(0.041)^2){3/2}}[/tex]
B = 0.0000211 T
or
[tex]B=2.11\times 10^{-5}\ T[/tex]
Hence, this is the required solution.
The ring's magnetic dipole moment is 3.287×10⁻⁴ A-m² and on-axis the magnetic field strength 4.10 cm from the ring is 7.92×10⁻⁹ T
What is magnetic field?
The magnetic field is the field in the space and around the magnet in which the magnetic field can be fill.
The diameter of the superconducting ring is 2.10 mm. Thus the area of this ring is,
[tex]A=\dfrac{\pi(0.0021)^2}{4}\\A=3.46\times10^{-6}\rm \; m[/tex]
- A) The ring's magnetic dipole moment
Magnetic dipole moment is the product of magnitude of current and the cross-sectional area. As the value of current is 95 A. Thus the magnetic dipole can be given as,
[tex]\mu=95\times3.46\times10^{-6}\\\mu=3.287\times10^{-4}\rm Am^2[/tex]
- B) The on-axis magnetic field strength 4.10 cm from the ring
For some distance r and diameter d, the formulae of magnetic field can be given as,
[tex]B=\dfrac{\mu_o Id^2}{8(\dfrac{d^2}{4}+x^2)^{3/4}}[/tex]
Put the values as,
[tex]B=\dfrac{4\pi \times10^{-7} (95)(0.0021)^2}{8(\dfrac{(0.0021)^2}{4}+(0.041)^2)^{3/4}}\\B=7.92\times10^{-9}\rm T[/tex]
Hence, the ring's magnetic dipole moment is 3.287×10⁻⁴ A-m² and on-axis the magnetic field strength 4.10 cm from the ring is 7.92×10⁻⁹ T.
Learn more about magnetic field here;
https://brainly.com/question/7802337
