Question

The magnetic fields in tesla in the two regions separated by the \(z = 0\) plane are given by \(\vec{B_1} = 3\hat{i} + 5\hat{j}\) and \(\vec{B_2} = 8\hat{i} + 3\hat{j} + 5\hat{k}\). The magnitude of the surface current density at the interface between the two regions is \(\alpha \times 10^6\, \text{A/m}\). Given the permeability of free space \(\mu_0 = 4\pi \times 10^{-7}\, \text{N/A}^2\), the value of \(\alpha\) is ............... . (Round off to 2 decimal places)

Hint:

At the interface of two magnetic media, discontinuity in the tangential component of \(\vec{B}\) corresponds to a surface current density \(\vec{K} = \frac{1}{\mu_0} (\hat{n} \times (\vec{B_2} - \vec{B_1}))\).

Answer 1

Correct Answer: 2.86 - 2.88

Step 1: Identify components

At $z = 0$:

• Normal direction: $\hat{z}$
• Tangential directions: $\hat{x}$ and $\hat{y}$

Region 1:

• Normal: $B_{1z} = 5$ T
• Tangential: $\vec{B}_{1t} = 3\hat{x}$ T

Region 2:

• Normal: $B_{2z} = 5$ T
• Tangential: $\vec{B}_{2t} = \hat{x} + 3\hat{y}$ T

Step 2: Calculate surface current density

The discontinuity in tangential $\vec{H}$ gives surface current: $$\vec{K} = \hat{n} \times (\vec{H}_2 - \vec{H}_1)$$

where $\hat{n} = \hat{z}$ (normal from region 1 to region 2).

Since $\vec{H} = \frac{\vec{B}}{\mu_0}$:

$$\vec{H}_{1t} = \frac{3\hat{x}}{\mu_0}$$

$$\vec{H}_{2t} = \frac{\hat{x} + 3\hat{y}}{\mu_0}$$

$$\vec{H}{2t} - \vec{H}{1t} = \frac{-2\hat{x} + 3\hat{y}}{\mu_0}$$

Step 3: Calculate surface current

$$\vec{K} = \hat{z} \times \frac{-2\hat{x} + 3\hat{y}}{\mu_0}$$

Using $\hat{z} \times \hat{x} = \hat{y}$ and $\hat{z} \times \hat{y} = -\hat{x}$:

$$\vec{K} = \frac{-2\hat{y} - 3\hat{x}}{\mu_0} = \frac{-3\hat{x} - 2\hat{y}}{\mu_0}$$

Step 4: Calculate magnitude

$$|\vec{K}| = \frac{\sqrt{9 + 4}}{\mu_0} = \frac{\sqrt{13}}{\mu_0}$$

$$= \frac{\sqrt{13}}{4\pi \times 10^{-7}} = \frac{3.606}{4\pi \times 10^{-7}}$$

$$= \frac{3.606}{1.257 \times 10^{-6}} = 2.868 \times 10^6 \text{ A/m}$$

Therefore: $\alpha = 2.87$

Answer: 2.87