Answer:
[tex]\mu=180\times 10^{-3}A-m^2[/tex]
Explanation:
Given that,
Area of the loop, [tex]A=10^{-3}\ m^2[/tex]
Current flowing in the wire, I = 180 A
We need to find the magnetic moment of the current loop. It is given by :
[tex]\mu=I\times A[/tex]
[tex]\mu=180\times 10^{-3}[/tex]
[tex]\mu=180\times 10^{-3}A-m^2[/tex]
So, the magnetic moment of the current loop is [tex]180\times 10^{-3}A-m^2[/tex]. Hence, this is the required solution.
