Answer:
[tex]M=0.113\ Am^2[/tex]
Explanation:
Given that,
Radius of the circular loop, r = 10 cm = 0.1 m
Current flowing in the loop, I = 3.6 A
Uniform magnetic field, B = 12 T
To find,
The magnetic dipole moment of the loop.
Solution,
Let M is the magnitude of magnetic dipole moment of the loop. We know that the product of current flowing and the area of cross section. Its formula is given by :
[tex]M=I\times A[/tex]
A is the area of circular wire
[tex]M=I\times \pi r^2[/tex]
[tex]M=3.6\ A \times \pi (0.1\ m)^2[/tex]
[tex]M=0.113\ Am^2[/tex]
Therefore, the magnetic dipole moment of the loop is [tex]0.113\ Am^2[/tex]. Hence, this is the required solution.
