Question

Let \(\mu_{r 1}=2\) in region 1, defined by \(2 x+3 y-4 z>1\), while \(\mu_{r 2}=5\) in region 2 where \(2 x+3 y-4 z<1\). In region 1, \(\mathbf{H}_{1}=50 \mathbf{a}_{x}-30 \mathbf{a}_{y}+\) \(20 \mathbf{a}_{z} \mathrm{~A} / \mathrm{m} .\) Find \((a) \mathbf{H}_{N 1} ;(b) \mathbf{H}_{t 1} ;(c) \mathbf{H}_{22} ;(d) \mathbf{H}_{N 2} ;(e) \theta_{1}\), the angle between \(\mathbf{H}_{1}\) and \(\mathbf{a}_{N 21} ;(f) \theta_{2}\), the angle between \(\mathbf{H}_{2}\) and \(\mathbf{a}_{N 21}\).

Short answer

In this exercise, we are given the magnetic field in region 1, \(\mathbf{H_1}\), and need to determine several components and angles in regions 1 and 2. Follow these steps to find the required values: 1. Find the normal vector to the boundary surface (\(\mathbf{a}_{N21}= 2\mathbf{a}_x + 3\mathbf{a}_y - 4\mathbf{a}_z\)) and its magnitude (\(| \mathbf{a}_{N21} |= \sqrt{29}\)). 2. Find the normal component of \(\mathbf{H_1}\) (\(\mathbf{H}_{N1}\)) and the tangential component (\(\mathbf{H}_{t1}\)). 3. Using the boundary conditions, find \(|H_{N2}|\) and \(\mathbf{H}_{22}\). 4. Calculate the angles \(\theta_1\) and \(\theta_2\) between the magnetic fields and the normal vector using the dot product. By taking these steps, we can find the normal and tangential components of the magnetic field in each region and the angles between the fields and the normal vector.

Step-by-step solution

Find the normal vector

The normal vector to the boundary surface, \(\mathbf{a}_{N21}\), can be found using the coefficients of the given equation: \(2x + 3y - 4z = 1\) \(\mathbf{a}_{N21} = 2\mathbf{a}_x + 3\mathbf{a}_y - 4\mathbf{a}_z\) We can also find the magnitude of the normal vector, \(| \mathbf{a}_{N21} |\). \(| \mathfile{fontsize{12pt}\mid a_{N21} \mid} = \sqrt{2^2+3^2+(-4)^2} = \sqrt{29}\)

Find \(\mathbf{H}_{N1}\) and \(\mathbf{H}_{t1}\)

The normal component of \(\mathbf{H_1}\) is given by the dot product of \(\mathbf{H_1}\) and \(\mathbf{a}_{N21}\) divided by the magnitude of \(\mathbf{a}_{N21}\). \(\mathbf{H}_{N1} = \frac{\mathbf{H_1} \cdot \mathbf{a}_{N21}}{| \mathbf{a}_{N21} |}\) \(\mathbf{H}_{t1} = \mathbf{H_1} - \mathbf{H}_{N1}\)

Find \(\mathbf{H}_{22}\) and \(\mathbf{H}_{N2}\)

Using the boundary conditions on the magnetic field, we have: \(|H_{N1}|\mu_{r1} = |H_{N2}|\mu_{r2}\), so \(|H_{N2}| = \frac{H_{N1}\mu_{r1}}{\mu_{r2}}\) Since \(\mathbf{H}_{t1} = \mathbf{H}_{t2}\), we can find \(\mathbf{H}_{22}\) as follows: \(\mathbf{H}_{22} = \mathbf{H}_{t1} + \mathbf{H}_{N2}\)

Find \(\theta_1\) and \(\theta_2\)

The angles between the magnetic fields and the normal vector can be obtained using the dot product: \(\cos\theta_1 = \frac{\mathbf{H_1}\cdot\mathbf{a}_{N21}}{|\mathbf{H_1}|}\) \(\cos\theta_2 = \frac{\mathbf{H_{22}}\cdot\mathbf{a}_{N21}}{|\mathbf{H_{22}}|}\) Then, we can find \(\theta_1\) and \(\theta_2\) by taking the inverse of the cosine function applied to the found values.