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6 August, 02:26

A magnet 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?

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  1. 6 August, 04:24
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    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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