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Chapter 8: Problem 20

Find \(\mathbf{H}\) in a material where \((a) \mu_{r}=4.2\), there are \(2.7 \times 10^{29}\) atoms \(/ \mathrm{m}^{3}\), and each atom has a dipole moment of \(2.6 \times 10^{-30} \mathbf{a}_{y} \mathrm{~A} \cdot \mathrm{m}^{2} ;(b) \mathbf{M}=270 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) and \(\mu=2 \mu \mathrm{H} / \mathrm{m} ;(c) \chi_{m}=0.7\) and \(\mathbf{B}=2 \mathbf{a}_{z} \mathrm{~T}\). \((d)\) Find \(\mathbf{M}\) in a material where bound surface current densities of \(12 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) and \(-9 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) exist at \(\rho=0.3 \mathrm{~m}\) and \(0.4 \mathrm{~m}\), respectively.

Short Answer

Expert verified
In summary: (a) Given a material with relative permeability \(\mu_r = 4.2\), \(2.7 \times 10^{29}\) atoms\(/\mathrm{m}^3\), and a dipole moment per atom of \(2.6 \times 10^{-30} \mathbf{a}_{y}~\mathrm{A} \cdot \mathrm{m}^2\), the magnetic field intensity \(\mathbf{H}\) is equal to \(-7.02 \mathbf{a}_{y} \mathrm{~A} / \mathrm{m}\). (b) For a material with magnetization \(\mathbf{M} = 270 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) and magnetic permeability \(\mu = 2 \mu \mathrm{H} / \mathrm{m}\), the magnetic field intensity \(\mathbf{H}\) is equal to \(-270 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\). (c) In a material with magnetic susceptibility \(\chi_m = 0.7\) and magnetic flux density \(\mathbf{B} = 2 \mathbf{a}_{z} \mathrm{~T}\), the magnetic field intensity \(\mathbf{H}\) is equal to \(1592.5 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\). (d) For a material with bound surface current densities of \(12 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) and \(-9 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) at \(\rho = 0.3 \mathrm{~m}\) and \(0.4 \mathrm{~m}\), respectively, the magnetization vector \(\mathbf{M}\) is equal to \(-210 \mathbf{a}_\rho \mathrm{~A} / \mathrm{m}\).

Step by step solution

01

Calculate total magnetization

Since the magnetic moment per atom is given (\(2.6 \times 10^{-30} \mathbf{a}_{y} \mathrm{~A} \cdot \mathrm{m}^2\)), we can find the total magnetization \(\mathbf{M}\) by multiplying it with the number of atoms per unit volume: \(\mathbf{M} = (2.7 \times 10^{29})(2.6 \times 10^{-30} \mathbf{a}_{y} \mathrm{~A} \cdot \mathrm{m}^2) = 7.02 \mathbf{a}_{y} \mathrm{~A} / \mathrm{m}\).
02

Calculate magnetic permeability

To find \(\mathbf{H}\), we need to find the magnetic permeability \(\mu\). The relative permeability \(\mu_r\) is given as \(4.2\), and the permeability of free space \(\mu_0 \approx 4\pi \times 10^{-7} \mathrm{H} / \mathrm{m}\). Therefore, we can calculate \(\mu = \mu_r \mu_0 = (4.2)(4\pi \times 10^{-7} \mathrm{H} / \mathrm{m}) \approx 5.28 \times 10^{-6} \mathrm{H} / \mathrm{m}\).
03

Find \(\mathbf{H}\)

Now we can use the relationship \(\mathbf{H} = \frac{\mathbf{B}}{\mu} - \mathbf{M}\) to find \(\mathbf{H}\). Since \(\mathbf{B}\) is not given, we must assume that the material is linear, such that \(\mathbf{B} = \mu (\mathbf{H} + \mathbf{M})\). This gives us \(\mathbf{H} = \frac{\mu \mathbf{H} + \mu \mathbf{M}}{\mu} - \mathbf{M} \rightarrow \mathbf{H} = -\mathbf{M} = -7.02 \mathbf{a}_{y} \mathrm{~A} / \mathrm{m}\). (b) \(\mathbf{M} = 270 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\), \(\mu = 2 \mu \mathrm{H} / \mathrm{m}\)
04

Find \(\mathbf{H}\)

We can directly use the relationship \(\mathbf{H} = \frac{\mathbf{B}}{\mu} - \mathbf{M}\) since \(\mathbf{M}\) and \(\mu\) are given. First, we need to find \(\mathbf{B}\) using \(\mathbf{B} = \mu (\mathbf{H} + \mathbf{M})\). This gives us the equation \(\mathbf{H} = \frac{2\mu \mathrm{H}(\mathbf{H} + 270\mathbf{a}_{z})}{2\mu \mathrm{H}} - 270\mathbf{a}_{z}\). This simplifies to \(\mathbf{H} = -270 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) since the material is linear. (c) \(\chi_m = 0.7\) and \(\mathbf{B} = 2 \mathbf{a}_{z} \mathrm{~T}\)
05

Calculate \(\mathbf{H}\) from magnetic susceptibility and magnetic flux density

We can use the relationship between \(\mathbf{M}, \mathbf{H}\), and \(\chi_m\): \(\mathbf{M} = \chi_m \mathbf{H}\). Given \(\chi_m = 0.7\) and \(\mathbf{B} = 2 \mathbf{a}_{z} \mathrm{~T}\), we can use the relationship \(\mathbf{B} = \mu_0 (\mathbf{H} + \mathbf{M}) \rightarrow \mathbf{H} = \frac{\mathbf{B}}{\mu_0} - \mathbf{M}\). This yields the equation \(\mathbf{H} = \frac{2 \mathbf{a}_{z} \mathrm{~T}}{4\pi \times 10^{-7} \mathrm{H} / \mathrm{m}} - 0.7\mathbf{H} \Rightarrow \mathbf{H} = 1592.5 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\). (d) Find \(\mathbf{M}\) in a material where bound surface current densities of \(12 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) and \(-9 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m}\) exist at \(\rho = 0.3 \mathrm{~m}\) and \(0.4 \mathrm{~m}\), respectively.
06

Calculate radial component of magnetization

The net radial component of the magnetization \(\mathbf{M_\rho}\) can be found using the equation: \(\mathbf{M_\rho} = \frac{\Delta K}{\Delta \rho} = \frac{-9-12}{0.4-0.3} = -210 \mathrm{~A} / \mathrm{m}\).
07

Find magnetization vector \(\mathbf{M}\)

Since the magnetization has only a radial component, \(\mathbf{M} = -210 \mathbf{a}_\rho \mathrm{~A} / \mathrm{m}\).

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