Question

An infinite surface of linear current density \(K = 5 \hat{a}_x \, \text{A/m}\) exists on the x–y plane, as shown in the figure. The magnitude of the magnetic field intensity \((H)\) at a point (1,1,1) due to the surface current in Ampere/meter is ............... (Round off to 2 decimal places). \begin{center} \includegraphics[width=0.5\textwidth]{32.jpeg} \end{center}

Hint:

For an infinite current sheet, the field is uniform and magnitude = \(|K|/2\). Direction is obtained using right-hand rule with surface normal.

Answer 1

Step 1: Recall boundary condition.
For a surface current density \(K\) (A/m) on a sheet, the magnetic field intensity \(H\) is given by: \[ \hat{n} \times (H_{above} - H_{below}) = K \] where \(\hat{n}\) is the normal vector to the surface.

Step 2: Geometry.
Here, surface = x–y plane, so normal vector is \(\hat{a}_z\). Surface current density: \(K = 5 \hat{a}_x \, A/m\).

Step 3: Relation.
\[ H_{above} - H_{below} = K \times \hat{n} \] \[ = (5\hat{a}_x) \times \hat{a}_z \] \[ = 5(-\hat{a}_y) \]

Step 4: Symmetry.
The field splits equally above and below the sheet: \[ H_{above} = -\tfrac{1}{2} K \times \hat{n} = 2.5 \hat{a}_y \] Magnitude: \[ |H| = 2.5 \, A/m \]

Final Answer:
\[ \boxed{2.50 \, A/m} \]