Thank you for visiting A 99 0 A current circulates around a 1 70 mm diameter superconducting ring Part A What is the ring s magnetic dipole moment Express. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
Part A: The magnetic dipole moment is a measure of the ring's ability to generate a magnetic field. The magnetic dipole moment of the ring is 2.23 x 10^-7 A m^2. The magnetic field strength on the axis of the ring, at a distance of 5.00 cm from the ring, is 9.25 x 10^-4 T. It is given by the formula:
Magnetic dipole moment = (pi) * (radius)^2 * (current).
Here, radius = 0.85 mm
= 0.85 x 10^-3 m
and current = 99.0 A.
So, the magnetic dipole moment is given by: Magnetic dipole moment = (pi) * (0.85 x 10^-3 m)^2 * (99.0 A)
= 2.23 x 10^-7 A m^2. Hence, the magnetic dipole moment of the ring is 2.23 x 10^-7 A m^2.
Part B: The magnetic field strength on the axis of the ring can be determined using the formula:
Magnetic field strength = (mu_0) * (current) / (2 * radius),
where mu_0 is the permeability of free space. Here,
current = 99.0 A,
radius = 0.85 mm
= 0.85 x 10^-3 m,
and the distance from the ring is
5.00 cm = 0.05 m.
So, the magnetic field strength is given by:
Magnetic field strength = (4 * pi * 10^-7 T m/A) * (99.0 A) / (2 * 0.85 x 10^-3 m
= 9.25 x 10^-4 T.
Hence, the magnetic field strength on the axis of the ring, at a distance of 5.00 cm from the ring, is 9.25 x 10^-4 T.
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