Explanation:
Given that,
Length of the rod, l = 5.31 cm = 0.0531 m
Diameter of the rod, d = 1.34 cm
Radius of the rod, r = 0.0067 m
Magnetization in the rod, [tex]M=6.46\times 10^3\ A/m[/tex]
We need to find the magnetic dipole moment of the rod. The magnetic dipole moment is given by:
[tex]\mu=M\times V[/tex]
V is the volume of the cylindrical rod
[tex]\mu=M\times\pi r^2l[/tex]
[tex]\mu=6.46\times 10^3\times \pi (0.0067)^2\times 0.0531[/tex]
[tex]\mu=0.049\ J/T[/tex]
So, the magnetic dipole moment of the rod is 0.049 J/T. Hence, this is the required solution
