Answer:
The magnetic dipole moment = 0.0216 J/T
Explanation:
Magnetic dipole moment: The mathematical expression for the magnetic dipole moment of a cylindrical rod is as shown below.
μ = MV ................. Equation 1
Where μ = magnetic dipole moment, M = magnetic field, V = volume of the cylindrical rod.
But
V = πr²L............ Equation 2
Where r = radius of the cylindrical rod, L = length of the rod.
Substituting equation 2 into equation 1
μ = Mπr²L....................... Equation 3.
given: M = 5.50×10³ A/m, r = d/2 = 1/2 = 0.5 cm = 0.005 m, L = 5 cm = 0.05 m
Constant: π = 3.143.
Substituting these values into equation 3
μ = 5.5×10³×3.143×(0.005)²×0.05
μ = 0.0216 J/T
Thus the magnetic dipole moment = 0.0216 J/T
