Amagnet in the form of a cylindrical rod has a length of5.00 cm and a diameter of 1.00 cm. it has a uniform magnetization of5.50 x 10^3 a/m. what is its magnetic dipole moment?

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

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