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

Step 1: Use boundary condition for magnetic fields.
The surface current density \(\vec{K}\) at the interface is given by: \[ \mu_0 \vec{K} = \hat{n} \times (\vec{B_2} - \vec{B_1}) \] Here, \(\hat{n}\) is the normal to the surface (along \(\hat{k}\)).

Step 2: Compute difference in magnetic fields.
\[ \vec{B_2} - \vec{B_1} = (8 - 3)\hat{i} + (3 - 5)\hat{j} + (5 - 0)\hat{k} = 5\hat{i} - 2\hat{j} + 5\hat{k} \]

Step 3: Compute cross product.
\[ \hat{n} = \hat{k} \] \[ \hat{k} \times (\vec{B_2} - \vec{B_1}) = \hat{k} \times (5\hat{i} - 2\hat{j} + 5\hat{k}) = (2\hat{i} + 5\hat{j}) \] \[ |\vec{K}| = \frac{1}{\mu_0} |\hat{n} \times (\vec{B_2} - \vec{B_1})| = \frac{1}{4\pi \times 10^{-7}} \sqrt{2^2 + 5^2} = \frac{5.385}{1.2566 \times 10^{-6}} = 4.28 \times 10^6 \] \[ $\Rightarrow$ \alpha = 4.28 \] Rounded to two decimals: \( \alpha = 4.28 \).

Step 4: Conclusion.
Hence, the surface current density magnitude is \(4.28 \times 10^6\, \text{A/m}\) and \(\alpha = 4.28.\)