Respuesta :
Answer:
Magnetic dipole moment is 0.0683 J/T.
Explanation:
It is given that,
Length of the rod, l = 7.3 cm = 0.073 m
Diameter of the cylinder, d = 1.5 cm = 0.015 m
Magnetization, [tex]M=5.3\times 10^3\ A/m[/tex]
The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :
[tex]M=\dfrac{\mu}{V}[/tex]
[tex]\mu=M\times \pi r^2\times l[/tex]
Where
r is the radius of rod, r = 0.0075 m
[tex]\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m[/tex]
[tex]\mu=0.0683\ J/T[/tex]
So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.
Answer:
[tex]0.068317J/T[/tex]
Explanation:
Magnetization is defined as the magnetic moment per unit volume means
[tex]Magnetization=\frac{Magnetic\ moment}{Volume}[/tex]
So magnetic moment = magnetization × volume
We have given magnetization = [tex]5.3\times 10^3A/m[/tex]
Diameter=1.5 cm ,so radius [tex]=\frac{1.5}{2}=0.75cm\ =0.0075m[/tex]
Area [tex]A =\pi r^2=3.14\times 0.0075^2=1.766\times 10^{-4}m^2[/tex]
Length = 7.3 cm =0.073 m
So volume [tex]=1.766\times10^{-4}\times 0.073=1.289\times 10^{-5}m^3[/tex]
Now magnetic moment = magnetization × volume [tex]=5.30\times 10^3\times 1.289\times 10^{-5}=0.068317J/T[/tex]
